I felt that right in the feels... This is legitimate push back from all the math puzzles us engineers and coders keep giving 'out of the box' or unintended answers for to wind him up ;) fair play sir.
My first time here, and I have nothing but respect and admiration for this presentation. The "other guy" who famously does these problems and refers to himself as Mr. Online Geometer or something 😉 would have taken twice as long, used three times the words, and explained one quarter of the solving strategies. This presentation is superior, and I learned two things I had forgotten since high school geometry 50 years ago or maybe never really knew: the chord-chord bit, and the circumference deal of (AxBxC) / 4r^2. I turn 65 in a month and take particular pleasure whenever I can increase my knowledge regardless of its direct applicability in my life. BTW, I guessed 21 1/4 cm within 60 seconds, and solved using right triangles and simultaneous equations, but that took longer. I knew chords would be significant but didn't know the formula. The final solution is brilliant. Cheers from SoCal. Go Dodgers.
Who is this "other guy" you're talking about? I've seen quite a few famous math channels on youtube, but no one does geometry problems as much as Presh.
Complete the perpendicular that is currently 16, let the additional length be x. Subtend two radii to the edges of the circle where the lines meet the circle. Form two equations: r^2 = 21^2 + x^2 and r^2 = (16+x)^2 + 9^2.
Didn't ANYONE ELSE use a variation of the master method like I did and constructed two intersecting chiod formulas together the answer .more organic inthink
@@leif1075I agree the last method is the least elegant because it requires the least intuition and the most memorization of niche formulas. The chord method was the neatest in my opinion
That's the first point addressed in the video. Pradesh preemptively said that this is not how you solve it, so I'm wondering if you watched the video at all. 🤣
@@sambena And now, imagine these are real measurements, in inches, feet, meters, or whatever, for the construction of a tunnel, and you have to be sure that some piece of equipment will fit between the last vertical segment and the arc at the right of the figure. That "a little over 21" is as useless as can be, and that's why we use maths and don't rely on approximations. 😀
If we take the leftmost point of 9cm line segment as an origin, x axis going right and y axis going down, we get: Point (9, 0) lies on a circle. Point (21,16) lies on a circle as well. x-coordinate of a center of a circle is 0. General formula for a circle is (x - C_x)^2 + (y - C_y)^2 = R^2. We can replace C_x with 0 right away, getting x^2 + (y - C_y)^2 = R^2 Plug both points into this equation: 9^2 + (0 - C_y)^2 = R^2 21^2 + (16 - C_y)^2 = R^2 Right parts are equal, so we can equate left parts: 9^2+(0-C_y)^2 = 21^2 + (16 - C_y)^2 81 + C_y^2 = 441 + 256 - 32C_y + C_y^2 | subtract C_y^2 on both sides 81 = 441 + 256 - 32C_y | bring 81 to the right side and C_y to the left 32C_y = 441 + 256 - 32 32C_y = 616 C_y = 616/32 = 19.25 Notice that distance between (C_x, C_y) and point (9, 0) is a radius. R = sqrt((0-9)^2+(19.25-0)^2) = sqrt(81 + 370.5625) = sqrt(451.5625) = 21.25
Another approach: the center lies on the perpendicular bisector of the hypotenuse of the right triangle with legs 12&16, which is a scaled 3-4-5 triangle so its perpendicular has slope 3/4 and passes through a point 15cm right of the vertical radius, hence is 3/4*15=11.25 units above the horizontal radius, so the 12cm line is 3.25cm above the horizontal radius. Knowing this distance we can calculate the radius using the Pythagorean Theorem: r^2=9^2+19.25^2 (or r^2=3.25^2+21^2 from the other point on the circle).
You can build a pythagorean triple on any odd number like this. Just take half of the square, plus/minus half: 13^2=169 169/2=84.5 Therefore, (13,84,85) is a valid triple :)
@@6c3333 It's hard to believe that always works! 3^2=9, 9/2=4.5, +/- .5 = (3, 4, 5). Today, I learned about power chords and inscribed triangle circumradii. I thought I was done for the day, but noooo. You gave me this gem as well. Thank you!
Well done. A lot of creativity and insight. Another method is to use the equation of the circle x^2 + y^2 = r2. We have two points of contacts. If we let the center be (0,0) and the y distance between the center and the first line be a, we have two equations 1) 21^2 + a^2 = r^2 2) 9^2 + (16+a)^2 = r^2 We can use comparison to solve for a. We get 3,25. Then we can come back to eq 1 to solve for r.
Very Easy just continue the vertical of 16 unit and tell that extra length as x. So we get the 1st equation 9²+(16+x)²=r². Now drop vertical from rightmost edge of 12 unit length, again that vertical length which was dropped will be x so we get 2nd equation (9+12)²+x²=r². On solving we get x=3.25 and r=21.25
My solution before watching: Place the measured segments on graph paper, with the y-axis on the vertical radius of the quarter circle and the x-axis on the 12cm segment. Form a segment from the top end of the 16cm segment to the right end of the 12cm segment. Form the perpendicular bisector of that segment. The bisector and the y-axis meet at the center, at the point (0,-3.25). The radius is the distance from that center to either end of the constructed segment, 21.25 cm.
Always keen on looking at your videos. Method 3 can end in another solution: The intersecting chords have legs with lengths of 12; 30; 16; 45/2 With the formula 4r^2=a^2+b^2+c^2+d^2 you get 1806,25 for 4r^2, which gives you the 21,25 for r again
I solved it with trigonometry r.cosα=9 => r.sinα=√(r^2-9^2) r.cosβ=21. => r.sinβ=√(r^2-21^2) r.sinα-r.sinβ=16 => √(r^2-9^2)-√(r^2-21^2)=16 We make it power 2 and it s 2nd degree equation where x=r^2
Funny thing is that's how I _thought_ Presh was going to solve this. I couldn't be bothered to do the calculation, so I watched anyway. Now I'm blown away!
I used trig too and now I feel silly after seeing a pure geometry method. Though I like how trig gets us straight to the radius without much transformation.
As a geometry teacher I sometimes find your geometry videos humbling. I can quickly solve videos out there, but some of your methods surprise and challenge me. (Students/former students: Helping a teacher understand what you are thinking is the historical and fundamental reason we show our work!)
Very true! Thank you for what you do. Teachers (especially good and passionate ones) are increasingly underappreciated (and increasingly rare) in the world. As you said, showing your work allows the teacher to understand what you are thinking. It's easy to say that getting the right answer is most important, but what if the answer isn't right? There isn't really a good way to figure out what went wrong and remediate the source of the confusion. As an added bonus, sometimes the teacher can learn something, too! When I was learning geometry (maybe 30 years ago), I was in grade 10, about 14 years old. My teacher was maybe 28. I'm looking back at my 28-year-old self and am amazed by how much I _didn't_ know! Not that teachers are ignorant; far from it! But it's humbling to find something you didn't know, or didn't think of, and opportunities to learn and grow are forever abundant if we're willing to look for them. I think Neil deGrasse Tyson said it best (paraphrased): It's one thing to realize that we don't yet know the answers to many of our questions, but imagine the questions we don't even know to ask!
@@sabinrawr wonderful reply! My late wife was an Army officer. She not only flew the CH-47 helicopter and put forth maximal effort to be the very best pilot, she always brought that dedication to everything she did-especially leadership. She liked to say knowledge falls into three broad categories. First comes knowing that we know something (it’s the basis of the Dunning-Kruger Effect). Then we start recognizing our limitations becoming aware of things that we don’t understand very deeply. Finally, we become aware that there is far more to be known than we can ever hope to not only grasp but even to be aware of. In short, “You don’t know what you don’t know.”
As an electrical engineer I went directly to make an equation with loveley sin and cos and found another way of solving it. Let's say there is a xy coordinate system. I then mapped a function of y(x) that descibes the circle. First, just dependend on the anlge of the current position of this quarter circle, we get y(alpha)=r*sin(alpha). For x then we get y(x)=r*sin(arccos(x/r)). We know that y(21) - y(9) = 16. By solving this equation with the little trick of sin(arccos(x)) = sqrt(1-x^2) we get that r = 21.25. A little more complicated ^^ Cheers
Just started watching but thanks for showing the image on a dark background. Watching this b4 going to sleep and its much easier on my eyes that way, lol
I solved this with calculus. Let R be the radius and f(x) = sqrt(r^2 - x^2) The image shows that the avarage slope in the section from x=9 to x=21 is -16/12, so -16/12 = [ f(21) - f(9) ] / (21 - 9) Therefor -16 = f(21) - f(9) So: -16=sqrt(r^2 - 21^2) - sqrt(r^2 - 9^2) I solved this equation by building the square twice and so some algebraic standard stuff and got r = 21.25
I just call that little piece below the 16 chord X, then 16(2X+16) = 360 from intersecting chords theorem. Solve for X (3.25) Then use right angle triange , 9squared plus 19.25 squared = 451.5625: find sq root = 21.25 QED
- Connect the two points being on the circle - Construct the center vertical to this line - The intersection between this center vertical and the vertical line crossing the vertical line at the left end point of the 9 cm line is the midpoint of the circle.
First I mirrored the drawing in the vertical line to get two more points that the circle passes through. No more than one circle can pass through four different points, so it's not indeterminate. Then I took the perpendicular bisector of the line segment (21,0)-(9,16) and intersected it with the y-axis. The center is (0,-3.25), and I computed the radius with the distance formula.
I like it that it's counter-intuitive that zig-zag lines like those can fit ONLY into a circle that size -- even though the space above the 9 section and the space below the 12 section seem they could be different with a different sized circle.
What a wonderfully clear exposition! I do have a little quibble, though - It involves the difference between an equation (has an equal sign in it) and an expression (which doesn't). At 2:45, I would say that the two expressions on the left are each equal to r^2, so they are equal to each other. In my day, I called this the Comparison method, but it seems that today it is seen as a variety of the Substitution method. ;-)
With analytical geometry: x^2 + y^2 = r^2 if the center is in the leftmost bottom point. Let the vertical distance from the center to "9" is y. Then for the two points on the circle we have 9^2 + y^2 = r^2 (9 + 12)^2 + (y - 16)^2 = r^2 Together their solution gives y = 19.25. If put y in the first eq. you obtain r = 21.25.
Time for some lateral thinking. - Grab a screenshot of the figure. - Apply the Erode effect with Paint Shop Pro, to make the lines 1 pixel thin. - Measure 135 pixels for the 9 cm line, and 315 pixels for the radius - Write its ratio as a proportion: 135:9 = 315:x - Solve the proportion: x=315*9/135 = 21
you can do it pretty simple if you use the contact points to make a line, then, grab the oposite pendient and use it in the middlepoint, that line always goes to the center, after that you just need to evaluate the x coordinate where the center is, you can know it because of the distance that are showed on the drawing
Got a weird system of equations: r = 21/cos(θ) r = -9/cos(θ+2arctan(4/3)) I just used the law of sines on some triangles, I believe you get that r=+-85/4 but of course r is positive!
My solution - not sure which example it fits into - I guess it's closest to the first method? Consider the radius at the two points the lines are in contact with the circle : 1) r = sqrt(9^2 + b^2) for the upper left point also 2) r = sqrt(21^2 + a^2) for the lower right point and the difference between y axis a and y axis value b is 16 3) b = a +16 sub 3 into 1 and put equations equal to each other 9^2 + (a+16)^2 = 21^2 + a^2 expand brackets and solve for a a^2 + 32a + 16^2 - a^2 = 21^2 - 9^2 32a = 21^2 - 9^2 - 16^2 a = (21^2 - 9^2 - 16^2 )/32 a = 3.25 sub y1 into 2) to get the radius r = sqrt(21^2 + 3.25^2) r = 21.25
I used the first (triangle) method. I took Geometry 50 years ago. I don't remember the last two methods. I was blown away at both the Chord-Chord rule and the Circumradius rule. ‼
Add method 5: assign coordinates as you did in method 2. The segment from (9,a) to (21,a-16) is a chord of the circle and its midpoint is (15, a-8). The perpendicular from the midpoint goes through the center of the circle, (0,0) so we must have that the vector from (9,a) to (21,a-16) is perpendicular to the vector from (0,0) to (15,a-8). The first vector is (12,-16) and the second is (15,a-8). They are perpendicular iff their dot product is 0. The dot product is 12 x 15 - 16 (a-8) = 308 - 16 a, which is zero when a = 77/4.
I started out constructing the 2 radii in the first solution, but then switched to focusing on angles in that isosceles triangle (2 radii and a chord) instead. You can get 2 equations for the angle, one if the form of acos(..) and another in the form of asin(A) + asin(B). Simplifying the 2nd into one asin(X) and using sin^2 + cos^2, you can solve the equation you get in r^4. Extremely roundabout, but still works :P
First: The radius is greater than 21 cm. Now I have to try to solve the problem. Yes. The problem is solvable: r = 21,25. r^2= 9^2+ 16+y2)^2 r^2= 21^2+y2^2 We have an aquation system of two equations with two variables. Solve for y2: y2=104/34. Insert y2=104/32 in any of the equations and r^2= 21,25=21+1/4.
We can construct a radius going to the end of the 9 centimeter length. We can then draw a length from the end of the 16 centimeter length down to the bottom of the quarter circle, and call that length h. This gives us the equation 9²+(16+h)²=r². By creating another copy of the radius and rotating it so that the endpoint matches with the 12 centimeter length, we get the equation 21²+h²=r². Since both equations share a common equivalence, we can set them equal to each other. 9²+(16+h)²=21²+h². Solving for h… 81+256+32h+h²=441+h² 337+32h+h²=441+h² 337+32h=441 32h=104 h=13/4 …and plugging in the solved variable… 21²+(13/4)²=r² 441+169/16=r² 7225/16=r² …we get that r is *85/4 centimeters* or *21.25 centimeters* :)
Before watching solution: The most interesting question is whether the radius is even uniquely constrained. A circle that passes through 3 points is completely determined, both in center location and radius. The zig-zag defines two points, but is that enough? As it turns out, it is. The zigzag clearly defines two points that are on the circle -- the corner between the 9cm horizontal segment and the 16cm vertical segment. Now we need one more, and the easiest one I can find is this: The top horizontal is perpendicular to the vertical radius, so the point 9cm to the left on that extended segment is also on the circle. That defines 3 points, meaning that the circle is completely determined. Actually determining it is a much less interesting problem, but the sketch of how to construct it is to use the (-9,9) segment as one chord, the diagonal to the other two points as another chord, then constructing the two perpendicular bisectors of each chord, which will intersect at the center. Then draw the circle with that center and radius equal to the distance to any of the outer points. The trick is that the diagram only shows a quarter of the circle, and the key to solving it (at least this way) involves another arc of the circle which isn't drawn.
Its been 4 yrs I started watching your videos, they are just awesome and brain teasing.Specially I love the logic puzzles you provide. There is also a book called 'What is the name of this book' by Raymond Smyullan.I haven't read it completely but I found it interesting for puzzles and logic.
Using the Small Angel Approximation, Cos 0 = 1, so the the Radius = 21 Once you have the length of the 16 extended Chord, you can work out the height above the diameter of the extended 12+9+9+12 Chord as H = (L - 2x16) / 2 With H and the length 12+9, you can use trig to work out the Angle Theta of the hypotenuse at the center of the circle from Tan Theta = O/A = H / 21 Theta = ArcTan (H / 21) And then you can use trig again 21 = r Cos Theta So r = 21 / Cos Theta
From the thumbnail (assuming the image is depicting a quarter disk): Let a be the distance between the line segment of length 12, and the bottom horizontal line of the quarter disk. Define horizontal x-axis (pointing to the right), coinciding with the line segment of length 12 ; and vertical y-axis (pointing up), coinciding with left vertical side of quarter disk. Equation of the (quarter) circular arc in this frame of reference becomes: x² + (y+a)² = r² where r is the radius of the circular arc. Two points on the circle arc are (x,y) = (9, 16) and (x,y) = (9+12, 0) = (21,0) , which must meet the circle equation: 9² + (16+a)² = r² [eq. 1] 21² + (0+a)² = r² [eq. 2] Combine eq. 1 and eq. 2 : 9² + (16+a)² = 21² + (0+a)² 9² + 16² + 32a + a² = 21² + a² 9² + 16² + 32a = 21² 32a = 21² - 9² - 16² 32a = (21 - 9)(21 + 9) - 16² 32a = (12)*(30) - 16² 4a = (12)*(30)/8 - 16²/8 4a = (3)*(15) - 16*2 4a = 45 - 32 4a = 13 a = 13/4 Plug result back into eq. 2 : 21² + (0+13/4)² = r² 441 + 169/16 = r² 451 + 9/16 = r² 7216 + 9 = 16r² 7225 = 16r² 289*25 = 16r² 17² * 5² = 4² * r² r = 17 * 5 / 4 = 85/4 = 21.25
I have two questions, pls someone explain it to me: 1. timestamp 2:54; I tried every single possible logic in my head, in the binomials, where does "32h" came from? 2. timestamp 3:28; I also can't find the result to "7225" from the equation "21^2 + (13/4)^2".
1. Just expand (16+h)^2 and look at the first-order term in h, what do you get? 2. Just add these 2 numbers as common fractions with a (common) denominator of 16. What is the numerator of the sum?
With the help of GPT-4o: The radius of the circle can be found using the following steps: 1. Equation of the Perpendicular Bisector: The line equation y=(x−6)⋅12/16+8 represents the perpendicular bisector of the hypotenuse of the right triangle with sides 12 cm and 16 cm. 2. Finding the Circle Center (xM, yM): xM = −9 Substitute xM into the line equation to get yM: yM = (−9−6)⋅12/16+8 = −3.25 3. Calculating the Radius r: Using the Pythagorean theorem: r² = 9² + (16 − yM)² Thus, the radius of the circle is 21.25 cm.
Let’s take a trip back in time to solve this problem. The first means is to use the Greek idea of chords, but we are going to go much deeper. 9 + 12 = 21. 21 = halfchord of 42. We use intersecting chord theory to get the chord of the line segment 16. A*B = C*D where A,B and C,D are the length of 2 line segments, respectively created by two intersecting chords. 42 - 12 = 30. 30* 12 = 360 360/16 = 22.5 [360/8 = 45] 45/2 + 32/2 = 77/2 . The halfchord of 77/2 = 77/4 Radius = SQRT(9^2 + 77^2/16) = SQRT(81*16/16 + 77^2/16) = 1/4 * SQRT (1296 + 5929) = SQRT(7225)/4 = positive solution of +/- 85/4 = 21.5 16 and 12 are sides of a triangle the remaining side is a chord sides of a right triangle divisible by 3 or 4 and a another side is a divisor of both this one happens to be 4*(3-4-5), the common divisor is 4 and so the chord is 4 * 5 so the chord of some other angle is 20 with respect to the radius. All of a sudden we find our self being such into the past and human civilization is laid out before us. Ok So a bisector bisects this chord at 10 units. This bisector from the origin always is orthogonal to the chord. A chord of 20, which slopes -16 down per 12, its bisector with a slope 3 parts up per 4 parts right through the origin. If the bisector has slope 3/4 then the intercept is which we know has a hypotenuse of 5 parts for every 4. It seems however that the Babylonian scribes are whispering off in the distance, let’s run to see if we can hear what they are saying [running, running] we need a running back! Let’s summon one! So we now have a mystical right triangle, it’s time to eat some magic mushrooms and eat some ergot of rye. Because Now everything we have is imaginary. Lets call the origin 0 let’s call the the intersection of “9” and “16” i and the intersection of the chord of Len 20 and its bisector j, the halfchord is thus ij in length and if we know the bisect we can solve for oi which happens to also be a radius. The length of the bisector is also oj, “The Juice”, he’s dead though so let’s bring him back to life cause he’s a running back. ..[“Nichole …”, yeah it must be the shrooms]. Since the bisector is a halfway point on a line defined by two orthogonal segments it travels halfway orthogonal to both segments so j has an x value of Xj = Xi + 12/2 = 9 + 6 = 15 So The Juice is a running along and once he has traveled 15 along X axis he has to stop and say how sorry he is to Nichole [“Nichole I didn’t mean to …”] . What is his position along Y. Yj = 3/4 * Xj = 3 * 15 / 4 = 45/4. So how far has OJ run until his soul was overwhelmed by guilt? OJ has run SQRT (15^2 + 15^2*3^2/4^2) = 1/4 *SQRT (15^2*4^2+ 15^2*3^2) = 15/4 SQRT (4^2 + 3^2 = 5^2) = 75/4 yards Ok now we know where that netherworld escapee is we send out the kicking team and ⚡️kick⚡️ him back to 𒆳. So now we don’t need a running back anymore, we need to throw a Hail Mary pass and see if the babylonian spirits accept our answer. Recalling that a bisector cuts a chord to form its halfchords (10/each) then they are orthogonal. The solution is appearing through the clouds of heaven, either that or I am about to have a stroke. Chanting with the shrooms “Hail Mary, mother of god, please let SQRT (10^2 + 75^2/4^2) equal the radius” as the magic football flies through the air. Weeeeeeeeee. In flight however the football has magical writing appear on its sides then to be replaced SQRT (4^2 * 5^2 * 2 ^2 / 4^2 + 3^2 * 5^2 * 5^2 / 4^2 ) 1/4 * SQRT (5^2 * 4^2 * 2^2 + 5^2 * 3^2 * 5^2) 5/4 * SQRT( 4^2 * 2^2 + 3^2 * 5^2) 5/4 * SQRT (8^2 + 15^2) 5/4 * SQRT(64 + 225) 5/4 * SQRT (289) a 1965 ford galaxy appears momentarily and replaces the pigskin and a portal opens. 85/4 appears on the rear license plate, as the car vanishes though the portal The front license plate magically remains behind. When we turn it over it reads 𒑱𒑰 𒌋𒑱 with a picture of the Plimpton 322 cuneiform tablet next to it with some writing that says “if we are remembered we still live”
There is another solution to this problem which is way outside the box. I present this because I think this is how the Babylonians mathematically calculated their approximation of the Square root of 2, which was so important to Pythagoras’s mathematics cult that he killed one of his acolytes for revealing it. It’s not a solution but an approximation based on rectilinear methods. Though I need to caution that as far as we currently know the first wheel is the ox wagons created by the Sumerians. So they may have known at least some trig. If we have a 3-4-5 triangle we can guess the Chord 90° is between 5/3= 1.6666 and 5/4 = 1.25. The average of the two is (20+ 15)/(12*2) = 35/24 = 1 + 11/24 The next triangle is 20-21-29. We have chords of 29/21 and 29/20 which give 1.38 and 1.45, the average is 2.83/2 = 1.415. A third is 696, 697, 985. 985/696 > chord 90° > 985/697, average 1.41421. Squared = 2.0000030 So let’s apply this logic to our problem and just forget everything we ever learned about geometry except that there are orthogonal things, and squares. We can assume they knew something like the Pythagorean formula. Let’s begin this problem arguing from the absurd, what if the short leg was a segment on the X axis. Then (x1, y1) = 9,16 and x2,y2 = (21,0) Accordingly every point on the circle is equal distance from the center. SQRT (9^2+16^2) = SQRT (81+256 = 337) = 18.3. We must add height to the first point on the chord. If we superpose the quadrant inverted across the diagonal onto itself we get 2 @ 12-16-20 triangles. 9, z (original point) and z - 16, 21 [21 is the sum of non-redundant horizontal segements] Are two adjacent point on the top (i and j’) since r must be at least 21 then R^2 > 441 then 9^2 + Z^2 > 441 : Z^2 > 441 - 81 : Z^2 > 360]. Z > 19 19 - 16 = 3. 3^2 + 21^2 = 450 450 - 81 = 369 = z ^2 thus z > 19.2 19.2 - 16 = 32/10. 32^2/10^2 + 210^2/10^2 = 1/100 (1024 + 44100 = 45124) = 451.24 451.24 - 81 = 370.24 = z^2 = 19.2462 This process is repeated until the coordinates settle on as 9, 19.25 and 21, 3.25 when these, about 3 more cycles at their level of precession The radius was thus 21.25. r-16 must be >5 then 9^2 + Z^2 = >5^2 + 21^2 = 441 + >25 = >466 = 81 + Z^2 Thus >381 = Z^2 or z > 19.5 or 39/2 . This is the first approximation of z From this we can estimate the the radius SQRT(18^2/2^2 + 39^2/2^2) = 1/2 SQRT(3^2*3^2*2^2 + 3^2*13^2) = 3/2 SQRT(36 + 169) 3/2 SQRT(205) = 30/2 SQRT(2.05) = 15 * 1.4317 = 21.476 not a bad first approximation. If this is the case the 19.5 is the estimate of y at one end of the chord, the estimate of y at the other end is 19.5 - 16 since 16 is the height of the triangle. The other end is 21^2 + 3.5 ^2 Thus est. r^2 = 7^2*6^2/2^2 + 7^2/2^2 = (7/2)^2 * 37 = 21.29^2
I would use generic equation for a circle (x-h)^2+(y-k)^2=r^2 and put it's middle into the point [0;0], so it will be x^2+y^2=r^2. After that, I would say, that we have two points [9;y] and [21;y-16] which are sitting on the circle. Next step would be two equations 9^2+y^2=r^2 and 21^2+(y-16)^2=r^2 (which are just the two points put into the circle equation). By solving them I get that y = 19.25 Solving the equation for circle I get that r= 21.25
When you see a circle geometry problem, the first thing you should do is draw connections from the center to important points on the edge. Then the little "h" in method one starts becoming obvious.
Let O be the center of the quadrant where the vertical and horizontal radii meet. Let A, B, C, D respectively be the left and right ends of the 9 cm segment, and the left and right ends of the 12 cm segment. Extend the segment of 12 cm length to the left until it joins the vertical radius at point E. length(EO)= sqrt (r^2- (AB+CD)^2) =sqrt(r^2-441). Also, (EO+AE)^2 + AB^2 =OB^2 => (sqrt(r^2-441)+16)^2+9^2=r^2 => r=21.25
I did it the first way, but I'm lying down so I had to do it in my head. It was fairly tough to do the arithmetic to get h = 13/4, but not too bad. But then, 21^2 + (13/4)^2 = r^2 was a challenge. I figured it would be easier without the fraction, so I multiplied through: 84^2 + 169 = (4r)^2. But what's 84^2? Here's where I though it was cool: 169 = 84 + 85, so 84^2 + 169 = 84^2 + 84 + 85 = 84*85 + 85 = 85^2. Therefore, r = 85/4. It's just arithmetic, but I thought it was pretty cool.
I had another approach. (16+h) is the geometric median of (9+R) and (9-R). So, (16+h)^2= R^2-81. 256+32h+h^2=R^2-81. => R^2=337+32h+h^2. And (12+9) is the geometric median of (R+h) and (R-h). So, 441=R^2-h^2. => R^2=441+h^2. Using these 2 equations, 337+32h+h^2=441+h^2, so h=104/32=13/4.
Didn't watch, solved it with analytic geometry. x²+y²=r² Two points on the circle: A(21,a) B(9,b) 21²+a²=r² 9²+b²=r² 16+a=b From these 3 equations: r=21.25
@@robertveith6383 I know that. Inradius is twice the area divided by the perimeter, the circumradius is the produce of the sides over two times the area. Properly parentheses are required, I agree, but I was simply thanking for the reminder of that. I wasn't attempting to express it rigorously, just referentially.
my method was i feel a little simpler. just consider the equation for a circle, sqrt(r^2-x^2) and notice that we have 2 points on the circle. the slope between them is -16/12 so the slope of our circle will be the same over 9 to 9+12. plug this all in and the -1/12 cancels. substituting R^2 for r^2-9^2 helped me manipulate but its unnecessary. its a single equation that can be solved with algebra and after doing so you end up with an answer.
Because of the question "Is it Possible?" I first drew out the zig-zag to scale then drew the chord between the points to be on the circle and bisected it, constructed the radial line thru the center area and dropped the 9 offset line to locate the center of the circle. The circle drawn fit so I knew it was going to be possible. Did Pythagoras and similar triangles to work out the dimensions and got the same 21.25 answer. If there's a hard way I'll find it.
I did this in CAD to get the radius in a few seconds. This is also a means of finding the centre by choosing a suitable point as origin 0,0 and then getting the equations of the two lines and solving them to see where they cross.
Hey, I love your videos. Can you make a video on a tetration problem? More specifically anti tetration. Suppose a tetration x tetrated to 5 = 2. X =? Is there a general formula?
Jesus loves you ❤️ Please repent and turn to Him and receive Salvation before it is too late. The end times written about in the Bible are already happening in the world. Jesus is the son of God and He died for our sins on the cross and God raised Him from the dead on the third day. Jesus is waiting for you with open arms but time is running out. Please repent and turn to Him before it is too late. Accept Jesus into your heart and invite Him to be Lord and Saviour of your life and confess and believe that Jesus is Lord, that He died for your sins on the cross and that God raised Him from the dead. Confess that you are a sinner in need of God's Grace and ask God to forgive you for all your sins through Jesus. Jesus loves you. Nothing can compare to how He loves you. When He hung on that cross, He thought of you. As they tore open His back, He thought of your prayer time with Him. As the thorns dug into His head, He thought of you spending time reading the Bible. As the spears went into His side, He imagined embracing you in heaven. Please repent and turn to Jesus now before it's too late. His heart longs to be with you but time is running out.
We know its more than 21... for the radius... The other notion is that any time your given a length in any object you can sqaure it nad replicate it to make a grid and it can be subdivided and increased in size to which you can calculate exact points, however in this case you can use it to get closer and closer to the radii, but the 9 is the key imo to get an exact count. The easiest way is taht there is only one exact radii. so a mirror arc setup would for a limitation in doing the boxing of the others into various quadralaters, ratehr than approaching a infnite sum, you would be appraoching a finite sum. Thats jsut my thoughts...
3:01 i dont understand something - you name the hypothenusa of both triangles r - and both r's are not equal slopes/degrees. But you make them = equal, and use it in your calculation. So i am in the middle of dismantling how you solve these riddles
So I did this another way entirely because I didn’t think to use y as my second unknown. I drew in three chords: one of them by doubling the 9cm segment to get an 18cm chord (bc a radius is a perp bisector of a chord), one from the left end of that to the end of the 12 cm segment which by pthm is 34, and one from the ends of the 12 and 16 cm segments which by pthm is 20cm Then I used angle/arc relationships. The three chords form an inscribed triangle, so I used law of cosines to get one of the inscribed angles. Doubled that the get the intercepted arc, transferred that to a central angle that intercepted one of my chords. Having an isosceles triangle with two radii and one chord, and knowing the vertex angle and the length of the chord, I bisected the triangle and solved the remaining right triangle using basic trig. I got 21.25… the long way 😂
Sir is there any theorem about intersecting chords Like AB. BC =DE.EF As there may be chords of different lengths. If one chord is 20 (12+8) And second is 8(5+3) It means they should be perpendicular
I'm not sure what you're getting at. First of all, it's not possible to have two intersecting chords of length 20 and 8 that intersect in such a way as to be divided into length 12+8 and 5+3, since 12*8 ≠ 5*3. But it is possible to have chords of length 20 and 8 that intersect in such a way that the chord of length 8 is divided into segments of length 5 and 3, while the chord of length 20 is divided into segments of length 10+√85 and 10−√85, sine 5 * 3 = (10+√85) * (10−√85) = 15. Second, chords do not have to be perpendicular for this to work. It just happens that in the example shown in this video they are perpendicular, which then allows us to use Pythagorean theorem.
Used a^2 + b^2 = c^2, where c is the radius 9^2 + (x+16)^2 = r^2 21^2 + x^2 = r^2 Solved for x (3.25) then substituted 'x' into one of the equations to solve for 'r' (21.25). Then used the second equation to confirm 'x', and therefore 'r', is correct (first time I forgot to add in 9^2 and got different answers 👍).
I extended the 9cm length to the left to the other side of the circle, then solved for the circumcentre of the triangle formed with the other two points on the circumference.... nice!
Jesus loves you ❤️ Please repent and turn to Him and receive Salvation before it is too late. The end times written about in the Bible are already happening in the world. Jesus is the son of God and He died for our sins on the cross and God raised Him from the dead on the third day. Jesus is waiting for you with open arms but time is running out. Please repent and turn to Him before it is too late. Accept Jesus into your heart and invite Him to be Lord and Saviour of your life and confess and believe that Jesus is Lord, that He died for your sins on the cross and that God raised Him from the dead. Confess that you are a sinner in need of God's Grace and ask God to forgive you for all your sins through Jesus. Jesus loves you. Nothing can compare to how He loves you. When He hung on that cross, He thought of you. As they tore open His back, He thought of your prayer time with Him. As the thorns dug into His head, He thought of you spending time reading the Bible. As the spears went into His side, He imagined embracing you in heaven. Please repent and turn to Jesus now before it's too late. His heart longs to be with you but time is running out.
This is pretty much a "do it in your head" kind of exercise. All you need to memorise is an assortment of smallish Pythagorean triples, in this case 13:84:85.
Ok, approximate wheel with chambers with alot semi circles that equals 9 Cm's and it's radius is 21 but the solve is a little over 21 and it could be 23 or 24.
If you can see that tan (x) = 8/15 where 2x is the angle that corresponds to a chord of length 20, then it is much easier to see the solution (and much more elegant ;). As you do not name your points (A, B, C, O etc.) it would be tiresome for me to explain how you could see that. 8, 15, 17 is a Pythagorean triple BTW. So, 17 * 5 / 4 yields the desired result.
@@Grecks75 If you take the left-symmetric quarter circle together with the given one, and try to see an inscribed angle; x corresponding to the chord/arc in question you may see that tan(x) is 16/30.
Without doing any analysis other than "it's longer than 21 and definitely shorter than 22" I thought "It's going to be around 21.25" I was shocked when it was _exactly_ 21.25
This time, for once in my life, I resist the temptation. Yes, it's always fun solving some mathematics problems. But this time I really need to do some thing else.
Could someone explain to me what is the reason for the Last part where he said that the side of the triangle is 20. Is that a theorem or somethung like that?
It's a special case of the Pythagorean Theorem: the 3-4-5 triangle. We know the height is 16 (i.e. 4x) and the base is 12 (i.e. 3x), and the hypotenuse is 5x, which is 20.
I tried it two ways: - Geometrical construction: the perpendicular bisector right two points intersects the perpendicular to the left point in the circle midpoint. Technically this can be converted into algebra. - Direct algebraic solution: basically the first approach. It got kinda complex to calculate with variables for the three lengths, but not particularly hard with these concrete numbers. - I did not get the chord-chord solution myself, however the algebraic term (2 * 9 + 12)^2 I had while doing the math definitely hints towards at least mirroring the picture to the left - and if I had thought more I might have at least found the last solution from that, which happens to use these very terms too.
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
“This is not an engineering exercise we can approximate…”
Well played, Presh.
I felt that right in the feels... This is legitimate push back from all the math puzzles us engineers and coders keep giving 'out of the box' or unintended answers for to wind him up ;) fair play sir.
First Fundamental Theorem of Engineering:
Near enough is good enough.
oof!
....and engineers use Pythagoras, it drives Presh crazy
which makes this utterly useless for the real world, where we could also skip all this and just use a ruler
My first time here, and I have nothing but respect and admiration for this presentation. The "other guy" who famously does these problems and refers to himself as Mr. Online Geometer or something 😉 would have taken twice as long, used three times the words, and explained one quarter of the solving strategies. This presentation is superior, and I learned two things I had forgotten since high school geometry 50 years ago or maybe never really knew: the chord-chord bit, and the circumference deal of (AxBxC) / 4r^2. I turn 65 in a month and take particular pleasure whenever I can increase my knowledge regardless of its direct applicability in my life.
BTW, I guessed 21 1/4 cm within 60 seconds, and solved using right triangles and simultaneous equations, but that took longer. I knew chords would be significant but didn't know the formula. The final solution is brilliant.
Cheers from SoCal. Go Dodgers.
Who is this "other guy" you're talking about? I've seen quite a few famous math channels on youtube, but no one does geometry problems as much as Presh.
As much i love Presh i got here from a guy who solved and explained this problem in 3:30.
@@x7inferno476 To be fair, Presh did it in 3:45 including the estimate section, and then did it three other ways
Complete the perpendicular that is currently 16, let the additional length be x. Subtend two radii to the edges of the circle where the lines meet the circle. Form two equations: r^2 = 21^2 + x^2 and r^2 = (16+x)^2 + 9^2.
Same :D
same solution here
same sir
you stole my answer, cheers
Same
I felt attacked with the "this is not an engineering problem" hahahaha. Love your content
I also gave a wry smile at that point (mainly because that's exactly how I approach this kind of question!)
I've seen someone else solve this problem on UA-cam already, but im glad I watched this video for the alternative methods. Great work!
The last method is certainly the most elegant. It all comes down to the pythagorean theorem in the end though.
It does, but Presh would work himself into a tizzy if he had to say Pythagoras
Not sure how many people know the r = abc/(4S) formula. To me it is more elegant if no esoteric formula's are used.
Didn't ANYONE ELSE use a variation of the master method like I did and constructed two intersecting chiod formulas together the answer
.more organic inthink
@@richardbloemenkamp8532 Must know formula for national exams for my country at least
@@leif1075I agree the last method is the least elegant because it requires the least intuition and the most memorization of niche formulas. The chord method was the neatest in my opinion
A little over 21 (12+9) from the thumbnail.
That's the first point addressed in the video. Pradesh preemptively said that this is not how you solve it, so I'm wondering if you watched the video at all. 🤣
@@sambena yeah, and his answer of a little over 21 is absolutely insufficient
@@sambena And now, imagine these are real measurements, in inches, feet, meters, or whatever, for the construction of a tunnel, and you have to be sure that some piece of equipment will fit between the last vertical segment and the arc at the right of the figure. That "a little over 21" is as useless as can be, and that's why we use maths and don't rely on approximations. 😀
@@fixups6536 Depends on your expertise. I'm a geologist by training. +/-100,000 is a reasonable margin of error.
Close enough. Full credit.
If we take the leftmost point of 9cm line segment as an origin, x axis going right and y axis going down, we get:
Point (9, 0) lies on a circle.
Point (21,16) lies on a circle as well.
x-coordinate of a center of a circle is 0.
General formula for a circle is (x - C_x)^2 + (y - C_y)^2 = R^2. We can replace C_x with 0 right away, getting x^2 + (y - C_y)^2 = R^2
Plug both points into this equation:
9^2 + (0 - C_y)^2 = R^2
21^2 + (16 - C_y)^2 = R^2
Right parts are equal, so we can equate left parts:
9^2+(0-C_y)^2 = 21^2 + (16 - C_y)^2
81 + C_y^2 = 441 + 256 - 32C_y + C_y^2 | subtract C_y^2 on both sides
81 = 441 + 256 - 32C_y | bring 81 to the right side and C_y to the left
32C_y = 441 + 256 - 32
32C_y = 616
C_y = 616/32 = 19.25
Notice that distance between (C_x, C_y) and point (9, 0) is a radius.
R = sqrt((0-9)^2+(19.25-0)^2) = sqrt(81 + 370.5625) = sqrt(451.5625) = 21.25
Another approach: the center lies on the perpendicular bisector of the hypotenuse of the right triangle with legs 12&16, which is a scaled 3-4-5 triangle so its perpendicular has slope 3/4 and passes through a point 15cm right of the vertical radius, hence is 3/4*15=11.25 units above the horizontal radius, so the 12cm line is 3.25cm above the horizontal radius. Knowing this distance we can calculate the radius using the Pythagorean Theorem: r^2=9^2+19.25^2 (or r^2=3.25^2+21^2 from the other point on the circle).
That's how I did it. It was simpler for me that any of the methods he demonstrated
and 13^2 + 84^2 = 36^2 + 77^2 = 85^2
I think I like this method best of all 🏆
Also, this problem highlights an interesting pythagorean tripple:
84² + 13² = 85²
I don't see that one too often.
You can build a pythagorean triple on any odd number like this. Just take half of the square, plus/minus half: 13^2=169 169/2=84.5 Therefore, (13,84,85) is a valid triple :)
@@6c3333 It's hard to believe that always works! 3^2=9, 9/2=4.5, +/- .5 = (3, 4, 5). Today, I learned about power chords and inscribed triangle circumradii. I thought I was done for the day, but noooo. You gave me this gem as well. Thank you!
@@6c3333how do you prove that?
@@6c3333 That's incredibly elegant, is this something known since pythagorean or discovered afterwards?
Well done.
A lot of creativity and insight.
Another method is to use the equation of the circle x^2 + y^2 = r2. We have two points of contacts. If we let the center be (0,0) and the y distance between the center and the first line be a, we have two equations
1) 21^2 + a^2 = r^2
2) 9^2 + (16+a)^2 = r^2
We can use comparison to solve for a. We get 3,25. Then we can come back to eq 1 to solve for r.
This is the same as the 2nd method except you picked the y coordinate of the x=21 point as the referente, while Presh choose the x=9 point.
The circumradius method shown blew my mind. Makes me want to learn more about it.
Very Easy just continue the vertical of 16 unit and tell that extra length as x. So we get the 1st equation 9²+(16+x)²=r².
Now drop vertical from rightmost edge of 12 unit length, again that vertical length which was dropped will be x so we get 2nd equation (9+12)²+x²=r².
On solving we get x=3.25 and r=21.25
My solution before watching: Place the measured segments on graph paper, with the y-axis on the vertical radius of the quarter circle and the x-axis on the 12cm segment. Form a segment from the top end of the 16cm segment to the right end of the 12cm segment. Form the perpendicular bisector of that segment. The bisector and the y-axis meet at the center, at the point (0,-3.25). The radius is the distance from that center to either end of the constructed segment, 21.25 cm.
Always keen on looking at your videos.
Method 3 can end in another solution:
The intersecting chords have legs with lengths of 12; 30; 16; 45/2
With the formula 4r^2=a^2+b^2+c^2+d^2 you get 1806,25 for 4r^2, which gives you the 21,25 for r again
I solved it with trigonometry
r.cosα=9 => r.sinα=√(r^2-9^2)
r.cosβ=21. => r.sinβ=√(r^2-21^2)
r.sinα-r.sinβ=16
=> √(r^2-9^2)-√(r^2-21^2)=16
We make it power 2 and it s
2nd degree equation where x=r^2
Funny thing is that's how I _thought_ Presh was going to solve this. I couldn't be bothered to do the calculation, so I watched anyway. Now I'm blown away!
I used trig too and now I feel silly after seeing a pure geometry method. Though I like how trig gets us straight to the radius without much transformation.
Thank you for providing multiple ways some of the problem. I always like those videos best!
As a geometry teacher I sometimes find your geometry videos humbling. I can quickly solve videos out there, but some of your methods surprise and challenge me. (Students/former students: Helping a teacher understand what you are thinking is the historical and fundamental reason we show our work!)
Very true! Thank you for what you do. Teachers (especially good and passionate ones) are increasingly underappreciated (and increasingly rare) in the world.
As you said, showing your work allows the teacher to understand what you are thinking. It's easy to say that getting the right answer is most important, but what if the answer isn't right? There isn't really a good way to figure out what went wrong and remediate the source of the confusion. As an added bonus, sometimes the teacher can learn something, too!
When I was learning geometry (maybe 30 years ago), I was in grade 10, about 14 years old. My teacher was maybe 28. I'm looking back at my 28-year-old self and am amazed by how much I _didn't_ know! Not that teachers are ignorant; far from it! But it's humbling to find something you didn't know, or didn't think of, and opportunities to learn and grow are forever abundant if we're willing to look for them. I think Neil deGrasse Tyson said it best (paraphrased): It's one thing to realize that we don't yet know the answers to many of our questions, but imagine the questions we don't even know to ask!
@@sabinrawr wonderful reply! My late wife was an Army officer. She not only flew the CH-47 helicopter and put forth maximal effort to be the very best pilot, she always brought that dedication to everything she did-especially leadership.
She liked to say knowledge falls into three broad categories. First comes knowing that we know something (it’s the basis of the Dunning-Kruger Effect). Then we start recognizing our limitations becoming aware of things that we don’t understand very deeply. Finally, we become aware that there is far more to be known than we can ever hope to not only grasp but even to be aware of. In short, “You don’t know what you don’t know.”
As an electrical engineer I went directly to make an equation with loveley sin and cos and found another way of solving it. Let's say there is a xy coordinate system. I then mapped a function of y(x) that descibes the circle. First, just dependend on the anlge of the current position of this quarter circle, we get y(alpha)=r*sin(alpha). For x then we get y(x)=r*sin(arccos(x/r)). We know that y(21) - y(9) = 16. By solving this equation with the little trick of sin(arccos(x)) = sqrt(1-x^2) we get that r = 21.25. A little more complicated ^^ Cheers
Just started watching but thanks for showing the image on a dark background. Watching this b4 going to sleep and its much easier on my eyes that way, lol
Really nice selection of methods shown and well explained.
I solved this with calculus.
Let R be the radius and f(x) = sqrt(r^2 - x^2)
The image shows that the avarage slope in the section from x=9 to x=21 is -16/12, so
-16/12 = [ f(21) - f(9) ] / (21 - 9)
Therefor
-16 = f(21) - f(9)
So:
-16=sqrt(r^2 - 21^2) - sqrt(r^2 - 9^2)
I solved this equation by building the square twice and so some algebraic standard stuff and got r = 21.25
First time I've heard about chord-chord, and I've come across circumradius once a long time ago. Great tools to have in my knowledge now
I just call that little piece below the 16 chord X, then 16(2X+16) = 360 from intersecting chords theorem. Solve for X (3.25)
Then use right angle triange , 9squared plus 19.25 squared = 451.5625: find sq root = 21.25 QED
I used the chord-chord method; it was the first and simplest that stood out to me.
- Connect the two points being on the circle
- Construct the center vertical to this line
- The intersection between this center vertical and the vertical line crossing the vertical line at the left end point of the 9 cm line is the midpoint of the circle.
First I mirrored the drawing in the vertical line to get two more points that the circle passes through. No more than one circle can pass through four different points, so it's not indeterminate. Then I took the perpendicular bisector of the line segment (21,0)-(9,16) and intersected it with the y-axis. The center is (0,-3.25), and I computed the radius with the distance formula.
I like it that it's counter-intuitive that zig-zag lines like those can fit ONLY into a circle that size -- even though the space above the 9 section and the space below the 12 section seem they could be different with a different sized circle.
Very thorough, very elegant solutions!
What a wonderfully clear exposition! I do have a little quibble, though - It involves the difference between an equation (has an equal sign in it) and an expression (which doesn't). At 2:45, I would say that the two expressions on the left are each equal to r^2, so they are equal to each other. In my day, I called this the Comparison method, but it seems that today it is seen as a variety of the Substitution method. ;-)
With analytical geometry:
x^2 + y^2 = r^2 if the center is in the leftmost bottom point.
Let the vertical distance from the center to "9" is y. Then for the two points on the circle we have
9^2 + y^2 = r^2
(9 + 12)^2 + (y - 16)^2 = r^2
Together their solution gives y = 19.25. If put y in the first eq. you obtain r = 21.25.
1:03 "“This is not just an engineering exercise where we can approximate…”
Time for some lateral thinking.
- Grab a screenshot of the figure.
- Apply the Erode effect with Paint Shop Pro, to make the lines 1 pixel thin.
- Measure 135 pixels for the 9 cm line, and 315 pixels for the radius
- Write its ratio as a proportion: 135:9 = 315:x
- Solve the proportion: x=315*9/135 = 21
you can do it pretty simple if you use the contact points to make a line, then, grab the oposite pendient and use it in the middlepoint, that line always goes to the center, after that you just need to evaluate the x coordinate where the center is, you can know it because of the distance that are showed on the drawing
Got a weird system of equations:
r = 21/cos(θ)
r = -9/cos(θ+2arctan(4/3))
I just used the law of sines on some triangles, I believe you get that r=+-85/4 but of course r is positive!
My solution - not sure which example it fits into - I guess it's closest to the first method?
Consider the radius at the two points the lines are in contact with the circle :
1) r = sqrt(9^2 + b^2) for the upper left point
also
2) r = sqrt(21^2 + a^2) for the lower right point
and the difference between y axis a and y axis value b is 16
3) b = a +16
sub 3 into 1 and put equations equal to each other
9^2 + (a+16)^2 = 21^2 + a^2
expand brackets and solve for a
a^2 + 32a + 16^2 - a^2 = 21^2 - 9^2
32a = 21^2 - 9^2 - 16^2
a = (21^2 - 9^2 - 16^2 )/32
a = 3.25
sub y1 into 2) to get the radius
r = sqrt(21^2 + 3.25^2)
r = 21.25
I love when Presh gets all excited when he says "and that's the answer!" because sometimes I'm excited, too.
I used the first (triangle) method. I took Geometry 50 years ago. I don't remember the last two methods. I was blown away at both the Chord-Chord rule and the Circumradius rule. ‼
Add method 5: assign coordinates as you did in method 2. The segment from (9,a) to (21,a-16) is a chord of the circle and its midpoint is (15, a-8). The perpendicular from the midpoint goes through the center of the circle, (0,0) so we must have that the vector from (9,a) to (21,a-16) is perpendicular to the vector from (0,0) to (15,a-8). The first vector is (12,-16) and the second is (15,a-8). They are perpendicular iff their dot product is 0. The dot product is 12 x 15 - 16 (a-8) = 308 - 16 a, which is zero when a = 77/4.
In first method value for h must be 5.125 (41/8) instead of 3.25 (13/4) and the radius must be 21.616 approximately
I started out constructing the 2 radii in the first solution, but then switched to focusing on angles in that isosceles triangle (2 radii and a chord) instead. You can get 2 equations for the angle, one if the form of acos(..) and another in the form of asin(A) + asin(B). Simplifying the 2nd into one asin(X) and using sin^2 + cos^2, you can solve the equation you get in r^4. Extremely roundabout, but still works :P
First: The radius is greater than 21 cm. Now I have to try to solve the problem.
Yes. The problem is solvable: r = 21,25.
r^2= 9^2+ 16+y2)^2
r^2= 21^2+y2^2
We have an aquation system of two equations with two variables. Solve for y2: y2=104/34.
Insert y2=104/32 in any of the equations and r^2= 21,25=21+1/4.
We can construct a radius going to the end of the 9 centimeter length. We can then draw a length from the end of the 16 centimeter length down to the bottom of the quarter circle, and call that length h. This gives us the equation 9²+(16+h)²=r². By creating another copy of the radius and rotating it so that the endpoint matches with the 12 centimeter length, we get the equation 21²+h²=r². Since both equations share a common equivalence, we can set them equal to each other. 9²+(16+h)²=21²+h². Solving for h…
81+256+32h+h²=441+h²
337+32h+h²=441+h²
337+32h=441
32h=104
h=13/4
…and plugging in the solved variable…
21²+(13/4)²=r²
441+169/16=r²
7225/16=r²
…we get that r is
*85/4 centimeters*
or
*21.25 centimeters*
:)
Before watching solution:
The most interesting question is whether the radius is even uniquely constrained. A circle that passes through 3 points is completely determined, both in center location and radius. The zig-zag defines two points, but is that enough? As it turns out, it is.
The zigzag clearly defines two points that are on the circle -- the corner between the 9cm horizontal segment and the 16cm vertical segment. Now we need one more, and the easiest one I can find is this: The top horizontal is perpendicular to the vertical radius, so the point 9cm to the left on that extended segment is also on the circle. That defines 3 points, meaning that the circle is completely determined.
Actually determining it is a much less interesting problem, but the sketch of how to construct it is to use the (-9,9) segment as one chord, the diagonal to the other two points as another chord, then constructing the two perpendicular bisectors of each chord, which will intersect at the center. Then draw the circle with that center and radius equal to the distance to any of the outer points.
The trick is that the diagram only shows a quarter of the circle, and the key to solving it (at least this way) involves another arc of the circle which isn't drawn.
Its been 4 yrs I started watching your videos, they are just awesome and brain teasing.Specially I love the logic puzzles you provide. There is also a book called 'What is the name of this book' by Raymond Smyullan.I haven't read it completely but I found it interesting for puzzles and logic.
Using the Small Angel Approximation, Cos 0 = 1, so the the Radius = 21
Once you have the length of the 16 extended Chord, you can work out the height above the diameter of the extended 12+9+9+12 Chord as H = (L - 2x16) / 2
With H and the length 12+9, you can use trig to work out the Angle Theta of the hypotenuse at the center of the circle from Tan Theta = O/A = H / 21
Theta = ArcTan (H / 21)
And then you can use trig again 21 = r Cos Theta
So r = 21 / Cos Theta
From the thumbnail (assuming the image is depicting a quarter disk):
Let a be the distance between the line segment of length 12, and the bottom horizontal line of the quarter disk. Define horizontal x-axis (pointing to the right), coinciding with the line segment of length 12 ; and vertical y-axis (pointing up), coinciding with left vertical side of quarter disk.
Equation of the (quarter) circular arc in this frame of reference becomes:
x² + (y+a)² = r²
where r is the radius of the circular arc.
Two points on the circle arc are (x,y) = (9, 16) and (x,y) = (9+12, 0) = (21,0) , which must meet the circle equation:
9² + (16+a)² = r² [eq. 1]
21² + (0+a)² = r² [eq. 2]
Combine eq. 1 and eq. 2 :
9² + (16+a)² = 21² + (0+a)²
9² + 16² + 32a + a² = 21² + a²
9² + 16² + 32a = 21²
32a = 21² - 9² - 16²
32a = (21 - 9)(21 + 9) - 16²
32a = (12)*(30) - 16²
4a = (12)*(30)/8 - 16²/8
4a = (3)*(15) - 16*2
4a = 45 - 32
4a = 13
a = 13/4
Plug result back into eq. 2 :
21² + (0+13/4)² = r²
441 + 169/16 = r²
451 + 9/16 = r²
7216 + 9 = 16r²
7225 = 16r²
289*25 = 16r²
17² * 5² = 4² * r²
r = 17 * 5 / 4 = 85/4 = 21.25
The circle passes through (12,0),(0,16),(-18,16). Straight forward.
r = sqrt(21^2+3.25^2)
That is some serious brute force action but I love it
R^2 = 21^2 + y^2 R^2 = 9^2 + (16 + y)^2
Subtracting both equations gives
0 = (16 + y)^2 - 360 - y^2
360 = 256 + 32y
y = 3.25
Putting back into (1) gives
R = 21.25 cm
I have two questions, pls someone explain it to me:
1. timestamp 2:54; I tried every single possible logic in my head, in the binomials, where does "32h" came from?
2. timestamp 3:28; I also can't find the result to "7225" from the equation "21^2 + (13/4)^2".
1. Just expand (16+h)^2 and look at the first-order term in h, what do you get?
2. Just add these 2 numbers as common fractions with a (common) denominator of 16. What is the numerator of the sum?
The second method is extraordinary, I loved it!
Very very interesting to see such diverse methods. Thank you.
I constructed a formula for the length of the left hand vertical which reduced down to sqrt(r^2-81)-sqrt(r^2-441)=16 and that works out at 21.25.
With the help of GPT-4o:
The radius of the circle can be found using the following steps:
1. Equation of the Perpendicular Bisector:
The line equation y=(x−6)⋅12/16+8 represents the perpendicular bisector of the hypotenuse of the right triangle with sides 12 cm and 16 cm.
2. Finding the Circle Center (xM, yM):
xM = −9
Substitute xM into the line equation to get yM:
yM = (−9−6)⋅12/16+8 = −3.25
3. Calculating the Radius r:
Using the Pythagorean theorem:
r² = 9² + (16 − yM)²
Thus, the radius of the circle is 21.25 cm.
I just made the left hand vertical x, then worked out r^2 for both diagonals. Simple.
r^2 = x^2 + 21^2 and r^2 = (x + 16)^2 + 9^2
Let’s take a trip back in time to solve this problem. The first means is to use the Greek idea of chords, but we are going to go much deeper.
9 + 12 = 21. 21 = halfchord of 42. We use intersecting chord theory to get the chord of the line segment 16. A*B = C*D where A,B and C,D are the length of 2 line segments, respectively created by two intersecting chords.
42 - 12 = 30. 30* 12 = 360
360/16 = 22.5 [360/8 = 45] 45/2 + 32/2 = 77/2 . The halfchord of 77/2 = 77/4
Radius = SQRT(9^2 + 77^2/16) = SQRT(81*16/16 + 77^2/16) = 1/4 * SQRT (1296 + 5929) = SQRT(7225)/4 = positive solution of +/- 85/4 = 21.5
16 and 12 are sides of a triangle the remaining side is a chord sides of a right triangle divisible by 3 or 4 and a another side is a divisor of both this one happens to be 4*(3-4-5), the common divisor is 4 and so the chord is 4 * 5 so the chord of some other angle is 20 with respect to the radius. All of a sudden we find our self being such into the past and human civilization is laid out before us.
Ok So a bisector bisects this chord at 10 units. This bisector from the origin always is orthogonal to the chord. A chord of 20, which slopes -16 down per 12, its bisector with a slope 3 parts up per 4 parts right through the origin.
If the bisector has slope 3/4 then the intercept is which we know has a hypotenuse of 5 parts for every 4. It seems however that the Babylonian scribes are whispering off in the distance, let’s run to see if we can hear what they are saying [running, running] we need a running back! Let’s summon one!
So we now have a mystical right triangle, it’s time to eat some magic mushrooms and eat some ergot of rye. Because Now everything we have is imaginary. Lets call the origin 0 let’s call the the intersection of “9” and “16” i and the intersection of the chord of Len 20 and its bisector j, the halfchord is thus ij in length and if we know the bisect we can solve for oi which happens to also be a radius.
The length of the bisector is also oj, “The Juice”, he’s dead though so let’s bring him back to life cause he’s a running back. ..[“Nichole …”, yeah it must be the shrooms]. Since the bisector is a halfway point on a line defined by two orthogonal segments it travels halfway orthogonal to both segments so j has an x value of
Xj = Xi + 12/2 = 9 + 6 = 15
So The Juice is a running along and once he has traveled 15 along X axis he has to stop and say how sorry he is to Nichole [“Nichole I didn’t mean to …”] . What is his position along Y.
Yj = 3/4 * Xj = 3 * 15 / 4 = 45/4. So how far has OJ run until his soul was overwhelmed by guilt?
OJ has run SQRT (15^2 + 15^2*3^2/4^2) = 1/4 *SQRT (15^2*4^2+ 15^2*3^2) = 15/4 SQRT (4^2 + 3^2 = 5^2) = 75/4 yards
Ok now we know where that netherworld escapee is we send out the kicking team and ⚡️kick⚡️ him back to 𒆳. So now we don’t need a running back anymore, we need to throw a Hail Mary pass and see if the babylonian spirits accept our answer. Recalling that a bisector cuts a chord to form its halfchords (10/each) then they are orthogonal. The solution is appearing through the clouds of heaven, either that or I am about to have a stroke.
Chanting with the shrooms “Hail Mary, mother of god, please let SQRT (10^2 + 75^2/4^2) equal the radius” as the magic football flies through the air. Weeeeeeeeee.
In flight however the football has magical writing appear on its sides then to be replaced
SQRT (4^2 * 5^2 * 2 ^2 / 4^2 + 3^2 * 5^2 * 5^2 / 4^2 )
1/4 * SQRT (5^2 * 4^2 * 2^2 + 5^2 * 3^2 * 5^2)
5/4 * SQRT( 4^2 * 2^2 + 3^2 * 5^2)
5/4 * SQRT (8^2 + 15^2)
5/4 * SQRT(64 + 225)
5/4 * SQRT (289) a 1965 ford galaxy appears momentarily and replaces the pigskin and a portal opens.
85/4 appears on the rear license plate, as the car vanishes though the portal
The front license plate magically remains behind.
When we turn it over it reads
𒑱𒑰 𒌋𒑱 with a picture of the Plimpton 322 cuneiform tablet next to it with some writing that says
“if we are remembered we still live”
There is another solution to this problem which is way outside the box. I present this because I think this is how the Babylonians mathematically calculated their approximation of the Square root of 2, which was so important to Pythagoras’s mathematics cult that he killed one of his acolytes for revealing it. It’s not a solution but an approximation based on rectilinear methods. Though I need to caution that as far as we currently know the first wheel is the ox wagons created by the Sumerians. So they may have known at least some trig.
If we have a 3-4-5 triangle we can guess the Chord 90° is between 5/3= 1.6666 and 5/4 = 1.25.
The average of the two is (20+ 15)/(12*2) = 35/24 = 1 + 11/24
The next triangle is 20-21-29. We have chords of 29/21 and 29/20 which give 1.38 and 1.45, the average is 2.83/2 = 1.415.
A third is 696, 697, 985. 985/696 > chord 90° > 985/697, average 1.41421. Squared = 2.0000030
So let’s apply this logic to our problem and just forget everything we ever learned about geometry except that there are orthogonal things, and squares. We can assume they knew something like the Pythagorean formula.
Let’s begin this problem arguing from the absurd, what if the short leg was a segment on the X axis. Then (x1, y1) = 9,16 and x2,y2 = (21,0)
Accordingly every point on the circle is equal distance from the center. SQRT (9^2+16^2) = SQRT (81+256 = 337) = 18.3. We must add height to the first point on the chord.
If we superpose the quadrant inverted across the diagonal onto itself we get 2 @ 12-16-20 triangles.
9, z (original point) and z - 16, 21 [21 is the sum of non-redundant horizontal segements] Are two adjacent point on the top (i and j’) since r must be at least 21 then R^2 > 441 then 9^2 + Z^2 > 441 : Z^2 > 441 - 81 : Z^2 > 360]. Z > 19
19 - 16 = 3. 3^2 + 21^2 = 450
450 - 81 = 369 = z ^2 thus z > 19.2
19.2 - 16 = 32/10. 32^2/10^2 + 210^2/10^2 = 1/100 (1024 + 44100 = 45124) = 451.24
451.24 - 81 = 370.24 = z^2 = 19.2462
This process is repeated until the coordinates settle on as 9, 19.25 and 21, 3.25 when these, about 3 more cycles at their level of precession
The radius was thus 21.25.
r-16 must be >5 then 9^2 + Z^2 = >5^2 + 21^2 = 441 + >25 = >466 = 81 + Z^2
Thus >381 = Z^2 or z > 19.5 or 39/2 . This is the first approximation of z
From this we can estimate the the radius SQRT(18^2/2^2 + 39^2/2^2) = 1/2 SQRT(3^2*3^2*2^2 + 3^2*13^2) = 3/2 SQRT(36 + 169)
3/2 SQRT(205) = 30/2 SQRT(2.05) = 15 * 1.4317 = 21.476
not a bad first approximation.
If this is the case the 19.5 is the estimate of y at one end of the chord, the estimate of y at the other end is 19.5 - 16 since 16 is the height of the triangle. The other end is 21^2 + 3.5 ^2
Thus est. r^2 = 7^2*6^2/2^2 + 7^2/2^2 = (7/2)^2 * 37 = 21.29^2
I would use generic equation for a circle (x-h)^2+(y-k)^2=r^2 and put it's middle into the point [0;0], so it will be x^2+y^2=r^2.
After that, I would say, that we have two points [9;y] and [21;y-16] which are sitting on the circle.
Next step would be two equations 9^2+y^2=r^2 and 21^2+(y-16)^2=r^2 (which are just the two points put into the circle equation).
By solving them I get that y = 19.25
Solving the equation for circle I get that r= 21.25
When you see a circle geometry problem, the first thing you should do is draw connections from the center to important points on the edge. Then the little "h" in method one starts becoming obvious.
Wow third method is really interesting
On that cords method, you wind up with 7225/4=4r^2.... that would simplify to r^2=7225 or r=85, not sure where it went wrong
Let O be the center of the quadrant where the vertical and horizontal radii meet. Let A, B, C, D respectively be the left and right ends of the 9 cm segment, and the left and right ends of the 12 cm segment. Extend the segment of 12 cm length to the left until it joins the vertical radius at point E. length(EO)= sqrt (r^2- (AB+CD)^2) =sqrt(r^2-441). Also, (EO+AE)^2 + AB^2 =OB^2 =>
(sqrt(r^2-441)+16)^2+9^2=r^2 => r=21.25
I did it the first way, but I'm lying down so I had to do it in my head. It was fairly tough to do the arithmetic to get h = 13/4, but not too bad. But then, 21^2 + (13/4)^2 = r^2 was a challenge. I figured it would be easier without the fraction, so I multiplied through: 84^2 + 169 = (4r)^2. But what's 84^2? Here's where I though it was cool: 169 = 84 + 85, so 84^2 + 169 = 84^2 + 84 + 85 = 84*85 + 85 = 85^2. Therefore, r = 85/4. It's just arithmetic, but I thought it was pretty cool.
It sure is, it's the Pythagorean triple of (13, 84, 85)!
I had another approach. (16+h) is the geometric median of (9+R) and (9-R). So, (16+h)^2= R^2-81. 256+32h+h^2=R^2-81. => R^2=337+32h+h^2. And (12+9) is the geometric median of (R+h) and (R-h). So, 441=R^2-h^2. => R^2=441+h^2. Using these 2 equations, 337+32h+h^2=441+h^2, so h=104/32=13/4.
This was an excellent example.
Good work
Didn't watch, solved it with analytic geometry.
x²+y²=r²
Two points on the circle:
A(21,a)
B(9,b)
21²+a²=r²
9²+b²=r²
16+a=b
From these 3 equations:
r=21.25
Nice! You should still watch, though. He uses some surprising methods to solve it in some very approachable ways.
@@sabinrawr I watched it afterwards. I agree.
I had forgotten that old chestnut R = abc/4A, Thanks for the reminder!
When you are dividing by the expression 4A, and you are writing in this horizontal style, the equation is R = abc/(4A).
@@robertveith6383 I know that. Inradius is twice the area divided by the perimeter, the circumradius is the produce of the sides over two times the area. Properly parentheses are required, I agree, but I was simply thanking for the reminder of that. I wasn't attempting to express it rigorously, just referentially.
my method was i feel a little simpler. just consider the equation for a circle, sqrt(r^2-x^2) and notice that we have 2 points on the circle. the slope between them is -16/12 so the slope of our circle will be the same over 9 to 9+12. plug this all in and the -1/12 cancels. substituting R^2 for r^2-9^2 helped me manipulate but its unnecessary. its a single equation that can be solved with algebra and after doing so you end up with an answer.
He didn't just solve the problem, he dominates the whole problem
Because of the question "Is it Possible?" I first drew out the zig-zag to scale then drew the chord between the points to be on the circle and bisected it, constructed the radial line thru the center area and dropped the 9 offset line to locate the center of the circle. The circle drawn fit so I knew it was going to be possible. Did Pythagoras and similar triangles to work out the dimensions and got the same 21.25 answer. If there's a hard way I'll find it.
I did this in CAD to get the radius in a few seconds. This is also a means of finding the centre by choosing a suitable point as origin 0,0 and then getting the equations of the two lines and solving them to see where they cross.
It^s solvable because you have at least 3 fixed points on the circumference (by symmetry).
Teach or remind me all the time. Thank you.
your chord-chord method is redoing Pythagoras in an equiform triangle with twice the sides
Hey, I love your videos. Can you make a video on a tetration problem? More specifically anti tetration. Suppose a tetration x tetrated to 5 = 2. X =? Is there a general formula?
Jesus loves you ❤️ Please repent and turn to Him and receive Salvation before it is too late. The end times written about in the Bible are already happening in the world. Jesus is the son of God and He died for our sins on the cross and God raised Him from the dead on the third day. Jesus is waiting for you with open arms but time is running out. Please repent and turn to Him before it is too late. Accept Jesus into your heart and invite Him to be Lord and Saviour of your life and confess and believe that Jesus is Lord, that He died for your sins on the cross and that God raised Him from the dead. Confess that you are a sinner in need of God's Grace and ask God to forgive you for all your sins through Jesus.
Jesus loves you. Nothing can compare to how He loves you. When He hung on that cross, He thought of you. As they tore open His back, He thought of your prayer time with Him. As the thorns dug into His head, He thought of you spending time reading the Bible. As the spears went into His side, He imagined embracing you in heaven. Please repent and turn to Jesus now before it's too late. His heart longs to be with you but time is running out.
before watching, 21.25 using equation of a circle X^2 + Y^2 = R^2
We know its more than 21... for the radius... The other notion is that any time your given a length in any object you can sqaure it nad replicate it to make a grid and it can be subdivided and increased in size to which you can calculate exact points, however in this case you can use it to get closer and closer to the radii, but the 9 is the key imo to get an exact count. The easiest way is taht there is only one exact radii. so a mirror arc setup would for a limitation in doing the boxing of the others into various quadralaters, ratehr than approaching a infnite sum, you would be appraoching a finite sum. Thats jsut my thoughts...
3:01 i dont understand something - you name the hypothenusa of both triangles r - and both r's are not equal slopes/degrees. But you make them = equal, and use it in your calculation. So i am in the middle of dismantling how you solve these riddles
We don't care about the degree, they're both radius of the circle, and both highlighted triangles are right triangles.
So I did this another way entirely because I didn’t think to use y as my second unknown.
I drew in three chords: one of them by doubling the 9cm segment to get an 18cm chord (bc a radius is a perp bisector of a chord), one from the left end of that to the end of the 12 cm segment which by pthm is 34, and one from the ends of the 12 and 16 cm segments which by pthm is 20cm
Then I used angle/arc relationships. The three chords form an inscribed triangle, so I used law of cosines to get one of the inscribed angles. Doubled that the get the intercepted arc, transferred that to a central angle that intercepted one of my chords.
Having an isosceles triangle with two radii and one chord, and knowing the vertex angle and the length of the chord, I bisected the triangle and solved the remaining right triangle using basic trig.
I got 21.25… the long way 😂
Sir is there any theorem about intersecting chords
Like AB. BC =DE.EF
As there may be chords of different lengths.
If one chord is 20 (12+8)
And second is 8(5+3)
It means they should be perpendicular
I'm not sure what you're getting at. First of all, it's not possible to have two intersecting chords of length 20 and 8 that intersect in such a way as to be divided into length 12+8 and 5+3, since 12*8 ≠ 5*3.
But it is possible to have chords of length 20 and 8 that intersect in such a way that the chord of length 8 is divided into segments of length 5 and 3, while the chord of length 20 is divided into segments of length 10+√85 and 10−√85, sine 5 * 3 = (10+√85) * (10−√85) = 15.
Second, chords do not have to be perpendicular for this to work. It just happens that in the example shown in this video they are perpendicular, which then allows us to use Pythagorean theorem.
I tried method 3 before watching this, but couldn't figure out the value of y. Thanks 🙏
Used a^2 + b^2 = c^2, where c is the radius
9^2 + (x+16)^2 = r^2
21^2 + x^2 = r^2
Solved for x (3.25) then substituted 'x' into one of the equations to solve for 'r' (21.25). Then used the second equation to confirm 'x', and therefore 'r', is correct (first time I forgot to add in 9^2 and got different answers 👍).
Just one question,
Where /how to make this amazing animation
Like tool , software please suggest me
I extended the 9cm length to the left to the other side of the circle, then solved for the circumcentre of the triangle formed with the other two points on the circumference.... nice!
Jesus loves you ❤️ Please repent and turn to Him and receive Salvation before it is too late. The end times written about in the Bible are already happening in the world. Jesus is the son of God and He died for our sins on the cross and God raised Him from the dead on the third day. Jesus is waiting for you with open arms but time is running out. Please repent and turn to Him before it is too late. Accept Jesus into your heart and invite Him to be Lord and Saviour of your life and confess and believe that Jesus is Lord, that He died for your sins on the cross and that God raised Him from the dead. Confess that you are a sinner in need of God's Grace and ask God to forgive you for all your sins through Jesus.
Jesus loves you. Nothing can compare to how He loves you. When He hung on that cross, He thought of you. As they tore open His back, He thought of your prayer time with Him. As the thorns dug into His head, He thought of you spending time reading the Bible. As the spears went into His side, He imagined embracing you in heaven. Please repent and turn to Jesus now before it's too late. His heart longs to be with you but time is running out.
How did you comment 3 days ago when video was posted just 1 minute ago?
What??
@@AhmadHasanOfficial Or was video privately sent to him 3 days ago?
@@NeverMatter Probably early access from Patreon or whatever
This is pretty much a "do it in your head" kind of exercise. All you need to memorise is an assortment of smallish Pythagorean triples, in this case 13:84:85.
Got it. Looks like I used "method 1". Cool puzzle.
Just use the rectangle property of tthe circle. The solution can be got in maybe 3 lines.
You did a moderately good job of hiding the fact that both approaches did the same math.
I found the correct answer.🥳
Unfortunately, I had to use CAD and still couldn't figure out the methodology.😭
Ok, approximate wheel with chambers with alot semi circles that equals 9 Cm's and it's radius is 21 but the solve is a little over 21 and it could be 23 or 24.
Well done!
Knew about 3, 4, 5 right triangles, but 8, 15, 17? How long has that been true?
If you can see that tan (x) = 8/15 where 2x is the angle that corresponds to a chord of length 20, then it is much easier to see the solution (and much more elegant ;).
As you do not name your points (A, B, C, O etc.) it would be tiresome for me to explain how you could see that.
8, 15, 17 is a Pythagorean triple BTW. So, 17 * 5 / 4 yields the desired result.
I think I know which chord you're talking of, but how do you get to tan(x)=8/15? Can you drop me a hint?
@@Grecks75 If you take the left-symmetric quarter circle together with the given one, and try to see an inscribed angle; x corresponding to the chord/arc in question you may see that tan(x) is 16/30.
Im a Little foggy on that last one where did you get that formula
Without doing any analysis other than "it's longer than 21 and definitely shorter than 22" I thought "It's going to be around 21.25"
I was shocked when it was _exactly_ 21.25
This time, for once in my life, I resist the temptation. Yes, it's always fun solving some mathematics problems. But this time I really need to do some thing else.
Could someone explain to me what is the reason for the Last part where he said that the side of the triangle is 20. Is that a theorem or somethung like that?
It's a special case of the Pythagorean Theorem: the 3-4-5 triangle. We know the height is 16 (i.e. 4x) and the base is 12 (i.e. 3x), and the hypotenuse is 5x, which is 20.
I want ask a math question, what is the difference between "speed 10% up" and "used time 10% down"
There certainly is a difference: 1.1 does NOT equal 1/0.9, that's it!
Let (21,y) be on the circle.
Then r^2 = 21^2 + y^2 = 9^2 + (16 + y)^2, whence y = 104/32
So r = Sqrt (21^2 + y^2) = not pretty
Beauty of maths❤❤❤
I actually solved it with method 1. That was quite easy!
I tried it two ways:
- Geometrical construction: the perpendicular bisector right two points intersects the perpendicular to the left point in the circle midpoint. Technically this can be converted into algebra.
- Direct algebraic solution: basically the first approach. It got kinda complex to calculate with variables for the three lengths, but not particularly hard with these concrete numbers.
- I did not get the chord-chord solution myself, however the algebraic term (2 * 9 + 12)^2 I had while doing the math definitely hints towards at least mirroring the picture to the left - and if I had thought more I might have at least found the last solution from that, which happens to use these very terms too.