There is a slightly more elementary way to solve the first problem. The sides of similar triangles have the same ratio, so you just have (4-r)/r = r/(6-r) and just apply basic algebra to solve for r.
Puzzle 1 is FAR simpler than this video showed! The 4x6 right triangle has the same proportions as the right triangles formed in its corners. The semicircle's radius R forms right angles with the two sides (4 and 6) at the points of tangency and thus forms a square R x R at the lower left. So the upper-left right triangle has a left side of 4-R and a base of R, and those two sides are in the same proportions as 4 and 6 are. (4-R)/R = 4/6 = 2/3 4-R = (2/3)R 4 = (5/3)R 12/5 = R = 2.4
There's another solution that's even simpler (in my opinion). If you consider the point at the right angle to be the origin, then the hypotenuse is the line y=-(2/3)x+4, and the center of the circle, which is on this line, has coordinates (r,r). Substituting, we get r=-(2/3)r+4, and solving the equation we arrive at r=12/5.
For question one Take the right angled vertex as origin. Then the center if circle would be (r,r) and it lies on the line joining (0,4) and (6,0) which is 2x+3y=12 Substituting the point (r,r) gives us r=12/5
for the first puzzle i used the equation y=-4/6x+4, because the semi circle is tangent to the legs of the right triangle y=x and solve for x =-4/6x+4, x=12/5
Problem 1: Equation of hypotenuse: y/4 + x/6 = 1 Position of circle center: (x,y) = (r,r) Circle center is on hypotenuse: r/4 + r/6 = 1 ==> r = 1/(1/4 + 1/6) = 24/(6+4) = 24/10 = 2.4 Problem 2: Let O = leftmost point of the base, P = rightmost point of base, A, B, C are the peaks from left to right (so C is right 10 above P), while D and E are the local valleys from left to right. ( |AD| = 6 , |DB| = 5 , |BE| = 4 , |EC| = 3 ). Extend the 45-degree slope OA , as well as the vertical line PC ; let's name the intersection S. From C, draw a line perpendicular to OS ; let's call the intersection T. |CT| = |AD| + |BE| = 6+4 = 10 , and since triangle CST is an isosceles right triangle (= 90-45-45 triangle) we know |TS| = 10 and |CS| = 10*sqrt(2) . Since OPS is also an isosceles right triangle, |OP| = |PS| = |PC| + |CS| = 10 + 10*sqrt(2) = 24.142 (approximately) Problem 3: Let R = radius of large circle, and r = radius of small circle. Distance between two lines = 2R = 96 + 2r ==> r = R - 48 . Half of blue line segment and half of green line segment form a right triangle with hypotenuse (r+R) ==> 48^2 + R^2 = (R+r)^2 Substitute r = R - 48 yields 48^2 + R^2 = (R + R - 48)^2 48^2 + R^2 = (2R - 48)^2 48^2 + R^2 = 4*R^2 - 192R + 48^2 0 = 3*R^2 - 192R R^2 = 64R R = 0 OR R = 64 ==> radius of large circle is R = 64 Distance between centers of large circles is 2R = 2*64 = 128 .
Question 2 is even easier if you rearrange the segments so that all those sloping up are together, followed by the two that slope down before solving in a manner similar to the video.
I noticed the closing background music started to play after Presh said "That's the answer" for problem 1, but was abruptly stopped before it could continue playing more 😂😂😂😂😂 still can't stop laughing
For question 2 I found a different solution to get the same answer! I drew a straight line 90 degrees with the right face of 10, parallel with the base, and broke the top up into several triangles. First, the hypotenuse of the triangle with a leg of 3 is 3rt2. Then, the other leg is 3, leaving a tiny triangle with legs of 1 at the top of the side labeled 4. The hypotenuse of this will be 1rt2. Then, the side of length 5 will have 4 left below the line, and so the hypotenuse is 4rt2. Then, the side of 6 will have 2 above the line, leaving 2rt2. Lastly, I drew a line straight down from there, perpendicular with the base. Because the far right length is 10, this length is also 10 and this creates a right triangle with 10 as the legs. Therefore you get 10 + 10rt2. Super cool!
Exactly the solution I saw. Why complicate it when you can draw a rectangle and get Side X of the rectangle using the Triangles at the top at the same time creating an Equilateral Triangle at the far left.
The second problem can also be calculated by taking the "4" line and moving it to the corner where the "3" line and the "10" line meet. Move the "6" line to extend the "4 "line. This then creates a single triange that has a base of "10" and a height of "10" so the vertical height is 10*sqrt(2) ( + the provided 10) = 24.14. As the original triange is a "45, 45, right" triangle, the base = height. Therefore the distance (base) = 24.14.
For problem 3, once you create the right angle triangle, we know one side length must be 48 since it cuts the 96 measurement into two equal parts. Would that simplify the equations?
Yes, it does. My solution is a bit different (it starts with realizing that 96 + 2r = 2R because they are perfectly sandwiched between two parallel lines) but it also uses the hypotenuse of that triangle. (with r = small circle radius, R = big circle radius)
@@BenDRobinson I think so too. Variable names doesn't matter as long as it makes calculation easier. It's just that R and r for radii of a bigger and a smaller circle (respectively) feels more intuitive to me. One could name them x and y, p and q, etc. and still get the same answer, because the math is still the same.
@@Gamer-df6if I'm fond of R and r for such problems too, although once you start explaining it out loud I sometimes wish I'd chosen two different letters, since saying "big r" and "little r" gets tedious! In any case, my point wasn't about the choice of variable name, but answering the question of why he used a variable at all, instead of just writing in 48 as per the comment I was answering.
Just a note about problem #2. In the solution we never used the value "3" as the length of a mountain slope. So, I can create another problem just like this one but replacing "3" with let's say "7". Apparently, I'll get the same answer for a different picture. I mean to say that at the end we need to have some sort of verification that the given configuration is at all possible and the problem is valid.
Substituting 3 by 7 is still possible. It will affect the heights of the first two mountain peaks, if we count from left to right: the peaks will become lower.
@@-igor- no, I explain in a comment for more details, this 5+3 must to add 10, you can demonstrate this changing 5 for ‘a’ and 3 for ‘b’, them move this segments to top of triangle.
@@AldoRPX you can also slide "a" down-right, to be collinear with "b", yes? They must not add to 10. Their sum should be less than 14.142 (square root of 200).
I got the solution for problem 2 but first let's name some vertices: The one between the 3 legth line and the 10 length line is A, the one between the 3 length and the 4 length is B, the one between the 4 and the 5 is C, between the 5 and the 6 is D, the one between the 6 and the other long side that intersects x is E, the one between it and x is F and the last one is G. Solution: Draw AH intersecting FG so we have a 45° right angled triangle with a leg of the length 10 so using Pythagorean theorem we can find that AH has a length of √200 which is 10√2 and GH is equal to AG so it is also 10 then draw CI intersecting FG then draw HJ intersecting CI so we have a new rectangle and a new little right angled triangle and we already know that the side lengths of the rectangle are 4 and √200 then again draw IK intersecting EF so we have another rectangle and right angled triangle which has angle F with measure of 45° so also angle I in that triangle will be 45° so then we will have the two angles DIK and KIF with measures of 90° and 45° respectively so we can find angle JIH using 180-(90+45) which is 45 so JIH is a 45° right angled triangle which we can use Pythagoras theorem to get it's hypotenuse which is gonna be √32, now returning back to triangle KFI that we know is a 45° right angled triangle with a leg length of 6 so using Pythagorean theorem the hypotenuse equal √72, now we have x divided into three parts with legths of 10, √32 and √72 so by adding them we get that x equals 10+10√2 as the simplest form. So what makes me not so confident about this answer is the the results is not a number that looks so satisfying.
For question 1. I used the properties of tangents and related to squares and triangles. We can construct a line from the centre of the circle to the point of tangency, it will be a perpendicular line. After doing so for bothe the sides we can assume the quadrilateral thus formed as a square because the length of tangents from the same external point is the same. Let the side of the square be "x" . Now the triangle is divided into 3 parts. The smaller triangle at the top. A square at the middle and a smaller triangle at the bottom. Area of the upper triangle = (4x-x^2)/2 Area of the bottom triangle = (6x-x^2)/2 Area of the square = x^2 Sum of there areas = sum of area of the whole figure. Sum of there area =5x 5x = 12 x = 12/5 x= 2.4 On observing the length x (side of the square) we get that it is the radius of the circle too. Thus radius is 2.4 units.
For the small triangle-the horizontal leg is x, the vertical leg is 4-x. Triangle is therefore (4-x)(x)/2. He then distributed “(4-x)(x)” to get (4x-x^2)/2. Small triangle is the same way, but with 6. (He typoed there and missed the “/2”, but still included it properly in subsequent calculations.)
Why would you not immediately observe the side of the square as being the same as the radius of the circle? I mean, you're setting out to solve r right? So if the observation step is all the way here, at the end, then how can you justify making the square in the first place? It's a minor nitpick, but what I'm trying to say is, why introduce a variable "x" that is easily demonstrably equal to a variable that's already been labeled, namely r?
2번 문제에서 길이 5와 3은 필요 없습니다. 6+4=10이므로 밑변이 10이고 높이가 x인 직각삼각형 2개가 빗변을 공유하게 그릴 수 있습니다. 즉 이 직각삼각형의 한 각도는 (45+90)/2=67.5입니다. 따라서 x=10tan(67.5)입니다. 계산기가 없다면 다음과 같이 구할 수 있습니다. 67.5도를 45도와 22.5로 나누는 직선을 그으면 이 직각삼각형은 한 변이 10인 직각이등변 삼각형과 양쪽 각도가 22.5도인 이등변삼각형으로 나눌 수 있습니다. 즉 10+10root2입니다.
Question 1 is maybe easier just noticing that, once you draw the two radii, three are two similar triangles, one with height h=4-r and one with base b=6-r. As they are similar h/r=r/b. These equations are easily solved for r=2.4. Question 2 is maybe easier by extending the other way, to make one big 45-45-90 triangle. The height equals the base, which is x. You can then extend to construct an internal triangle which is again 45-45-90 where the base is x-10 and the sides are 6+4. Using pythag on this internal triangle you can find x. The lengths 5 and 3 by the way do not need to be defined. Question 3 feels like there should be a simple geometric way to demonstrate that the big circles are four times the diameter of the small circles and/or that the green and blue lines as drawn form 3-4-5 right angled triangles if the diamond is completed. But I did the solution given in the video.
Yeah, the way he did it was weird. I just counted 20 distance of 45 zig zag (18 literally, and then 2 to reach the starting y value again) and then a right isolecles with side length 10.
What is the mistake I did in problem 1? Please help. I made a mirror of the given triangle in order to complete the rectangle. Now inside the rectangle there is a full circle. As it fits inside the circle isn't the dianeter of that circle = a? Which means d=4, r =2? What did I do wrong. Please help.
No, there's no full circle inside the completed rectangle. If you actually draw it on paper, you'll see that the semicircle in the mirrored half doesn't connect at all - not even close - and so no circle is formed. Good idea, but just didn't work in this puzzle.
@@qc1okay Sorry I get it now, it's because the semicircle isn't in the middle of the triangle. Fliiping makes it not connected so. I get it now. Thank you for helpiong.
First problem: It means that the middle of the semi-circle is "r" distance away from both sides of the triangle at the same time. This creates a square with side of length "r". We can imagine that perpendicular sides of the triangle are at the X and Y axis. Then general equation for the hypotenuse would be then 4x+6y-24 and we know that the middle of the semi-circle is on this hypotenuse so 4x+6y-24 = 0 leads to 2x+3y-12 = 0 and that to 2r+3r-12 = 0. Which gives us 5r=12 and r=12/5 Second problem: Solved exactly the same way as described in video.
@mindyourdecisions the question #2 is wrong planted, this figure can’t exist. Segments 5 and 3 must to add 10, or 45º is a impossible angle for these measures. You can project 5+3 at the top of figure and you will find that the hypotenuse is 10sqrt(2)+18 When according to your solution it must to be 10sqrt(2)+20.😅
For problem 1, consider the point at the right angle to be the origin. Then the hypotenuse is the line y=-(2/3)x+4. The center of the circle, which is on this line, has coordinates (r,r). So we can find r by solving the equation r=-(2/3)r+4.
Mountains: I did (6+5+4+3)cos45° + 10 + sqrt(2), because the height has increased by sqrt(2) by the leftmost peak, and that gives the same answer (24.14214). I think the demonstrated approach is easier though!
The 3rd problem becomes trivial when you realize that because the hypothenusa (a+b) is just as much longer as the long side (a) as the short side (a-b) is shorter - all you have to do is find a known triangle with those properties, like the famous triangle with sides 5, 4 and 3. The sides of the triangle in the diagram are larger but have the same ratio. As we know that a - b = 48, then a = 4/3 * 48 = 64 and 2a = 128.
That was a bit of a gamble on your end, no? How do you easily prove there's no other pythagorean triple that has these properties without being congruent to a 3-4-5 triange?
@@tomdekler9280 For any hypotenusa of a right triangle there is a pair of sides that form a triple a,b,c that is in the same ratio as the 3,4,5 triple. If there were another triple with different ratios for the same hypotenusa, the two straight sides would differ from a and b by the same factor delta d to form a triangle (a-d),(b+d),c Suppose a² + b² = c² and (a-d)² + (b+d)² = c². Then (a-d)² + (b+d)² = a² + b² => a² -2ad + d² + b² +2bd + d²=a² + b² => 2d² -2ad +2bd = 0 => 2d * (d - a + b) = 0 => d = 0 or d = a - b. And that's the same triangle or the mirrored triange; if we fill in a=3, b=4, c=5 then d = a-b = -1, with sides 4,3,5. This holds true for all triangles with these properties. We know for any hypothenusa of any length there is a triangle with the same ratio as the 3,4,5 triangle, and the only other right triangle with these properties must be its mirror image.
For problem 3, I would prefer to start off with the more general case. Focusing on the same right triangle as in the video, one can quickly establish that the small circles have one quarter the radius of the large ones, and therefore that 96 is 3/4 of the answer.
@@tomdekler9280 as Presh labelled the triangle in the diagram, it has sides a-b,a and a+b. So by Pythagoras (a)^2 + (a-b)^2 = (a+b)^2. When you expand and simplify you get a^2 = 4ab, so dividing by a gives a=4b. A handy thing to remember with Pythagoras theorem - to avoid confusion with a,b we've already used, we'll use X,Y,Z: X^2 + Y^2 = Z^2 is that although we can't factorise X^2 + Y^2 (without complex numbers), if we look at it as: X^2 = Z^2 - Y^2 we CAN factorise that RHS. That is, we can always use the form: X^2 = (Z + Y)(Z - Y) In a case like this that speeds things up a bit because it gives us: a^2 = [ (a+b) + (a-b) ][ (a+b) - (a-b) ] = [2a][2b] = 4ab
First one, I used similar triangles, sides (6-r)/(r) and (r)/(4-r) being proportional. Second one, I saved a step by "unfolding" the zigzags to a single 8/10/H triangle.
Another way to solve semicircle in triangle is to look at either of the 2 small triangles formed by radii as drawn. Both of these are similar to the full triangle & therefore have legs in ratio a/b. For the left-hand one, vertical leg is (a/b)r. This plus r forms the "a" side of the big triangle. So a = r [1 + (a/b)] , r = a/[1 + (a/b)]. Multiplying through by "b" top & bottom, gives r = ab/(a+b). For the 4 circles, solving the triangle gives a = 4b. As soon as you have that, you can express everything in terms of b. Vertical distance between centers is 6b, horizontal distance between big circles centers is 8b. So answer = (8/6) * 96 = 128.
Problem 1: The point [r, r] is on the line running between [0, 4] and [6, 0]. The slope of that line is -2/3 and the y intercept us 4, so the equation of the line is y = -(2/3)x + 4. Substitute r for both x and y (because the point [r, r] is on the line) and solve: r = -(2/3)r + 4, (5/3)r = 4, r = 12/5.
For the first question, the center point of the circle must lay on the top right corner of the largest square to fit in the triangle. So i just took the equation of the hypotenuse (y=[-a/b]x+a), setting y=x to find the side lengths of the square=r.
Cool that in the last problem when the distance between small circles is 1100000_2 the distance between the large circles is 10000000_2. In other words if the distance between the small circles is the sum of two sequential base 2 numbers, the distance between the large circles will be the next base 2 number.
Problem 1 seems easier in a coordinate system with the very basic equation for a line. With the origin at the bottom-left, the equation for the line the hypotenuse lies upon is going to be y=mx+b, with m being rise/run = -4/6 = -2/3, and b = 4 (the y-intercept). r is going to be the point on this line where x and y are equal. Then it’s just very simple math: y = -2/3x +4; r = -2/3r + 4; 5/3r = 4; r = (4*3)/5 = 2.4. QED.
Dunno about you but 24/10 is easier to convert to a decimal than 12/5, so 12/5 is hardly "simplified". If you were intending to leave as an improper fraction, then 12/5 is simplified. If intending to leave as a mixed number then i would tend to first do 24/10 = 2 4/10 and then simplify 4/10 to 2/5 giving a final answer of 2 2/5.
For problem 3, it's much nicer to solve the general equation a² + (a-b)² = (a+b)² first: a = 4b. Then a - b = 48, 4b - b = 3b = 48, b = 16, a = 64, 2a = 128.
With problem 1, I put the origin of coordinates at the right angle. The top corner is (0, 4), the right hand corner is (6, 0), and the semicircle centre is (r, r). These points must fall on a straight line. A condition for this is the determinant equation | 0 4 1 | | 6 0 1 | = 0 | r r 1 | 4(r - 6) + 6r = 0 10r = 24 r = 12/5
I'm not the best at math so I have a few questions on this solution. Why is it 0,4 and 6,0 how did you get those numbers? Why does the third column have 1s. I'm assuming it's a 3x3 grid because we have 3 variables and 3 sides?
@@user-ui8my9zs7oThe sides are perpendicular so the points rest on the two axis with points of the form (a,0) for the x axis and (0,b) for the y axis The matrix column is 1 just because that is the matrix that describes twice the area of a triangle, that is just how it is defined
@@user-ui8my9zs7oI'll help bro. The x-axis represents the horizontal unit of length. In this case it's 6 units. Thus the co-ordinates are (6,0). Similarly the y-axis represents the vertical unit of length in this case it's 4 units. Thus the co-ordinates are (0,4).
Problem 2 is very simple: Let's denote the vertices of the polygon A, B, C, D, E, F and G (counterclockwise, starting from the bottom left). You extend the line "3"=CD until it intersects the base x at point H. Then BH = 10 because the angle at point C is 45°. The extension of GF to the bottom right intersects the first extension at point J. This means GJ = 6+4=10. But this is the projection of the horizontal distance AH with an angle of 45°, therefore AH = 10·√2. Therefore, x = AH + BH = 10·√2 + 10 = 10·(√2 + 1) ≈ 24,142....
For problem 1 it is slightly easier to calculate the tangent of the upper angle of the triangle which is b/a and also r/(a-r) which yields the same result!
MindYourDecisions, in #3 after substituting for b on the left side of the equations it can NOT be a(squared) "+" 48(squared) unless you have already squared 48 as in the substitution (before condensing, as you forgot to show your work like you did the other side) would be: a(squared) + (a - a - 48)(squared), or a(squared) + (- 48)(squared), leaving it as a(squared) - 48(squared).
In the first problem, notice the triangle breaks into two smaller triangles and a square with area r^2. The areas of the two smaller triangles are (4 - r) * r / 2 and (6 -r ) * r /2. Summing the areas of the two triangles and square results in the simple equation 5r = 12, or r = 12 / 5
No need to calculate areas. Just notice that the smaller triangles are similar to the initial one. So, by identical ratios of sides : (a-r)/r = a/b = r/(b-r). Any of these equalities will give you the result.
on 2, I just formed the 45-degree right triangle with a side of 10, and then figured that the "altitude" perpendicular to the hypotenuse was 6+4=10, so a line segment perpendicular to the leftmost angled side that hits the point 10 units from the right would be 10, and then since that makes a 45-45-90 right triangle, the part of the slope is also 10 which makes the hypotenuse (bottom) is the square root of 10^2 plus 10^2, so square root of 200 = 10 sqrt 2, plus 10. I probably explained that terribly, but it makes it so you don't have to figure out the two quadrilaterals separately and just do one additional triangle along the base.
For problem 3 there a way to do it without the triangles, it might take more steps but it’s easier to understand: Both circles (big ones) have equidistant centers so that becomes 2r or D. We know that D is the distance from both parallel lines. So D=2r+96 (or 4r+d with d being the distance between circles). Now we need to solve for d: -> first solve for r to then substitute: -> 2r+d=96 [save this step] -> 2r=96-d -> r=48-d -> substitute r in D (x is just to fill): -> 4r+d=x -> 4(48-d)+d=x -> 192-4d+d=x -> 192-3d=x -> -3d=-192 -> d=192/3 -> d=64 We now need to solve r: ->substitute d in r: -> 2r+64=96 -> 2r=96-64 -> 2r=32 -> r=16. Finally we substitute r and d in D. -> 4r+d=D -> 4(16)+64=D -> 64+64=D For style purposes: D=128. And since D is the distance between lines and between centers, then the problem is solved. As I said it took more steps but for someone who don’t want to go with a pitagorean approach. This is the way… it’d be cool if someone could check my way to solve the problem.
you can also use similar triangles to solve problbleme if we set two points at tangency, we can get two similar triangles to the large one. the smaller triangle will have the length of r and 4-r, after that, we can use the ratio of the triangles to slove for r, hence 4/6 = (4-r)/r, and finally got the answer 2.4 similarly, we can also use the triangle with the sided length of r and 6-r, and substute the ratio and get 4/6 = r/(6-r), and finally got the answer 2.4
I solved problem 2 differently. I made the mountain range into a giant 45/45/90 triangle, from the point the 3 line meets the 10 line I created a smaller 45/45/90 triangle that's hypotenuse was the rest of the length of the base and one of it's legs was parallel to the 6 and 4, which mean it was a 10/10/10√2 triangle giving me 10+10√2 for the length of the base. It took like a minute and a half to do. Which I guess is the same thing but one big triangle instead of 2 small ones.
128. (Problem 3) Hi all, looking at all the circles, you don't see the triangles, which serve a pythagoras on a silver platter. That, together with the highly factoriable 96, predicts a quadratic solution. I found 2 equations: (R is the radius larger circle, r used for smaller one) R^2 +48^2 = (R+r)^2 ;and 2R = 96+2r ;thus eliminating R, yields: 3r^2 +96r -2304 = 0 ;use positive solution of 16 (radius small circle) add 96+2r = 128.
For the first problem i just drew the radius perpendicular to the base of the triangle and doing so created a smaller triangle with the same angles as the large one. Then i solved for x using the property of similar triangles
Acertijo 1.- La figura propuesta se puede descomponer en un cuadrado r*r, y dos triángulos rectángulos semejantes cuyas hipotenusas son colineales y sus catetos valen, [(4-r), r] y [r, (6-r)] → (4-r)/r =r/(6-r)→ r=2,4 Acertijo 2.- La figura propuesta se puede inscribir en un cuadrado X*X; la ladera montañosa izquierda prolongada se cortará con el segmento lateral derecho de longitud =10 en el vértice superior derecho del cuadrado → Longitud del lado derecho del cuadrado = X = 10 + (6+4)√2 = 10+10√2 =10(1+√2) =24.14 Acertijo 3.- Radio R=(96/2)+r =48+r → (2r+48)² - (48+r)² =48² → r=16 → R=48+16=64 → 2R=128 Curiosos acertijos. Gracias y un saludo cordial.
I solved the mountain problem in a weird way. so, if we know hypotenuse of an isosceles right triangle, side length will be hyp/root 2 so to calculate the height of the of the tallest peak, considered it as 10 - 3/r2 + 4/r2 - 5/r2 + 6/r2 making it 10 + 2/r2 or 10 + r2 next total width is 10 +2/r2 + 6/r2 + 5/r2 + 4/r2 + 3/r2 = 10 + 20/r2 = 10 + 10r2
At 7:40, why say the vertical leg of the triangle is “a-b”?? You’re given that the distance between the small circle centers is 96, and you know via symmetry that the line connecting the centers of the large circles bisects the 96. So we know this leg of the triangle is 48. Putting “a-b” instead unnecessarily complicates the subsequent algebra. When in doubt, use the K.I.S.S, method.
Hi Presh, I hope if I keep making this comment it will make a change. Every time you make a video with multiple problems, I end up having to watch the first problem, jump to the first solution, jump back to the second problem, jump to the second solution, etc. A video is in the wrong order if people have to jump around like that, so is there any way you can give us problem, solution, problem, solution?
I had issues.. I was trying to take the curvature of the earth into my calculations for the mountins one. . but since the diagram didn't have the units my numbers didn't work.
Another solution to problem 1, though not so simple: Name the vertices of the triangle A,B,C (counterclockwise), point A at the right angle at the bottom left. Hypothenuse a = √(6²+4²) = √52 = 2√13. Let the center of the semicircle on the hypothenuse a be M. Then due to proportionality CM/r = a/6, which gives CM = (r·√13)/3 Pythagoras applies with r² + (4-r)² = CM² in the small triangle at the top left. This leads to the quadratic equation 5r² - 72r + 144 = 0, and therefore r = 2.4 or r = 12. The latter is rejected because this is not possible according to the dimensions.
The second problem’s solution becomes more direct if you first extend the “6” line until it intersects the next line. The “6” length is now 10 units long. You can now follow the same process as in he video, but you now have a single triangle to determine the 10rt2, rather than needing to add two roots together.
First extending the 6 line till it reaches the base does not make it 10 in length as its peak is higher than the 10 AND because it becomes the hypotenuse of Line X and Line Y (the line between left most 45 angle and the peak of Line 6). However, by drawing a line between the peak of Line 10 straight through to Line Y (at 90 degrees of Line 10's peak) you can then calculate the hypotenuse of each of those triangles and get the overall value of X. The new line causes Line Y to then have 2 45 degree angles which then means the 2 sides created are equal length of 10.
@@dracconis69When I said to extend the “6” line until it intersects the next line, I was looking at a later iteration of Presh’s solution-my apologies. Allow me to rephrase: first extend the “3” line until it reaches the base. THEN extend the “6” line until it reaches what is now the “next” line, which is the extension of the “3” line.
@@tanaysingh7913 that has to do with similar triangles. in the video, look at the triangle that is made thanks to the upper side of the r x r square. it has the same proportions as the original triangle (you can google those theorems to understand why, not necessarily Thale's, that one is excessive here) and its legs are a - r and r, again according to the drawing. therefore the proportions a / b and (a - r) / r are equal.
From the thumbnail? No. It's not proven or given that the angles of the mountain range are 90°. It just looks like they are but we can't rely on a sketch. If the angle at the mountain tops is 89, the base is longer and of they are 91, the base is shorter. They could even be 180 (at least the ones in the middle)
Suggestion: How about this problem. Find the radius, ρ, of a circle inscribed between a circle of radius r, and a concentric ellipse (a = r, b < a), for any angle θ, given that they are both centred at origin.
What was the Reason for him to change the Equation in Problem 1 from r=a*b/a+b to 1/r= 1/a+1/b when hes not even using it for the solution? I just dont get it, also dont understand how he got there
The approach I used to do the first problem in my head just from looking at the thumbnail: Draw a line from the centre of the semi circle to the tangency point on the vertical side. The smaller triangle created in the upper left corner shares all three angles so must be similar to the large triangle. It has dimensions (4-r) height by r length. Equating the ratio of side lengths between the small and large triangles, (4-r)/r = 4/6 and solve for r = 2.4
The definition of boundary implies that two countries (or states, provinces, whatever) share it. Could you be a bit more specific? I just don't understand your problem.
@@Friendly-Neighborhood-Asexual see yourself Ukraine and Russia, Israel and Palestine, India and Pakistan, India and China, North and South Korea, South China Sea. Are the most sensitive borders in the world where order is maintained by exerting power from all over the world.
@@Pravin.ShidoreYou prove yourself wrong by the very fact that you had to qualify your comment by saying “most sensitive borders”. If we instead look at “least sensitive borders”, you’ll find ones like USA and Canada, or Minnesota and Wisconsin, or my property and my neighbor’s property. All of these are “shared”, none involve the “dominance” you incorrectly claim is necessary.
because all those ideas are interconnected in geometry. if you study all these properties of triangles rigorously, you'll understand the specific method is arbitrary here. if it's easier for you to think in similar triangles, go for it. some people prefer thinking about areas when it suits the problem at hand.
@@laincoubert7236 It's not a question of it being easier for that one commenter to "think". It's that 99.99% of math-oriented people would find it far easier to solve (4-R)/R = 4/6 than to solve it the far, far more complicated way this video did. I also apply my 99.99% figure to the likelihood that the maker of this video will NOT admit this, as supported by his track record of almost never admitting these sorts of oversights. His channel is probably the best on UA-cam, but it would be so much better if he had follow-up videos showing how "crowd-solving" often beats him.
@@qc1okay again, this is not a far easier way to think in, it's just a different one. why? because both of these solutions can be understood by a middle schooler in their first year of geometry. why? because both solutions share the same foundation of geometry (axioms, theorems, etc). like we're still in territory that is not complex. i mean, you can prove pythagoras theorem literally in a million different ways. if this channel were to unironically prove it with, i don't know, calculus, then your objection would be understandable. but on this level, i see it as a matter of preference. i can easily see a group of students that would go for the details and instantly look at similar triangles and get your equation, but i can also see a group that struggles with aligning those triangles in their head so they zoom out and think about areas in that sketch. personally, i'm with you in the first camp, but i also wish i was familiarized with other types of perception in school because you need that versatility to be good at solving various exam problems. generally speaking, i agree with you that this channel often provides us with weird solutions and that they're not accepting any feedback lol. to me this is not one of those cases though maybe i didn't really articulate my point that well, but what i'm trying to say is similar to a 3blue1brown video about alice and bob approaching the same problem in different ways. highly recommend it btw, regardless of our discussion here.
Thanks!
There is a slightly more elementary way to solve the first problem. The sides of similar triangles have the same ratio, so you just have (4-r)/r = r/(6-r) and just apply basic algebra to solve for r.
That was my strategy! I agree that area was a little faster, but I would consider area to be the more elementary approach than similar triangles.
You can do the equality with the bar triangle, even faster! (a-r)/r = a/b
(4-R)/R = 4/6 is an even simpler way, comparing the upper-left corner's right triangle with the large triangle.
i realized this after using arctan theta to get 4/6 and then needing to use the tangent again to solve for the radius. silly me!
My way, as Frank Sinatra said.
Puzzle 1 is FAR simpler than this video showed! The 4x6 right triangle has the same proportions as the right triangles formed in its corners. The semicircle's radius R forms right angles with the two sides (4 and 6) at the points of tangency and thus forms a square R x R at the lower left. So the upper-left right triangle has a left side of 4-R and a base of R, and those two sides are in the same proportions as 4 and 6 are.
(4-R)/R = 4/6 = 2/3
4-R = (2/3)R
4 = (5/3)R
12/5 = R = 2.4
Ahhhhhhhhh…..
What does 4-R mean
@@itsame5272 four minus r
Other commenters now have suggested R/(6-R) = 4/6 as well as (4-R)/R = R/(6-R), but I think (4-R)/R = 4/6 is the simplest.
There's another solution that's even simpler (in my opinion). If you consider the point at the right angle to be the origin, then the hypotenuse is the line y=-(2/3)x+4, and the center of the circle, which is on this line, has coordinates (r,r). Substituting, we get r=-(2/3)r+4, and solving the equation we arrive at r=12/5.
For question one
Take the right angled vertex as origin. Then the center if circle would be (r,r) and it lies on the line joining (0,4) and (6,0) which is 2x+3y=12
Substituting the point (r,r) gives us r=12/5
Wow,
for the first puzzle i used the equation y=-4/6x+4, because the semi circle is tangent to the legs of the right triangle y=x and solve for x =-4/6x+4, x=12/5
I went different ways as in the video, but luckily found the correct solutions though.
Nice questions, fun to figure out. 😀
Problem 1:
Equation of hypotenuse: y/4 + x/6 = 1
Position of circle center: (x,y) = (r,r)
Circle center is on hypotenuse: r/4 + r/6 = 1 ==> r = 1/(1/4 + 1/6) = 24/(6+4) = 24/10 = 2.4
Problem 2:
Let O = leftmost point of the base, P = rightmost point of base, A, B, C are the peaks from left to right (so C is right 10 above P), while D and E are the local valleys from left to right. ( |AD| = 6 , |DB| = 5 , |BE| = 4 , |EC| = 3 ).
Extend the 45-degree slope OA , as well as the vertical line PC ; let's name the intersection S. From C, draw a line perpendicular to OS ; let's call the intersection T. |CT| = |AD| + |BE| = 6+4 = 10 , and since triangle CST is an isosceles right triangle (= 90-45-45 triangle) we know |TS| = 10 and |CS| = 10*sqrt(2) . Since OPS is also an isosceles right triangle, |OP| = |PS| = |PC| + |CS| = 10 + 10*sqrt(2) = 24.142 (approximately)
Problem 3:
Let R = radius of large circle, and r = radius of small circle.
Distance between two lines = 2R = 96 + 2r ==> r = R - 48 .
Half of blue line segment and half of green line segment form a right triangle with hypotenuse (r+R) ==>
48^2 + R^2 = (R+r)^2
Substitute r = R - 48 yields
48^2 + R^2 = (R + R - 48)^2
48^2 + R^2 = (2R - 48)^2
48^2 + R^2 = 4*R^2 - 192R + 48^2
0 = 3*R^2 - 192R
R^2 = 64R
R = 0 OR R = 64
==> radius of large circle is R = 64
Distance between centers of large circles is 2R = 2*64 = 128 .
Thank you, sir.
Very informative explanation.
Oh great work❤🇧🇩🇧🇩❤
Great collection of puzzles.
Question 2 is even easier if you rearrange the segments so that all those sloping up are together, followed by the two that slope down before solving in a manner similar to the video.
I did trig hehe
I noticed the closing background music started to play after Presh said "That's the answer" for problem 1, but was abruptly stopped before it could continue playing more 😂😂😂😂😂
still can't stop laughing
lol, seems like it happened after the 2nd problem's solution, too.
For question 2 I found a different solution to get the same answer! I drew a straight line 90 degrees with the right face of 10, parallel with the base, and broke the top up into several triangles. First, the hypotenuse of the triangle with a leg of 3 is 3rt2. Then, the other leg is 3, leaving a tiny triangle with legs of 1 at the top of the side labeled 4. The hypotenuse of this will be 1rt2. Then, the side of length 5 will have 4 left below the line, and so the hypotenuse is 4rt2. Then, the side of 6 will have 2 above the line, leaving 2rt2. Lastly, I drew a line straight down from there, perpendicular with the base. Because the far right length is 10, this length is also 10 and this creates a right triangle with 10 as the legs. Therefore you get 10 + 10rt2. Super cool!
Exactly the solution I saw. Why complicate it when you can draw a rectangle and get Side X of the rectangle using the Triangles at the top at the same time creating an Equilateral Triangle at the far left.
The second problem can also be calculated by taking the "4" line and moving it to the corner where the "3" line and the "10" line meet. Move the "6" line to extend the "4 "line. This then creates a single triange that has a base of "10" and a height of "10" so the vertical height is 10*sqrt(2) ( + the provided 10) = 24.14. As the original triange is a "45, 45, right" triangle, the base = height. Therefore the distance (base) = 24.14.
I think you said this better than I did.
For problem 3, once you create the right angle triangle, we know one side length must be 48 since it cuts the 96 measurement into two equal parts. Would that simplify the equations?
Yes, it does. My solution is a bit different (it starts with realizing that 96 + 2r = 2R because they are perfectly sandwiched between two parallel lines) but it also uses the hypotenuse of that triangle.
(with r = small circle radius, R = big circle radius)
I presume Presh was mindful of that, but chose to use the variable 'a' to set up a more general view of things.
@@BenDRobinson I think so too.
Variable names doesn't matter as long as it makes calculation easier.
It's just that R and r for radii of a bigger and a smaller circle (respectively) feels more intuitive to me.
One could name them x and y, p and q, etc. and still get the same answer, because the math is still the same.
@@Gamer-df6if I'm fond of R and r for such problems too, although once you start explaining it out loud I sometimes wish I'd chosen two different letters, since saying "big r" and "little r" gets tedious! In any case, my point wasn't about the choice of variable name, but answering the question of why he used a variable at all, instead of just writing in 48 as per the comment I was answering.
Just a note about problem #2. In the solution we never used the value "3" as the length of a mountain slope. So, I can create another problem just like this one but replacing "3" with let's say "7".
Apparently, I'll get the same answer for a different picture. I mean to say that at the end we need to have some sort of verification that the given configuration is at all possible and the problem is valid.
The problem is wrong planted, 45º is impossible angle in this figure for this lengths, or 5 and 3 is not possible for that angle.
Substituting 3 by 7 is still possible. It will affect the heights of the first two mountain peaks, if we count from left to right: the peaks will become lower.
@@AldoRPX 5 + 3 = 8, and it is less than the hypotenuse of the yellow triangle which is √200, approx = 14.142. So, it is possible.
@@-igor- no, I explain in a comment for more details, this 5+3 must to add 10, you can demonstrate this changing 5 for ‘a’ and 3 for ‘b’, them move this segments to top of triangle.
@@AldoRPX you can also slide "a" down-right, to be collinear with "b", yes? They must not add to 10. Their sum should be less than 14.142 (square root of 200).
Thank you for your shareing Master
I got the solution for problem 2 but first let's name some vertices: The one between the 3 legth line and the 10 length line is A, the one between the 3 length and the 4 length is B, the one between the 4 and the 5 is C, between the 5 and the 6 is D, the one between the 6 and the other long side that intersects x is E, the one between it and x is F and the last one is G.
Solution: Draw AH intersecting FG so we have a 45° right angled triangle with a leg of the length 10 so using Pythagorean theorem we can find that AH has a length of √200 which is 10√2 and GH is equal to AG so it is also 10 then draw CI intersecting FG then draw HJ intersecting CI so we have a new rectangle and a new little right angled triangle and we already know that the side lengths of the rectangle are 4 and √200 then again draw IK intersecting EF so we have another rectangle and right angled triangle which has angle F with measure of 45° so also angle I in that triangle will be 45° so then we will have the two angles DIK and KIF with measures of 90° and 45° respectively so we can find angle JIH using 180-(90+45) which is 45 so JIH is a 45° right angled triangle which we can use Pythagoras theorem to get it's hypotenuse which is gonna be √32, now returning back to triangle KFI that we know is a 45° right angled triangle with a leg length of 6 so using Pythagorean theorem the hypotenuse equal √72, now we have x divided into three parts with legths of 10, √32 and √72 so by adding them we get that x equals 10+10√2 as the simplest form.
So what makes me not so confident about this answer is the the results is not a number that looks so satisfying.
For question 1.
I used the properties of tangents and related to squares and triangles.
We can construct a line from the centre of the circle to the point of tangency, it will be a perpendicular line. After doing so for bothe the sides we can assume the quadrilateral thus formed as a square because the length of tangents from the same external point is the same.
Let the side of the square be "x" .
Now the triangle is divided into 3 parts.
The smaller triangle at the top.
A square at the middle and a smaller triangle at the bottom.
Area of the upper triangle = (4x-x^2)/2
Area of the bottom triangle = (6x-x^2)/2
Area of the square = x^2
Sum of there areas = sum of area of the whole figure.
Sum of there area =5x
5x = 12
x = 12/5
x= 2.4
On observing the length x (side of the square) we get that it is the radius of the circle too.
Thus radius is 2.4 units.
I dont get the formula for the area of the triangle
For the small triangle-the horizontal leg is x, the vertical leg is 4-x. Triangle is therefore (4-x)(x)/2. He then distributed “(4-x)(x)” to get (4x-x^2)/2. Small triangle is the same way, but with 6. (He typoed there and missed the “/2”, but still included it properly in subsequent calculations.)
@@verkuilb i fixed it ! Sorry for the mistake
Why would you not immediately observe the side of the square as being the same as the radius of the circle?
I mean, you're setting out to solve r right? So if the observation step is all the way here, at the end, then how can you justify making the square in the first place?
It's a minor nitpick, but what I'm trying to say is, why introduce a variable "x" that is easily demonstrably equal to a variable that's already been labeled, namely r?
Hey I got it
I was denoting 4-x as 4 dash x rather then 4 minus x
2번 문제에서 길이 5와 3은 필요 없습니다. 6+4=10이므로 밑변이 10이고 높이가 x인 직각삼각형 2개가 빗변을 공유하게 그릴 수 있습니다. 즉 이 직각삼각형의 한 각도는 (45+90)/2=67.5입니다. 따라서 x=10tan(67.5)입니다.
계산기가 없다면 다음과 같이 구할 수 있습니다. 67.5도를 45도와 22.5로 나누는 직선을 그으면 이 직각삼각형은 한 변이 10인 직각이등변 삼각형과 양쪽 각도가 22.5도인 이등변삼각형으로 나눌 수 있습니다. 즉 10+10root2입니다.
Question 1 is maybe easier just noticing that, once you draw the two radii, three are two similar triangles, one with height h=4-r and one with base b=6-r. As they are similar h/r=r/b. These equations are easily solved for r=2.4.
Question 2 is maybe easier by extending the other way, to make one big 45-45-90 triangle. The height equals the base, which is x. You can then extend to construct an internal triangle which is again 45-45-90 where the base is x-10 and the sides are 6+4. Using pythag on this internal triangle you can find x. The lengths 5 and 3 by the way do not need to be defined.
Question 3 feels like there should be a simple geometric way to demonstrate that the big circles are four times the diameter of the small circles and/or that the green and blue lines as drawn form 3-4-5 right angled triangles if the diamond is completed. But I did the solution given in the video.
Yeah, the way he did it was weird. I just counted 20 distance of 45 zig zag (18 literally, and then 2 to reach the starting y value again) and then a right isolecles with side length 10.
What is the mistake I did in problem 1? Please help. I made a mirror of the given triangle in order to complete the rectangle. Now inside the rectangle there is a full circle. As it fits inside the circle isn't the dianeter of that circle = a? Which means d=4,
r =2?
What did I do wrong. Please help.
No, there's no full circle inside the completed rectangle. If you actually draw it on paper, you'll see that the semicircle in the mirrored half doesn't connect at all - not even close - and so no circle is formed. Good idea, but just didn't work in this puzzle.
@@qc1okay Sorry I get it now, it's because the semicircle isn't in the middle of the triangle. Fliiping makes it not connected so. I get it now. Thank you for helpiong.
Good job sir!
Recycling old videos?
I hear the exit music between problems 1&2 and 2&3.
Yeah i watched the video 4 hours ago of 2nd problem he posted years ago
Ouch, I don't think he wanted anybody to notice that. :-P
Thanks.
love the mission impossible reference
First problem:
It means that the middle of the semi-circle is "r" distance away from both sides of the triangle at the same time. This creates a square with side of length "r".
We can imagine that perpendicular sides of the triangle are at the X and Y axis. Then general equation for the hypotenuse would be then 4x+6y-24 and we know that the middle of the semi-circle is on this hypotenuse so
4x+6y-24 = 0 leads to 2x+3y-12 = 0 and that to 2r+3r-12 = 0. Which gives us 5r=12 and r=12/5
Second problem:
Solved exactly the same way as described in video.
What is the snow quality at the peaks? Good for skiing?
@mindyourdecisions the question #2 is wrong planted, this figure can’t exist. Segments 5 and 3 must to add 10, or 45º is a impossible angle for these measures. You can project 5+3 at the top of figure and you will find that the hypotenuse is 10sqrt(2)+18 When according to your solution it must to be 10sqrt(2)+20.😅
For problem 1, consider the point at the right angle to be the origin. Then the hypotenuse is the line y=-(2/3)x+4. The center of the circle, which is on this line, has coordinates (r,r). So we can find r by solving the equation r=-(2/3)r+4.
Mountains: I did (6+5+4+3)cos45° + 10 + sqrt(2), because the height has increased by sqrt(2) by the leftmost peak, and that gives the same answer (24.14214). I think the demonstrated approach is easier though!
The 3rd problem becomes trivial when you realize that because the hypothenusa (a+b) is just as much longer as the long side (a) as the short side (a-b) is shorter - all you have to do is find a known triangle with those properties, like the famous triangle with sides 5, 4 and 3. The sides of the triangle in the diagram are larger but have the same ratio.
As we know that a - b = 48, then a = 4/3 * 48 = 64 and 2a = 128.
That was a bit of a gamble on your end, no? How do you easily prove there's no other pythagorean triple that has these properties without being congruent to a 3-4-5 triange?
@@tomdekler9280 For any hypotenusa of a right triangle there is a pair of sides that form a triple a,b,c that is in the same ratio as the 3,4,5 triple. If there were another triple with different ratios for the same hypotenusa, the two straight sides would differ from a and b by the same factor delta d to form a triangle (a-d),(b+d),c
Suppose a² + b² = c² and (a-d)² + (b+d)² = c².
Then (a-d)² + (b+d)² = a² + b²
=> a² -2ad + d² + b² +2bd + d²=a² + b²
=> 2d² -2ad +2bd = 0
=> 2d * (d - a + b) = 0
=> d = 0 or d = a - b.
And that's the same triangle or the mirrored triange; if we fill in a=3, b=4, c=5 then d = a-b = -1, with sides 4,3,5.
This holds true for all triangles with these properties.
We know for any hypothenusa of any length there is a triangle with the same ratio as the 3,4,5 triangle, and the only other right triangle with these properties must be its mirror image.
@@barttemolder3405 That's a fun proof, I like it! Thanks!
@@tomdekler9280 Thanks for making me do it, I enjoyed it as well. But I still needed algebra to prove my trivial solution :)
Yes, I find it interesting how often these problems end up involving the 3-4-5 triangle. This one does so in a less contrived way than many.
For problem 3, I would prefer to start off with the more general case. Focusing on the same right triangle as in the video, one can quickly establish that the small circles have one quarter the radius of the large ones, and therefore that 96 is 3/4 of the answer.
how do you quickly establish that?
@@tomdekler9280 as Presh labelled the triangle in the diagram, it has sides a-b,a and a+b. So by Pythagoras (a)^2 + (a-b)^2 = (a+b)^2. When you expand and simplify you get a^2 = 4ab, so dividing by a gives a=4b.
A handy thing to remember with Pythagoras theorem - to avoid confusion with a,b we've already used, we'll use X,Y,Z:
X^2 + Y^2 = Z^2
is that although we can't factorise X^2 + Y^2 (without complex numbers), if we look at it as:
X^2 = Z^2 - Y^2
we CAN factorise that RHS. That is, we can always use the form:
X^2 = (Z + Y)(Z - Y)
In a case like this that speeds things up a bit because it gives us:
a^2 = [ (a+b) + (a-b) ][ (a+b) - (a-b) ] = [2a][2b] = 4ab
First one, I used similar triangles, sides (6-r)/(r) and (r)/(4-r) being proportional.
Second one, I saved a step by "unfolding" the zigzags to a single 8/10/H triangle.
Another way to solve semicircle in triangle is to look at either of the 2 small triangles formed by radii as drawn. Both of these are similar to the full triangle & therefore have legs in ratio a/b. For the left-hand one, vertical leg is (a/b)r. This plus r forms the "a" side of the big triangle.
So a = r [1 + (a/b)] , r = a/[1 + (a/b)]. Multiplying through by "b" top & bottom, gives r = ab/(a+b).
For the 4 circles, solving the triangle gives a = 4b. As soon as you have that, you can express everything in terms of b. Vertical distance between centers is 6b, horizontal distance between big circles centers is 8b. So answer = (8/6) * 96 = 128.
Problem 1:
The point [r, r] is on the line running between [0, 4] and [6, 0]. The slope of that line is -2/3 and the y intercept us 4, so the equation of the line is y = -(2/3)x + 4. Substitute r for both x and y (because the point [r, r] is on the line) and solve: r = -(2/3)r + 4, (5/3)r = 4, r = 12/5.
For the first question, the center point of the circle must lay on the top right corner of the largest square to fit in the triangle. So i just took the equation of the hypotenuse (y=[-a/b]x+a), setting y=x to find the side lengths of the square=r.
Cool that in the last problem when the distance between small circles is 1100000_2 the distance between the large circles is 10000000_2.
In other words if the distance between the small circles is the sum of two sequential base 2 numbers, the distance between the large circles will be the next base 2 number.
Problem 1 seems easier in a coordinate system with the very basic equation for a line. With the origin at the bottom-left, the equation for the line the hypotenuse lies upon is going to be y=mx+b, with m being rise/run = -4/6 = -2/3, and b = 4 (the y-intercept). r is going to be the point on this line where x and y are equal. Then it’s just very simple math: y = -2/3x +4; r = -2/3r + 4; 5/3r = 4; r = (4*3)/5 = 2.4. QED.
Dunno about you but 24/10 is easier to convert to a decimal than 12/5, so 12/5 is hardly "simplified".
If you were intending to leave as an improper fraction, then 12/5 is simplified.
If intending to leave as a mixed number then i would tend to first do 24/10 = 2 4/10 and then simplify 4/10 to 2/5 giving a final answer of 2 2/5.
For the second question, I rearrange all slopes to form a symmetrical kite, so I can calculate x by 10/tan(45/2) = 24.1421
The 3rd one was insane.
For problem 3, it's much nicer to solve the general equation a² + (a-b)² = (a+b)² first: a = 4b. Then a - b = 48, 4b - b = 3b = 48, b = 16, a = 64, 2a = 128.
With problem 1, I put the origin of coordinates at the right angle. The top corner is (0, 4), the right hand corner is (6, 0), and the semicircle centre is (r, r). These points must fall on a straight line. A condition for this is the determinant equation
| 0 4 1 |
| 6 0 1 | = 0
| r r 1 |
4(r - 6) + 6r = 0
10r = 24
r = 12/5
I'm not the best at math so I have a few questions on this solution. Why is it 0,4 and 6,0 how did you get those numbers? Why does the third column have 1s. I'm assuming it's a 3x3 grid because we have 3 variables and 3 sides?
Oh I think I see why you have 0,4 and 0,6 I'm guessing that's the coordinates. But why the 1?
@@user-ui8my9zs7oThe sides are perpendicular so the points rest on the two axis with points of the form (a,0) for the x axis and (0,b) for the y axis
The matrix column is 1 just because that is the matrix that describes twice the area of a triangle, that is just how it is defined
@@user-ui8my9zs7oI'll help bro.
The x-axis represents the horizontal unit of length. In this case it's 6 units. Thus the co-ordinates are (6,0).
Similarly the y-axis represents the vertical unit of length in this case it's 4 units. Thus the co-ordinates are (0,4).
But why 1? Does that imply that there's a 3d length? Or I guess I size on th z axis?
Problem 2 is very simple: Let's denote the vertices of the polygon A, B, C, D, E, F and G (counterclockwise, starting from the bottom left). You extend the line "3"=CD until it intersects the base x at point H. Then BH = 10 because the angle at point C is 45°. The extension of GF to the bottom right intersects the first extension at point J. This means GJ = 6+4=10. But this is the projection of the horizontal distance AH with an angle of 45°, therefore AH = 10·√2.
Therefore, x = AH + BH = 10·√2 + 10 = 10·(√2 + 1) ≈ 24,142....
For problem 1 it is slightly easier to calculate the tangent of the upper angle of the triangle which is b/a and also r/(a-r) which yields the same result!
Bro just make me a maths fan.......make more video like this bro....please
MindYourDecisions, in #3 after substituting for b on the left side of the equations it can NOT be a(squared) "+" 48(squared) unless you have already squared 48 as in the substitution (before condensing, as you forgot to show your work like you did the other side) would be: a(squared) + (a - a - 48)(squared), or a(squared) + (- 48)(squared), leaving it as a(squared) - 48(squared).
Problem 3 was on my school exams 25 years ago.
In the first problem, notice the triangle breaks into two smaller triangles and a square with area r^2. The areas of the two smaller triangles are (4 - r) * r / 2 and (6 -r ) * r /2. Summing the areas of the two triangles and square results in the simple equation 5r = 12, or r = 12 / 5
No need to calculate areas.
Just notice that the smaller triangles are similar to the initial one. So, by identical ratios of sides : (a-r)/r = a/b = r/(b-r).
Any of these equalities will give you the result.
on 2, I just formed the 45-degree right triangle with a side of 10, and then figured that the "altitude" perpendicular to the hypotenuse was 6+4=10, so a line segment perpendicular to the leftmost angled side that hits the point 10 units from the right would be 10, and then since that makes a 45-45-90 right triangle, the part of the slope is also 10 which makes the hypotenuse (bottom) is the square root of 10^2 plus 10^2, so square root of 200 = 10 sqrt 2, plus 10.
I probably explained that terribly, but it makes it so you don't have to figure out the two quadrilaterals separately and just do one additional triangle along the base.
For problem 3 there a way to do it without the triangles, it might take more steps but it’s easier to understand:
Both circles (big ones) have equidistant centers so that becomes 2r or D.
We know that D is the distance from both parallel lines.
So D=2r+96 (or 4r+d with d being the distance between circles).
Now we need to solve for d:
-> first solve for r to then substitute:
-> 2r+d=96 [save this step]
-> 2r=96-d
-> r=48-d
-> substitute r in D (x is just to fill):
-> 4r+d=x
-> 4(48-d)+d=x
-> 192-4d+d=x
-> 192-3d=x
-> -3d=-192
-> d=192/3
-> d=64
We now need to solve r:
->substitute d in r:
-> 2r+64=96
-> 2r=96-64
-> 2r=32
-> r=16.
Finally we substitute r and d in D.
-> 4r+d=D
-> 4(16)+64=D
-> 64+64=D
For style purposes: D=128. And since D is the distance between lines and between centers, then the problem is solved. As I said it took more steps but for someone who don’t want to go with a pitagorean approach. This is the way… it’d be cool if someone could check my way to solve the problem.
you can also use similar triangles to solve problbleme
if we set two points at tangency, we can get two similar triangles to the large one. the smaller triangle will have the length of r and 4-r, after that, we can use the ratio of the triangles to slove for r, hence 4/6 = (4-r)/r, and finally got the answer 2.4
similarly, we can also use the triangle with the sided length of r and 6-r, and substute the ratio and get 4/6 = r/(6-r), and finally got the answer 2.4
That's what I did. Really immediate.
Who has solved all this problems on self😊😊
things are getting easier lately, took me 3 to 4 minutes in total.
2/3
Got the radius 5 in the 1st question
I solved problem 2 differently. I made the mountain range into a giant 45/45/90 triangle, from the point the 3 line meets the 10 line I created a smaller 45/45/90 triangle that's hypotenuse was the rest of the length of the base and one of it's legs was parallel to the 6 and 4, which mean it was a 10/10/10√2 triangle giving me 10+10√2 for the length of the base. It took like a minute and a half to do.
Which I guess is the same thing but one big triangle instead of 2 small ones.
128. (Problem 3) Hi all, looking at all the circles, you don't see the triangles, which serve a pythagoras on a silver platter. That, together with the highly factoriable 96, predicts a quadratic solution.
I found 2 equations: (R is the radius larger circle, r used for smaller one)
R^2 +48^2 = (R+r)^2 ;and
2R = 96+2r ;thus eliminating R, yields:
3r^2 +96r -2304 = 0 ;use positive solution of 16 (radius small circle)
add 96+2r = 128.
For the first problem i just drew the radius perpendicular to the base of the triangle and doing so created a smaller triangle with the same angles as the large one. Then i solved for x using the property of similar triangles
Acertijo 1.- La figura propuesta se puede descomponer en un cuadrado r*r, y dos triángulos rectángulos semejantes cuyas hipotenusas son colineales y sus catetos valen, [(4-r), r] y [r, (6-r)] → (4-r)/r =r/(6-r)→ r=2,4
Acertijo 2.- La figura propuesta se puede inscribir en un cuadrado X*X; la ladera montañosa izquierda prolongada se cortará con el segmento lateral derecho de longitud =10 en el vértice superior derecho del cuadrado → Longitud del lado derecho del cuadrado = X = 10 + (6+4)√2 = 10+10√2 =10(1+√2) =24.14
Acertijo 3.- Radio R=(96/2)+r =48+r → (2r+48)² - (48+r)² =48² → r=16 → R=48+16=64 → 2R=128
Curiosos acertijos. Gracias y un saludo cordial.
I solved the mountain problem in a weird way.
so, if we know hypotenuse of an isosceles right triangle, side length will be hyp/root 2
so to calculate the height of the of the tallest peak, considered it as 10 - 3/r2 + 4/r2 - 5/r2 + 6/r2 making it 10 + 2/r2 or 10 + r2
next total width is 10 +2/r2 + 6/r2 + 5/r2 + 4/r2 + 3/r2 = 10 + 20/r2 = 10 + 10r2
I did a 20 step solution drawing countless figures and using trig lol.
Second one was tricky
At 7:40, why say the vertical leg of the triangle is “a-b”?? You’re given that the distance between the small circle centers is 96, and you know via symmetry that the line connecting the centers of the large circles bisects the 96. So we know this leg of the triangle is 48. Putting “a-b” instead unnecessarily complicates the subsequent algebra. When in doubt, use the K.I.S.S, method.
Hi Presh, I hope if I keep making this comment it will make a change. Every time you make a video with multiple problems, I end up having to watch the first problem, jump to the first solution, jump back to the second problem, jump to the second solution, etc. A video is in the wrong order if people have to jump around like that, so is there any way you can give us problem, solution, problem, solution?
I had issues.. I was trying to take the curvature of the earth into my calculations for the mountins one. . but since the diagram didn't have the units my numbers didn't work.
Another solution to problem 1, though not so simple:
Name the vertices of the triangle A,B,C (counterclockwise), point A at the right angle at the bottom left. Hypothenuse a = √(6²+4²) = √52 = 2√13. Let the center of the semicircle on the hypothenuse a be M. Then due to proportionality CM/r = a/6, which gives CM = (r·√13)/3
Pythagoras applies with r² + (4-r)² = CM² in the small triangle at the top left. This leads to the quadratic equation 5r² - 72r + 144 = 0, and therefore r = 2.4 or r = 12. The latter is rejected because this is not possible according to the dimensions.
i did every answer correctly - but with different ways of solving
for the first question i just used the law of similar triangle and deduced that 6-r/r = r/4-r and got r=12/5
The second problem’s solution becomes more direct if you first extend the “6” line until it intersects the next line. The “6” length is now 10 units long. You can now follow the same process as in he video, but you now have a single triangle to determine the 10rt2, rather than needing to add two roots together.
First extending the 6 line till it reaches the base does not make it 10 in length as its peak is higher than the 10 AND because it becomes the hypotenuse of Line X and Line Y (the line between left most 45 angle and the peak of Line 6). However, by drawing a line between the peak of Line 10 straight through to Line Y (at 90 degrees of Line 10's peak) you can then calculate the hypotenuse of each of those triangles and get the overall value of X. The new line causes Line Y to then have 2 45 degree angles which then means the 2 sides created are equal length of 10.
@@dracconis69When I said to extend the “6” line until it intersects the next line, I was looking at a later iteration of Presh’s solution-my apologies. Allow me to rephrase: first extend the “3” line until it reaches the base. THEN extend the “6” line until it reaches what is now the “next” line, which is the extension of the “3” line.
I used intercept (Thale's) theorem for problem 1. Since a/b = (a-r)/r, this was peace of cake to go further from there.
Sorry but can u tell me how a/b=a-r/r
@@tanaysingh7913 that has to do with similar triangles. in the video, look at the triangle that is made thanks to the upper side of the r x r square. it has the same proportions as the original triangle (you can google those theorems to understand why, not necessarily Thale's, that one is excessive here) and its legs are a - r and r, again according to the drawing. therefore the proportions a / b and (a - r) / r are equal.
@@laincoubert7236 thanks 🙂
This was in cat 2023 slot 3 just instead of triangle rectangle was given .
As easy as the solution to question 2 was, I actually found question 3 to be a lot easier for me than question 2 lol.
From the thumbnail? No. It's not proven or given that the angles of the mountain range are 90°. It just looks like they are but we can't rely on a sketch. If the angle at the mountain tops is 89, the base is longer and of they are 91, the base is shorter. They could even be 180 (at least the ones in the middle)
another solution for the problem1 using pythagoras theorem and algebric equation from 🇧🇩Bangladesh
length of the arms
4,6 and √(4²+6²)=2√13
{ dividing the third arm proportionaly to the horizontal and vertical line
2√13 * 4/(4+6)
= 2√13 * 4/10
= 2√13 * 2/5
= 4√13/5
2√13 * 6/(4+6)
= 2√13 * 6/10
= 2√13 * 3/5
= 6√13/5
2√13 = (4√13/5 + 6√13/5)
}
{
here ∆abc is similar to ∆xrz so
x:r = a:b
x:r = 4:6
x/r = 4/6
6x = 4r
x =4r/6
x =2r/3......(i)
}
{
now in ∆xrz
x² + r² = z²
x² + r² = (4√13/5)²
x² + r² = 4²*13/5²
x² + r² = 16*13/25
x² + r² = 208/25
(2r/3)² + r² = 208/25
2²r²/3² + r² = 208/25
4r²/9 + r² = 208/25
4y²/9 + 9r²/9= 208/25
(4r²+9r²)/9= 208/25
13r²/9= 208/25
13r²= (208*9)/25
13r²= 1872/25
r² = (1872/13)/25
r² = 144/25
r² = 12²/5²
r = 12/5 = 2.4
}
sorry for the poor notation
Suggestion:
How about this problem.
Find the radius, ρ, of a circle inscribed between a circle of radius r, and a concentric ellipse (a = r, b < a), for any angle θ, given that they are both centred at origin.
What was the Reason for him to change the Equation in Problem 1 from r=a*b/a+b to 1/r= 1/a+1/b when hes not even using it for the solution? I just dont get it, also dont understand how he got there
problem 1 is even easier, you know that (b-r)/r=b/a and can then solve from there
How can u pls tell me
Problem 1: r/(6-r) = (4-r)/r
EXACTLY how I solved it!
MUCH easier to do R/(6-R) = 4/6.
So annoying those endcard shows up before the video has finished.
Is it me or those stuff become easier
Mind Your Decisions
The approach I used to do the first problem in my head just from looking at the thumbnail:
Draw a line from the centre of the semi circle to the tangency point on the vertical side. The smaller triangle created in the upper left corner shares all three angles so must be similar to the large triangle. It has dimensions (4-r) height by r length. Equating the ratio of side lengths between the small and large triangles, (4-r)/r = 4/6 and solve for r = 2.4
I suggest a problem
May be I'm wrong with respective math but I'm politically correct. Only solution is dominance no two can share a boundary.
The definition of boundary implies that two countries (or states, provinces, whatever) share it. Could you be a bit more specific? I just don't understand your problem.
😂
What does this mean
@@Friendly-Neighborhood-Asexual see yourself Ukraine and Russia, Israel and Palestine, India and Pakistan, India and China, North and South Korea, South China Sea. Are the most sensitive borders in the world where order is maintained by exerting power from all over the world.
@@Pravin.ShidoreYou prove yourself wrong by the very fact that you had to qualify your comment by saying “most sensitive borders”. If we instead look at “least sensitive borders”, you’ll find ones like USA and Canada, or Minnesota and Wisconsin, or my property and my neighbor’s property. All of these are “shared”, none involve the “dominance” you incorrectly claim is necessary.
I have an even easier solution to solve these problems. Just use CAD took me a minute for each one no thought required :>
Everything reminds me of her
Maf!!!
Why do you need to go to areas in problem 1 ? Two similar right triangles are in front of you. 4-r/r = 4/6. r = 2.4
because all those ideas are interconnected in geometry. if you study all these properties of triangles rigorously, you'll understand the specific method is arbitrary here. if it's easier for you to think in similar triangles, go for it. some people prefer thinking about areas when it suits the problem at hand.
@@laincoubert7236 It's not a question of it being easier for that one commenter to "think". It's that 99.99% of math-oriented people would find it far easier to solve (4-R)/R = 4/6 than to solve it the far, far more complicated way this video did. I also apply my 99.99% figure to the likelihood that the maker of this video will NOT admit this, as supported by his track record of almost never admitting these sorts of oversights. His channel is probably the best on UA-cam, but it would be so much better if he had follow-up videos showing how "crowd-solving" often beats him.
No, that is wrong. (4 - r)/r = 4/6. You must use grouping symbols on the left-hand side of the equation.
@@qc1okay again, this is not a far easier way to think in, it's just a different one. why? because both of these solutions can be understood by a middle schooler in their first year of geometry. why? because both solutions share the same foundation of geometry (axioms, theorems, etc). like we're still in territory that is not complex.
i mean, you can prove pythagoras theorem literally in a million different ways. if this channel were to unironically prove it with, i don't know, calculus, then your objection would be understandable. but on this level, i see it as a matter of preference. i can easily see a group of students that would go for the details and instantly look at similar triangles and get your equation, but i can also see a group that struggles with aligning those triangles in their head so they zoom out and think about areas in that sketch. personally, i'm with you in the first camp, but i also wish i was familiarized with other types of perception in school because you need that versatility to be good at solving various exam problems.
generally speaking, i agree with you that this channel often provides us with weird solutions and that they're not accepting any feedback lol. to me this is not one of those cases though
maybe i didn't really articulate my point that well, but what i'm trying to say is similar to a 3blue1brown video about alice and bob approaching the same problem in different ways. highly recommend it btw, regardless of our discussion here.
problem 1 has a way easier way to solve
even 8th grader can
从小学开始，第一次奥数的题，我竟然都对了
😭