For something that is "super easy, barely an inconvenience", I think it's funny that you took 50 seconds to explain that the radius of a circle with diameter 1 is 1/2.
He is trying to catch as many "normal people" as he can. If he takes less time maybe regular people wouldn't even try to understand what's going on here.
Then it takes 50 seconds to figure out that the radius of the small circle bisects the bottom of the equilateral triangle. Rather than simply saying, "By symmetry...".
I simply calculated the height of both points (1/2 and 1/3 of the height of △, which is √3/2), then subtracted the right one from the left one, and then used Pythagoras to calculate x.
Then you would have to argue why you take 1/3 of the height. Don't get me wrong, this is totally valid, one can see that by symmetry, but it is not something one can just assume.
@@AndreasStanglPlus I see what you mean and ofc it would be required in a formal proof, but for me that's just basic geometry, like using Pythagoras without proving it
I love how fully half of this video is him explaining in a roundabout way, with unnecessary calculations, that the horizontal distances were all 1/2 on the bottom line.
Na, half the video was him PROVING that the bottom horizontal distances were all 1/2. Sure you and I know that just from looking at it, but now he has actual mathematical proof of that from which he can make further factual statements with no possibility of error.
@@crosshairs007 Just saying "By symmetry this line segment has length 0.5" is also a proof given it is a regular triangle. His method was just way more complicated than it needed to be.
@joshhickman You're absolutely right, there are a shit ton of people in the comments going crazy about how he should have solved it using only anytical geometry (and not trigonometry) whereas the simplest way to do it is just coordinate geometry. Of course you'd need to know the relation between the radius of a circle inscribed in an equilateral triangle and its side, but that's common knowledge.
Yeah I thought that too . But who tf needs an easy solution . We mathematicians and computer science engineers are known to make shit complicated and easier at the same time lol . It's more fun solving hard problems or alternate solutions
There is another way to solve the problem with coordinates: Let's consider the coordinate system having as origin the vertex at the bottom left of the square and as unit vectors the two sides resulting from this same vertex. Then the coordinates of the center of the great circle are those of the center of the square, i.e. (1/2, 1/2). Moreover, noting that the height of the triangle is √3/2 (in the video) and knowing that the orthocenter of the equilateral triangle is at a distance from its base equal to one third of its height, the coordinates of the center of the small circle are therefore (3/2, √3/6). The distance between the centers of the two circles is therefore according to Pythagoras: x = √((3/2-1/2)² + (√3/6-1/2)²) = √(1 + 1/12 + 1/4 - √3/6) = √((8-√3)/6) ◼️
true of many of these vids. Which makes it interesting to pause the vid and do the problem first (to see how easy it REALLY) is first, and then watch and laugh at how complex he makes it look :D
Easier way of calculating r: The height of an equilateral triangle is also its angle bisector. The intersection of all three angle bisectors is the incenter (the center of the inscribed circle). The length from the incenter to a vertex is twice the length of the incenter to the edge, thus r is 1/3 the height (which we know to be √3/2). Therefore r=√3/6
it would be a more frustrating youtube video that way: it would be a series of facts, and people who don't know them feel like "huh, so I'm no good if I don't know that". And the video would be less free standing. I think he knows what he's doing, and he's doing it well :)
The centroid of an ET is 1/3 from any side. That side is 1/3 root 3. Subtract 0.5. Use old Pythagy's Theory and Bob's your uncle! Neat puzzle, thanks, Doc! :D
Good job! For some reason I like the video very much. On the square, you can divide the diameter by 2 and get the radius directly. On the triangle - use the fact that height and mediana are one and same. Then you can use the medicenter divideds the mediana in ratio 1:2
Center of an inscribed circle IS the center of a regular polygon (the circles are distracters). Horizontal distance is easily seen as 1 (half of each figure's width). Vertical distance is 1/2 for the square, and scaled for the triangle by the altitude versus 1. Then you use difference of altitudes versus horizontal difference, which leads to the Pythagorean solve he ends with. Same answer, fewer extraneous lines.
It's very easy to find r, just need a bit of pythagorean theorem, if h is the height of the equilateral triangle: (h-r)^2 = r^2 + (1/2)^2 and set h = sqrt(3)/2
I tried a Law of Cosines approach and found it pretty straight forward, using the diagonal of the square from the center, the angle bisector of the triangle to the incenter, and the given segment as side C across the angle of 105 degrees. This is known since their is a 30 degree angle between the square and the triangle, as well as the 90 degree angle of the square, and the 30 degree angle of the angle bisector.
can also solve it by knowing that centroid of triangle is h/3 and the distance from centroid to the midpoint of any side is r = h/3 (for equilateral triangles)
The length of the segment can be obtained by pythagoras. The horizontal side of the triangle is ½+½. The vertical is the difference in height of both circles, so 1-√3/2 (the circumcenter of an equilateral triangle is at √3/2). So x²=1+(1-√3/2)²
I have watched a few of this mans videos and he always makes a meal of an easy problem. I did this in about 1 minute by noticing that the equilateral triangle when you drop a perpendicular is a 30, 60, 90 triangle, ie 1,2, root 3 triangle cut in half. therefore height is root 3/2. Therefore r = root 3 /6. Then apply pythag... simple!
Would love to notice the fact that in equal sided triangle center of inscribed cirle is in intersection of bisectrices which are also heights and medians. So that pieces on the down side of triangle could be calculated without cosine function.
Why are you using trigonometry for a pure geometry problem? If you know that the line from the center of the small circle and perpendicular to the "base" is passing through the top vertex then you know you can use that it is an angle bisector and it is a median which cuts the base in to 2 equal segments. And the medians in every triangle interaect at the centroid and divide each other in to segments whose ratio is 2:1. There is also the 30-60-90 triangle with the edges ratio of 1:sqrt(3):2. I know that trigonometry is faster since you can do the calculations faster with calculators but then it should have been called a trigonometry problem and not a geometry problem.
Engineer here. The problem describes two points in the plane and asks for their Euclidean distance. How does casting this as a geometry problem or a trigonometry problem help or make a difference?
@@peterklenner2563 you should watch his overkill videos first. This is a geometry problem and can be resolved with pure geometry. It's like solving 5+3 with geometric series and limits and taylor series and complex numbers. Of course we can solve with them but 5+3 is a simple algebric problem and we can solve it with simple tools. You can solve the problem also with 1. vectors 2. Cartesean axes (analytic geometry? I forgot the correct term) And probably with topology also. But my question was why do it that way? It is not more simpler then the one i suggested with pure geometry (he introduced the problem as a geometric(!) one and not as just a problem)
@@udic01 Mr. Penn approaches the problem quite straightforwardly. Which shortcuts would be possible? More to the point, how could the last step of invoking the Pythagorean theorem be avoided or simplified?
Having picked up the problem from the thumbnail, before watching the vid, what I'd do is: 1. Drop an apothem from each polygon's center to its base 2. Draw a horizontal (i.e., parallel to the bases) through the ∆ center to the square's apothem. You now have a rt ∆ whose: • hypotenuse is x, • horizontal leg is a segment of the line we drew, of length b=1 (sum of half of each polygon's side); • and whose vertical leg is the difference in the heights of the two polygons' centers (that is, the difference between their incircle radii), a = ½ - 1/(2√3) = (√3 - 1)/(2√3) Enter Pythagoras, stage left: x² = a² + b² = (√3 - 1)²/12 + 1 = 1 + (2 - √3)/6 = (8 - √3)/6 x = √(8 - √3) / √6 = √(4/3 - 1/(2√3)) = 1.02208522087863126046709003932782... There may be a simpler symbolic form of the answer than this, idk... Fred
The center of the triangle circle is 60degrees between a vertex of the triangle and the center of a side. That creates a 30/60/90 triangle. Half the base is the long side making the short side half the base divided by root 3.
you can also do the second part by letting the bottem left of the square be an Origin. Then use vectors and find the modulus between the centre of the two circles
Another way to solve is since you know that a line from a corner of an equilateral triangle to it's center will bisect it, and that all angles are 60°, and since the x-distance is .5, you can find the y value (from the base) from .5tan(30). After that, it's pretty straight forward.
8:18 Homework... A triangle ABC is to be drawn with AB = 10, BC = 7 and the angle at A equal to θ, where θ is a certain defined angle. Of the two possible triangles that could be drawn, the larger triangle has three times the area of the smaller one. What is the value of cosθ?
Another way to get the triangle radius without trigonometry; The radius of the circle is equal to its Y coordinate (assuming the bottom side is at Y 0). And that also happens to be the center of the triangle. And to get the center, just add up all of the vertexes and divide by 3. Since X is already known, we only need Y. And it also turns out that two vertexes are at Y 0. So (0 + 0 + h) / 3 = h / 3. And h, in equilateral triangle is sqrt(3) / 2. And (sqrt(3) / 2) / 3 = sqrt(3) / 6. And it is the same answer.
Yoo English is not my primary language, I didn't learn about sin,cos, tan yet... but I still understood everything... That's a solid proof that you're a rly good teacher!
My answer was way different. I had "X is the 24th letter of the modern English language, having etymology dating back to the ancient greek chi and is often used as a stand-in for the primary unknown in the analysis of mathematical systems."
Everyone is saying that “coordinate geometry would be easier”, but they end up doing exactly what he just did, saying that they would put a coordinate system in the left corner of the square, getting the positions of both centers and calculating the distance. However in the video he just did the same thing, got the horizontal distance between each point (x1-x2), got the vertical distance between each point (y1-y2) and apply Pythagoras, exactly the same. In some comments they calculate the height of the center of the circle inscribed in the triangle (y2) by the well know formula 1/3 of height of triangle (a * sqrt(3)/2), that really would be easier than using tryg, but he decided to use the most explicit method which was easily proved and didn’t require this formula memorization or any other kind of proof. And again, remembering and using this formula has nothing to do with coordinate systems being easier.
@@stormlight131 Using coordinate geometry (or 2D-vectors) will always result in the same calculations, but finding the distance between (0, 1/2) and (1, sqrt(3)/6) is a standard procedure (which is actually applied Pythagoras). You don't have to construct extra lines to make a rectangle and transpose the y-coordinate of the triangle's incentre in order to find a side of right-triangle, just to then use Pythagoras; coordinate geometry does that for you. Similarly, finding the modulus of the difference between two vectors will require the same calculations, without having to construct a right triangle to apply Pythagoras to. These tools are useful conveniences, but they can't simplify the arithmetic.
Nice solution, but I think a far easier one is by placing them both in a coordinate system. You set the points of the square at (-1, 0), (0, 0), (0, 1) and (-1, 1), the center of the inscribed circle is at S (-1/2, 1/2) which you can easily work out. You set the points of the triangle at (0, 0), (1, 0) and (1/2, sqrt(3)/2), the radius of the inscribed circle is sqrt(3)/6 and from that follows that the center of the circle is at point S1 (1/2, sqrt(3)/6). And then you just calculate the distance SS1 between the centers.
I tell my students that whenever a circle touches anything, draw a line from the center of the circle to the point of tangency. In this case, this resulted in five extra unnecessary lines.
Actually, only 3 unnecesary lines (the 2 dropped perpendiculars from the centres of the circles are indeed necessary to show/calculate the difference in height between the centres)
This was pretty easy to solve using coordinates. The center of the square is at (0.5, 0.5), and the center of the triangle is at (1.5, sqrt(3)/6. Just use the distance formula to take care of the rest.
Since getting back into higher-ish maths, I forgot that rationalizing denominators was convention since we're always told we don't have to do stuff like that, or even simplifying for that matter. Incidentally, it took me a second after finishing the video to figure out why the radius mattered until I realized it was the y-component of the right endpoint.
Don't forget something called the "law of cosine". Connect the centers of the two circles to the right down corner of the square. Then solve the triangle.
Nice. That's what I was about to comment as well. We get a triangle with two easily to find sides (one is half of the diagonal of the square: √2/2, another is the distance between one corner of the equilateral triangle and the center of it's circle: √3/3 and the angle is 105 degrees). According to the rule of cosine, the distance we're looking for is the square root of the summ of the squares of these two sides minus the duble of the product of the sides multiplied by the cosine of the angle between them.
I just went almost to the end of the video cause this problem is incredibly easy. And I don't understand why I am seeing cos and angles in a problem where you only need to apply Pythagoras 3 times. Two to get the height of the center of the theoretical triangle inside the circle (with sides of .5), and another one for the hypotenuse of the theoretical triangle between the center of the square and the triangle.
You don't need the circles AT ALL. X is just the line between the centerpoint of the square of the triangle. As such, if we put a coordinate grid at the bottom left square, the centerpoint for the square is at 0.5/0.5 and the center point of the triangle is defined by a right triangle starting at 0.5/0 with the right angle at 1/0 and a 30° (Pi/6) angle at the starting point. Use cotangents to get the y-value of the centerpoint of the triangle. Having both the start and end point of x, you simply need to use a² + b² = c², a being the difference of the x-values, b being the difference of the y-values.
Me too. I looked at the problem and solved in my head. My solution was wrong but I tried, and with this amazing feeling of not getting afraid of the question I went to sleep proud of myself.
I really don't know what you were doing with angular functions there. This was fully solfable by Pythagoras. I did it in my head, without writing it down.
Dr. Penn really did make it more difficult than it was as his answer sqrt [8-sqrt3/6] also =1.022. This could be done in a minute. I did it by finding the radius of the circle (in the equilateral triangle to be 0.28867 using sqrt(s-a)(s-b)(s-c)/s) and the radius of the circle in the square is 0.5. the difference between the two is 0.211324 (or the length of the .perpendicular line formed). So now we have 2 sides and need to find x using the Pythagorean theorem. so 1^2 + 0.211324^2= x^2 1 + 0.04465 =x^2 1.04464 = x^2 1.022 = x Answer, and his answer "sqrt (8-sqrt3/6)" = 8-1.732/6 = 6.627/6 = 1.04465 = (1.04465) sqrt= 1.022 is equivalent to mine '1.022. (18:17)
Mr Penn, how do we tackle problems like the following? Smallest positive integer that starts with 3016 and is divisible by 3017. Generalization: smallest positive integer that starts with "n" and is divisible by "n+1". Is it more easy when "n is odd and n+1 is even" or when "n+1 is odd and n is even"?
@@JSSTyger it is an idea for Mr. Penn (if he likes and have time) to develop a video on number theory and play with numbers. Here the reference math.stackexchange.com/questions/3894122/ Another to play with is math.stackexchange.com/questions/1367371
@@carlosgiovanardi8197 I think the number that starts with 3016 is 30160949, which is 3017 times 9997. This would mean 9997 = (3016*10^4+949)/(3016+1).
It is much easier if you remember that the point of intersection of the medians of a triangle divide them in 2 parts in which one is the double of the other. So you would know that in a equilateral triangle the height is the same of the median and that means that the height of the center of the circle is 1/3 of the height of the triangle, than you subtract that from 0,5 which is the height of the center of the square and you find that the horizontal distance from those two points is 1, so you do Pitagora and you find the same result: 1,02. Sorry for my English but I'm italian
@@izvarzone sorry but I don't understand what you mean with non-symmetrical, but if you mean that the triangle is not equilateral it would work quite the same only if the triangle is isosceles on the horizontal base (the only difference would be the horizontal distance between the two points which would be equal to half the side of the square plus half the side of the base of the triangle). If you mean that the side of the square isn't equal to the side of the triangle the idea would be the same but you have to use different measures.
height of left point: 1/2 Height of right point: 1/3 Horizontal distance between the two points:1/6 Vertical distance between the two points:1/2+1/2=1 1+(1/6)^2=37/36 d=sqrt(37)/6
Just use the fact that the incenter is the intersection of the bisectors, but since it’s an equilateral triangle, the. It’s also the centroid. Which gives that the radius is 1/3 of the median, the median being radical3/2. Now you have big R and small r, Draw a line segment from the incenter perpendicular to R, thus constructing a right triangle of sides x, 1, and R-r. Apply Pythagorean theorem, and you get x.
I did the calculation w/ pythagoras since I'm not comfortable w/ trig values and got the same result pretty smoothly, only the result itself is imo fairly unsatisfying, as it is not just a 'clean' value nor does it give any insght into this particular arrangement of shapes. well explained regardless.
Make 2 lines from the center of the triangle's bottom sides to the facing angles, the crossing point is the center of the triangle, give them length R, use R.cos(x) = 0.5 (half of the base). Then by finding R , use R.sin(x) = y ( the height of the center from the bottom). Now make a trangle connecting both centers with side length of (0.5 - y) and base of , the distance between the centers can be found using pythagoras ( X^2 = (0.5-y)^2 + (1^2) ). Same answer, but in a minute, and without using the fact that there are 2 circles, because they share the same centers anyways.
The solution seems straightforward. However, you could use properties of an equilateral (or even isosceles) triangle (median = height = bisectrix) and the radius of inscribed circle as well. So, no extra explanations would be needed. I like your videos very much and this one contrasts with them. Here is no zest that exist in your other videos: just a school problem that does not require some trick to be solved efficiently.
In a equilateral triangle all 4 significant points - centers of inscribed circle, center of circle around, orthocenter and center of mass - are the same point and it is situated on the third of a height, and the height is h=(a/2)*SQRT(3). So the radius of the circle is (a/6)*SQRT(3). The distance between the centers of the circles on y axis is a/2-(a/6)*SQRT(3) or (a=1) (3-SQRT(3))/6. Square that we get (12-6SQRT(3))/36 or (2-SQRT(3)/6. The distance between centers on x axis is 1/2+1/2=1. We add that and get (8-SQRT(3))/6. So x= SQRT[(8-SQRT(3))/6] . This is basic 5th grade stuff, anyone who deals in mathematics ever a tad should know by heart. No trigonometry needed.
Big secret: in an equilateral triangle, the heights are identical to the medians (and both angular and perpendicular bisectors) which intersect in the centroid at a ratio of 2:1. So the radius of the small circle is equal to 1/3 of the triangle's height which is √3/2. So the difference in x dimension is 1 (2x1/2) and in y dimension it is 1/2 - √3/6 (or 1/2 - 1/(2√3)). These are the legs of a rectangular triangle. And then use Pythagoras to solve for x which is the hypothenuse.
X=sqrt(17)/4=1.0308 Vertical distance between Square and Triangle Centers = 1/2 - h = 1/4 sin30. = h /(1/2) = 2h, solving h = 1/4 Horizontal distance between Square and triangle = 1/2 + 1/2 = 1 Applying pithagoras theorem we solle x
Center of square: (1/2,1/2) Center of triangle which is the center of gravity of the triangle. Using G=(A+B+C)/3 we find that it is (3/2 , sqrt(3)/6) the rest is Pythagoras
Why not just use the obtuse triangle formed by connecting two centers and the commun point of the triangle and the square? Using cosine(7*pi/12). Prof Penn really knew how to make things complicated
If you just see that the radius of the circle inside the triangle is 1/3 of the height it will be easier and fast to do just applying pythagoras equation.
This can be solved extremely easily just by noticing that the height of point in the triangle is 1/3 of the height of the triangle because the incenter = baricenter in an equilateral triangle.
It should have been written as a given that the triangle is equilateral. Otherwise the small circle can have any diameter between a point and 1 and it can still be inside that triangle with a side equal to 1 right?
He never said the bottom of the square and bottom of the triangle were on the same line. It would be interesting to develop a formula for the length of that line segment based on the angle between the bases of the two figures.
I solved this problem using the area of the triangle. The hard part was solving for the radius of the circle inside the triangle. First, I need the area of the triangle. The width is 1, and the height is one leg of a 30,60,90 triangle where the hypotenuse is 1 and the other leg is 1/2. So the height is 1/2 * sqrt(3) = sqrt(3)/2. So, the area of the triangle is sqrt(3)/(2^2). Now is the fun part. I drop the angle bisectors from each corner until I reach the center of the circle. Now I have 3 smaller triangles each with an area of r/2, and their sum is 3r/2. Solving for r: 3r/2 = sqrt(3)/(2^2) 3r = sqrt(3)/2 r = sqrt(3)/2*3 r = 1/2*sqrt(3). I solved for the distance the same way as him. This was a fun problem! Thanks!
what if you bisected the triangle to find height, then divided by two to find height of the center. then use the distance formula with x1= 0.5 , x2 = 1.5, Y1=0.5 , and y2 equal to the height of the center of triangle which is sqrt 0.75
I did it by recognizing that it was going to be more than 1 and less than 7/6 which i derived by estimating the radius/length for both and adding... good enough for me
That has to be the most round about way of determining that the center of an equilateral triangle lines up with the mid-point of one of its sides. Cos(pi/3) is just thinking too hard. Why not just say that its symmetrical so it must be half?
The point at which triangles and rectangles meet (0,0) The left center is (-0.5, 0.5) The right center is (0.5, 0.5×3*(1/2)÷3) Pythagorean theorem of two points.
“Super easy, barely an inconvenience”, what a great call out. Dr. Penn, you rock.
It made me laugh and do a spit-take. Awesome.
Hey could you explain what does this mean.
I really didn't understand. I just thought he just meant it because it was easy.
@@fahad-xi-a8260 It's from screen rant pich meeting series
@@ThePiotrekpecet It's also been turned into a song! ua-cam.com/video/Rbm5I11Km6g/v-deo.html
@@fahad-xi-a8260 Just search up george ryan or screen rant pitch meetings. Many people in the comments use it as a joke.
For something that is "super easy, barely an inconvenience", I think it's funny that you took 50 seconds to explain that the radius of a circle with diameter 1 is 1/2.
@Flewt Puns I just did.
He is trying to catch as many "normal people" as he can. If he takes less time maybe regular people wouldn't even try to understand what's going on here.
Then it takes 50 seconds to figure out that the radius of the small circle bisects the bottom of the equilateral triangle. Rather than simply saying, "By symmetry...".
I wouldn't think too much about it considering "super easy..." is a joke in the first place.
@Hand Solo there’s nothing wrong with explaining givens to make students become more familiar with cosine uses.
I simply calculated the height of both points (1/2 and 1/3 of the height of △, which is √3/2), then subtracted the right one from the left one, and then used Pythagoras to calculate x.
Same.
@@csababekesi-marton2393 so did I much less complicated
Same thing: saw the thumbnail, went to calculator, came back with the answer like that
Then you would have to argue why you take 1/3 of the height. Don't get me wrong, this is totally valid, one can see that by symmetry, but it is not something one can just assume.
@@AndreasStanglPlus I see what you mean and ofc it would be required in a formal proof, but for me that's just basic geometry, like using Pythagoras without proving it
I love how fully half of this video is him explaining in a roundabout way, with unnecessary calculations, that the horizontal distances were all 1/2 on the bottom line.
Na, half the video was him PROVING that the bottom horizontal distances were all 1/2. Sure you and I know that just from looking at it, but now he has actual mathematical proof of that from which he can make further factual statements with no possibility of error.
@@crosshairs007 Just saying "By symmetry this line segment has length 0.5" is also a proof given it is a regular triangle. His method was just way more complicated than it needed to be.
Am I crazy or is this way easier without any circles at all and just using coordinate geometry?
Coordinates are usually considered bashes , and not elegant solutions. That's why synthetic proofs are better.
@joshhickman You're absolutely right, there are a shit ton of people in the comments going crazy about how he should have solved it using only anytical geometry (and not trigonometry) whereas the simplest way to do it is just coordinate geometry. Of course you'd need to know the relation between the radius of a circle inscribed in an equilateral triangle and its side, but that's common knowledge.
you are totally right. the circeles are there to confuse you.
Yeah I thought that too . But who tf needs an easy solution . We mathematicians and computer science engineers are known to make shit complicated and easier at the same time lol . It's more fun solving hard problems or alternate solutions
Basically,
Coordinate geometry is what’s done here.
Coz the distance formula comes from pythagorean theorem as such
There is another way to solve the problem with coordinates:
Let's consider the coordinate system having as origin the vertex at the bottom left of the square and as unit vectors the two sides resulting from this same vertex. Then the coordinates of the center of the great circle are those of the center of the square, i.e. (1/2, 1/2). Moreover, noting that the height of the triangle is √3/2 (in the video) and knowing that the orthocenter of the equilateral triangle is at a distance from its base equal to one third of its height, the coordinates of the center of the small circle are therefore (3/2, √3/6).
The distance between the centers of the two circles is therefore according to Pythagoras:
x = √((3/2-1/2)² + (√3/6-1/2)²)
= √(1 + 1/12 + 1/4 - √3/6)
= √((8-√3)/6) ◼️
Coordinate Geometry: "I also exist. Why no one uses me? 😭"
@DELTA en some people avoid Coordinate Geometry to make questions harder.
Coordinate geometry: "Am I a joke to you?"
It's how I approached this one. Found the same distances, just viewed them as points on a plane, and then used the distance formula between them.
@Enrique Mtz LoL! And what would be the fun in that?!? Why not try hard to confuse the hell out of most folks?
yes, it's so quick with coordinate geometry.
I used the fact that rs=A, so r=A/s, and since the triangle is equilateral with a sidelength of 1, it was easy to put values in the places of A and s
You made this much more complicated than it needed to be.
He does that for a reason. Haven’t you seen his overkill series lol
true of many of these vids. Which makes it interesting to pause the vid and do the problem first (to see how easy it REALLY) is first, and then watch and laugh at how complex he makes it look :D
Not at all. I guess that's the best way to go about it.
Easier way of calculating r:
The height of an equilateral triangle is also its angle bisector. The intersection of all three angle bisectors is the incenter (the center of the inscribed circle). The length from the incenter to a vertex is twice the length of the incenter to the edge, thus r is 1/3 the height (which we know to be √3/2). Therefore r=√3/6
it would be a more frustrating youtube video that way: it would be a series of facts, and people who don't know them feel like "huh, so I'm no good if I don't know that". And the video would be less free standing. I think he knows what he's doing, and he's doing it well :)
The centroid of an ET is 1/3 from any side. That side is 1/3 root 3. Subtract 0.5. Use old Pythagy's Theory and Bob's your uncle! Neat puzzle, thanks, Doc! :D
Good job! For some reason I like the video very much.
On the square, you can divide the diameter by 2 and get the radius directly. On the triangle - use the fact that height and mediana are one and same. Then you can use the medicenter divideds the mediana in ratio 1:2
Center of an inscribed circle IS the center of a regular polygon (the circles are distracters). Horizontal distance is easily seen as 1 (half of each figure's width). Vertical distance is 1/2 for the square, and scaled for the triangle by the altitude versus 1. Then you use difference of altitudes versus horizontal difference, which leads to the Pythagorean solve he ends with. Same answer, fewer extraneous lines.
It's very easy to find r, just need a bit of pythagorean theorem, if h is the height of the equilateral triangle:
(h-r)^2 = r^2 + (1/2)^2
and set h = sqrt(3)/2
super easy barely an inconvenience! I see what you did there!
Could you explain what did he do there.
@@fahad-xi-a8260 It's a reference to Ryan George's genius Pitch Meeting series over on the Screenrant UA-cam channel.
I tried a Law of Cosines approach and found it pretty straight forward, using the diagonal of the square from the center, the angle bisector of the triangle to the incenter, and the given segment as side C across the angle of 105 degrees. This is known since their is a 30 degree angle between the square and the triangle, as well as the 90 degree angle of the square, and the 30 degree angle of the angle bisector.
can also solve it by knowing that centroid of triangle is h/3 and the distance from centroid to the midpoint of any side is
r = h/3 (for equilateral triangles)
i don't know how you make this seem so easy... when you explain ... it's like tetris block falling together in my head perfectly!!!
Loving these geometry problems. Been giving these questions to my classes of pupils. Good work.
The length of the segment can be obtained by pythagoras. The horizontal side of the triangle is ½+½. The vertical is the difference in height of both circles, so 1-√3/2 (the circumcenter of an equilateral triangle is at √3/2). So x²=1+(1-√3/2)²
Beautiful work. My new favorite channel. Thank you.
I have watched a few of this mans videos and he always makes a meal of an easy problem. I did this in about 1 minute by noticing that the equilateral triangle when you drop a perpendicular is a 30, 60, 90 triangle, ie 1,2, root 3 triangle cut in half. therefore height is root 3/2. Therefore r = root 3 /6. Then apply pythag... simple!
I always like to see a 'real-world' value, like 1.022, more comforting you haven't screwed something along the way!
Nice problem though, I enjoyed it.
Would love to notice the fact that in equal sided triangle center of inscribed cirle is in intersection of bisectrices which are also heights and medians. So that pieces on the down side of triangle could be calculated without cosine function.
We can use that centre of circle lies at the "centr of gravity" of the triangel . Then we have r=1/3*(sin pi/3)=sqrt(3)/6
Its better to say "center of geometry".
Centre of gravity, centre of mass, and centre of geometry are different things
Don't they coincide for an equilateral triangle?
Why are you using trigonometry for a pure geometry problem?
If you know that the line from the center of the small circle and perpendicular to the "base" is passing through the top vertex then you know you can use that it is an angle bisector and it is a median which cuts the base in to 2 equal segments.
And the medians in every triangle interaect at the centroid and divide each other in to segments whose ratio is 2:1.
There is also the 30-60-90 triangle with the edges ratio of 1:sqrt(3):2.
I know that trigonometry is faster since you can do the calculations faster with calculators but then it should have been called a trigonometry problem and not a geometry problem.
Yes so true,I did it the same way
I've wanted to say this before too. Trig is less elegant if you're using 30-60-90 triangles
Engineer here. The problem describes two points in the plane and asks for their Euclidean distance. How does casting this as a geometry problem or a trigonometry problem help or make a difference?
@@peterklenner2563 you should watch his overkill videos first.
This is a geometry problem and can be resolved with pure geometry.
It's like solving 5+3 with geometric series and limits and taylor series and complex numbers. Of course we can solve with them but 5+3 is a simple algebric problem and we can solve it with simple tools.
You can solve the problem also with 1. vectors
2. Cartesean axes (analytic geometry? I forgot the correct term)
And probably with topology also.
But my question was why do it that way?
It is not more simpler then the one i suggested with pure geometry (he introduced the problem as a geometric(!) one and not as just a problem)
@@udic01 Mr. Penn approaches the problem quite straightforwardly. Which shortcuts would be possible? More to the point, how could the last step of invoking the Pythagorean theorem be avoided or simplified?
Quick geometry problem series are good basic builders or introduction to olympiad geometry
Having picked up the problem from the thumbnail, before watching the vid, what I'd do is:
1. Drop an apothem from each polygon's center to its base
2. Draw a horizontal (i.e., parallel to the bases) through the ∆ center to the square's apothem.
You now have a rt ∆ whose:
• hypotenuse is x,
• horizontal leg is a segment of the line we drew, of length b=1 (sum of half of each polygon's side);
• and whose vertical leg is the difference in the heights of the two polygons' centers (that is, the difference between their incircle radii), a = ½ - 1/(2√3) = (√3 - 1)/(2√3)
Enter Pythagoras, stage left:
x² = a² + b² = (√3 - 1)²/12 + 1 = 1 + (2 - √3)/6 = (8 - √3)/6
x = √(8 - √3) / √6 = √(4/3 - 1/(2√3)) = 1.02208522087863126046709003932782...
There may be a simpler symbolic form of the answer than this, idk...
Fred
The center of the triangle circle is 60degrees between a vertex of the triangle and the center of a side. That creates a 30/60/90 triangle. Half the base is the long side making the short side half the base divided by root 3.
This was the only problem of yours I could actually solve! I used analytic geometry along with a couple of lines to find x = sqrt((8 - sqrt(3))/6)
you can also do the second part by letting the bottem left of the square be an Origin. Then use vectors and find the modulus between the centre of the two circles
Another way to solve is since you know that a line from a corner of an equilateral triangle to it's center will bisect it, and that all angles are 60°, and since the x-distance is .5, you can find the y value (from the base) from .5tan(30). After that, it's pretty straight forward.
I love how mathmatical answers are so untelling.
It's not about the destination but about the theory's that we made friends with along the way.
8:18
Homework...
A triangle ABC is to be drawn with AB = 10, BC = 7 and the angle at A equal to θ, where θ is a certain defined angle.
Of the two possible triangles that could be drawn, the larger triangle has three times the area of the smaller one.
What is the value of cosθ?
Nyc one
Is it sqrt(17)/5 ?
@@srijanbhowmick9570 I dunno... Did you just throw this number randomly? 😛
@@goodplacetostop2973 his ans is right try using cosine rule
is it sqrt of (51/75)
Another way to get the triangle radius without trigonometry;
The radius of the circle is equal to its Y coordinate (assuming the bottom side is at Y 0). And that also happens to be the center of the triangle. And to get the center, just add up all of the vertexes and divide by 3. Since X is already known, we only need Y. And it also turns out that two vertexes are at Y 0. So (0 + 0 + h) / 3 = h / 3. And h, in equilateral triangle is sqrt(3) / 2. And (sqrt(3) / 2) / 3 = sqrt(3) / 6. And it is the same answer.
Yoo English is not my primary language, I didn't learn about sin,cos, tan yet... but I still understood everything... That's a solid proof that you're a rly good teacher!
My answer was way different. I had "X is the 24th letter of the modern English language, having etymology dating back to the ancient greek chi and is often used as a stand-in for the primary unknown in the analysis of mathematical systems."
1:45 doesn't that automatically divide the triangle base in half?
At 4:47 you draw a XY axes and the origin in the left bottom corner of the square and easily find the distance of two point knowing their coordinates.
I like how he is teaching. Amazing
Everyone is saying that “coordinate geometry would be easier”, but they end up doing exactly what he just did, saying that they would put a coordinate system in the left corner of the square, getting the positions of both centers and calculating the distance. However in the video he just did the same thing, got the horizontal distance between each point (x1-x2), got the vertical distance between each point (y1-y2) and apply Pythagoras, exactly the same.
In some comments they calculate the height of the center of the circle inscribed in the triangle (y2) by the well know formula 1/3 of height of triangle (a * sqrt(3)/2), that really would be easier than using tryg, but he decided to use the most explicit method which was easily proved and didn’t require this formula memorization or any other kind of proof. And again, remembering and using this formula has nothing to do with coordinate systems being easier.
Glad someone said it. I thought I might be crazy for thinking the coordinate geometry method is identical to the given method.
@@stormlight131 Using coordinate geometry (or 2D-vectors) will always result in the same calculations, but finding the distance between (0, 1/2) and (1, sqrt(3)/6) is a standard procedure (which is actually applied Pythagoras). You don't have to construct extra lines to make a rectangle and transpose the y-coordinate of the triangle's incentre in order to find a side of right-triangle, just to then use Pythagoras; coordinate geometry does that for you.
Similarly, finding the modulus of the difference between two vectors will require the same calculations, without having to construct a right triangle to apply Pythagoras to. These tools are useful conveniences, but they can't simplify the arithmetic.
Haha didn't know you liked pitch meetings too, Dr. Penn! Loved the problem, also.
Everybody asks what is x, but no one asks how is x :(
I'll do you one better. Why is x?
This comment was expected, but it is funny ngl...
How do you know that no one asks that? Maybe many people do ask.
@@jesusthroughmary I'll do you one better. Y = X.
@@jesusthroughmary when is x
Nice solution, but I think a far easier one is by placing them both in a coordinate system. You set the points of the square at (-1, 0), (0, 0), (0, 1) and (-1, 1), the center of the inscribed circle is at S (-1/2, 1/2) which you can easily work out. You set the points of the triangle at (0, 0), (1, 0) and (1/2, sqrt(3)/2), the radius of the inscribed circle is sqrt(3)/6 and from that follows that the center of the circle is at point S1 (1/2, sqrt(3)/6). And then you just calculate the distance SS1 between the centers.
01:31 did he just...?
Yh he fucken did, could not be prouder
we know that phrase from the first movie!!
I tell my students that whenever a circle touches anything, draw a line from the center of the circle to the point of tangency.
In this case, this resulted in five extra unnecessary lines.
Actually, only 3 unnecesary lines (the 2 dropped perpendiculars from the centres of the circles are indeed necessary to show/calculate the difference in height between the centres)
This was pretty easy to solve using coordinates. The center of the square is at (0.5, 0.5), and the center of the triangle is at (1.5, sqrt(3)/6. Just use the distance formula to take care of the rest.
Since getting back into higher-ish maths, I forgot that rationalizing denominators was convention since we're always told we don't have to do stuff like that, or even simplifying for that matter. Incidentally, it took me a second after finishing the video to figure out why the radius mattered until I realized it was the y-component of the right endpoint.
Don't forget something called the "law of cosine". Connect the centers of the two circles to the right down corner of the square. Then solve the triangle.
Nice. That's what I was about to comment as well. We get a triangle with two easily to find sides (one is half of the diagonal of the square: √2/2, another is the distance between one corner of the equilateral triangle and the center of it's circle: √3/3 and the angle is 105 degrees). According to the rule of cosine, the distance we're looking for is the square root of the summ of the squares of these two sides minus the duble of the product of the sides multiplied by the cosine of the angle between them.
2:22 Did you just calculate cos(pi/3) for dividing by 2? You know you could just divide by 2 to find the length to the middle?
after 5:00 , we get coordinates of both center, thus distance b/w two point is ans.
I just went almost to the end of the video cause this problem is incredibly easy. And I don't understand why I am seeing cos and angles in a problem where you only need to apply Pythagoras 3 times. Two to get the height of the center of the theoretical triangle inside the circle (with sides of .5), and another one for the hypotenuse of the theoretical triangle between the center of the square and the triangle.
Excellent solution and explanation
If a equilateral triangle we can't we just directly use inradius formula a/2root3.. ???
You don't need the circles AT ALL. X is just the line between the centerpoint of the square of the triangle. As such, if we put a coordinate grid at the bottom left square, the centerpoint for the square is at 0.5/0.5 and the center point of the triangle is defined by a right triangle starting at 0.5/0 with the right angle at 1/0 and a 30° (Pi/6) angle at the starting point. Use cotangents to get the y-value of the centerpoint of the triangle.
Having both the start and end point of x, you simply need to use a² + b² = c², a being the difference of the x-values, b being the difference of the y-values.
Find small r by A/S formula for incentre. Then find the coordinates of second circle. Use distance formulas.
Super easy, barely an inconvenience.
few ways here and I used the angle bisector property, r is .5tan60/2
another way could be cos formula or coordinate
Nice, I could actually do this one in my head! Took a few minutes, but still worth it so that I could stay lying down and not get a piece of paper lol
Nice, did the same😊
Thought the same and so did this one in my head to try and keep my aging brain in shape!
+
Me too. I looked at the problem and solved in my head. My solution was wrong but I tried, and with this amazing feeling of not getting afraid of the question I went to sleep proud of myself.
I really don't know what you were doing with angular functions there. This was fully solfable by Pythagoras.
I did it in my head, without writing it down.
Dr. Penn really did make it more difficult than it was as his answer sqrt [8-sqrt3/6] also =1.022.
This could be done in a minute.
I did it by finding the radius of the circle (in the equilateral triangle to be 0.28867 using sqrt(s-a)(s-b)(s-c)/s)
and the radius of the circle in the square is 0.5. the difference between the two is 0.211324 (or the length of the .perpendicular line formed). So now we have 2 sides and need to find x using the Pythagorean theorem.
so 1^2 + 0.211324^2= x^2
1 + 0.04465 =x^2
1.04464 = x^2
1.022 = x Answer, and his answer "sqrt (8-sqrt3/6)" = 8-1.732/6 = 6.627/6 = 1.04465 = (1.04465) sqrt= 1.022 is equivalent to mine '1.022. (18:17)
I tried to build a x-y croodinate at the bottom left corner of that square, and using croodinates are fast too!
Mr Penn, how do we tackle problems like the following?
Smallest positive integer that starts with 3016 and is divisible by 3017.
Generalization: smallest positive integer that starts with "n" and is divisible by "n+1".
Is it more easy when "n is odd and n+1 is even" or when "n+1 is odd and n is even"?
a = (n*10^b+c)/(n+1) where a, b, and c are positive integers. There, I started it for you :)
@@JSSTyger it is an idea for Mr. Penn (if he likes and have time) to develop a video on number theory and play with numbers.
Here the reference math.stackexchange.com/questions/3894122/
Another to play with is math.stackexchange.com/questions/1367371
@@carlosgiovanardi8197 I think the number that starts with 3016 is 30160949, which is 3017 times 9997. This would mean 9997 = (3016*10^4+949)/(3016+1).
We can use trig to find that 1/2 of 1 is 1/2? That's brilliant!
Have you never heard of special triangles? 30-60-90=1-2-3^1/2
It is much easier if you remember that the point of intersection of the medians of a triangle divide them in 2 parts in which one is the double of the other. So you would know that in a equilateral triangle the height is the same of the median and that means that the height of the center of the circle is 1/3 of the height of the triangle, than you subtract that from 0,5 which is the height of the center of the square and you find that the horizontal distance from those two points is 1, so you do Pitagora and you find the same result: 1,02. Sorry for my English but I'm italian
but would that work with non-symmetric shapes?
@@izvarzone sorry but I don't understand what you mean with non-symmetrical, but if you mean that the triangle is not equilateral it would work quite the same only if the triangle is isosceles on the horizontal base (the only difference would be the horizontal distance between the two points which would be equal to half the side of the square plus half the side of the base of the triangle). If you mean that the side of the square isn't equal to the side of the triangle the idea would be the same but you have to use different measures.
You don't need cosines, in a equilateral triangles medians and altitudes of a triangle intersect at one point.
height of left point: 1/2
Height of right point: 1/3
Horizontal distance between the two points:1/6
Vertical distance between the two points:1/2+1/2=1
1+(1/6)^2=37/36
d=sqrt(37)/6
It's very very good . this is incredible bro
put O(0;0) at the point where square and triangle joint!
A(-R,+R)
B(+a/2,r)
find distance between A and B!
d=sqrt( 9/4 + (0.5-0.5/sqrt(3))^2)
Just use the fact that the incenter is the intersection of the bisectors, but since it’s an equilateral triangle, the. It’s also the centroid. Which gives that the radius is 1/3 of the median, the median being radical3/2.
Now you have big R and small r,
Draw a line segment from the incenter perpendicular to R, thus constructing a right triangle of sides x, 1, and R-r. Apply Pythagorean theorem, and you get x.
I did the calculation w/ pythagoras since I'm not comfortable w/ trig values and got the same result pretty smoothly, only the result itself is imo fairly unsatisfying, as it is not just a 'clean' value nor does it give any insght into this particular arrangement of shapes. well explained regardless.
Make 2 lines from the center of the triangle's bottom sides to the facing angles, the crossing point is the center of the triangle, give them length R, use R.cos(x) = 0.5 (half of the base).
Then by finding R , use R.sin(x) = y ( the height of the center from the bottom).
Now make a trangle connecting both centers with side length of (0.5 - y) and base of , the distance between the centers can be found using pythagoras ( X^2 = (0.5-y)^2 + (1^2) ).
Same answer, but in a minute, and without using the fact that there are 2 circles, because they share the same centers anyways.
Nice, easy to understand
The solution seems straightforward. However, you could use properties of an equilateral (or even isosceles) triangle (median = height = bisectrix) and the radius of inscribed circle as well. So, no extra explanations would be needed.
I like your videos very much and this one contrasts with them. Here is no zest that exist in your other videos: just a school problem that does not require some trick to be solved efficiently.
In a equilateral triangle all 4 significant points - centers of inscribed circle, center of circle around, orthocenter and center of mass - are the same point and it is situated on the third of a height, and the height is h=(a/2)*SQRT(3). So the radius of the circle is (a/6)*SQRT(3). The distance between the centers of the circles on y axis is a/2-(a/6)*SQRT(3) or (a=1) (3-SQRT(3))/6. Square that we get (12-6SQRT(3))/36 or (2-SQRT(3)/6. The distance between centers on x axis is 1/2+1/2=1. We add that and get (8-SQRT(3))/6. So x= SQRT[(8-SQRT(3))/6] . This is basic 5th grade stuff, anyone who deals in mathematics ever a tad should know by heart. No trigonometry needed.
Big secret: in an equilateral triangle, the heights are identical to the medians (and both angular and perpendicular bisectors) which intersect in the centroid at a ratio of 2:1. So the radius of the small circle is equal to 1/3 of the triangle's height which is √3/2.
So the difference in x dimension is 1 (2x1/2) and in y dimension it is 1/2 - √3/6 (or 1/2 - 1/(2√3)). These are the legs of a rectangular triangle. And then use Pythagoras to solve for x which is the hypothenuse.
2:10
I have no idea it is an equilateral triangle until you said that. No clue for that on the board at the beginning.
those two triangles you split the equilateral triangle into are congruent, not just similar
X=sqrt(17)/4=1.0308
Vertical distance between Square and Triangle Centers = 1/2 - h = 1/4
sin30. = h /(1/2) = 2h, solving h = 1/4
Horizontal distance between Square and triangle = 1/2 + 1/2 = 1
Applying pithagoras theorem we solle x
Center of square: (1/2,1/2)
Center of triangle which is the center of gravity of the triangle. Using G=(A+B+C)/3 we find that it is (3/2 , sqrt(3)/6)
the rest is Pythagoras
Why not just use the obtuse triangle formed by connecting two centers and the commun point of the triangle and the square? Using cosine(7*pi/12). Prof Penn really knew how to make things complicated
If you just see that the radius of the circle inside the triangle is 1/3 of the height it will be easier and fast to do just applying pythagoras equation.
There it is!
This can be solved extremely easily just by noticing that the height of point in the triangle is 1/3 of the height of the triangle because the incenter = baricenter in an equilateral triangle.
Easy-peasy. Solved it in my mind.
In coordinate plane. Distance of (0.5, 0.5) and (1.5, tan30×0.5). D^2=1^2+[0.5(1- √3/3)]^2..
When he busted out the trig to prove an altitude from a vertex to a side of an equilateral Δ bisects the side.
How do you know it’s an equilateral?
It should have been written as a given that the triangle is equilateral. Otherwise the small circle can have any diameter between a point and 1 and it can still be inside that triangle with a side equal to 1 right?
Man, this was total overkill. Good job!
Excellent! Thank you!
He never said the bottom of the square and bottom of the triangle were on the same line. It would be interesting to develop a formula for the length of that line segment based on the angle between the bases of the two figures.
you can solve it with the law of cosines and a sum of arcs
but still great job
I solved this problem using the area of the triangle.
The hard part was solving for the radius of the circle inside the triangle.
First, I need the area of the triangle. The width is 1, and the height is one leg of a 30,60,90 triangle where the hypotenuse is 1 and the other leg is 1/2. So the height is 1/2 * sqrt(3) = sqrt(3)/2. So, the area of the triangle is sqrt(3)/(2^2).
Now is the fun part. I drop the angle bisectors from each corner until I reach the center of the circle. Now I have 3 smaller triangles each with an area of r/2, and their sum is 3r/2.
Solving for r:
3r/2 = sqrt(3)/(2^2)
3r = sqrt(3)/2
r = sqrt(3)/2*3
r = 1/2*sqrt(3).
I solved for the distance the same way as him. This was a fun problem! Thanks!
what if you bisected the triangle to find height, then divided by two to find height of the center. then use the distance formula with x1= 0.5 , x2 = 1.5, Y1=0.5 , and y2 equal to the height of the center of triangle which is sqrt 0.75
use pythagorean to find hight. base is .5 and hypotenuse is 1.
I did it by recognizing that it was going to be more than 1 and less than 7/6 which i derived by estimating the radius/length for both and adding... good enough for me
Great Ryan George reference, Dr. Penn.
Me:okay,i see ,now i understand.
My brains:no,you don't
You made this way more complicates than it needed to be. No trig required, you can just use apothems and 30-60-90 triangles with some pythagorean
That has to be the most round about way of determining that the center of an equilateral triangle lines up with the mid-point of one of its sides. Cos(pi/3) is just thinking too hard. Why not just say that its symmetrical so it must be half?
The point at which triangles and rectangles meet (0,0)
The left center is (-0.5, 0.5)
The right center is (0.5, 0.5×3*(1/2)÷3)
Pythagorean theorem of two points.