a geometry problem inspired by Japanese Sangaku

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[deebd]

That "well known fact" can be easily derived as well, so no need to memorize it: using the center point of the inscribed circle, the original triangle can be partitioned in 3 new triangles, and their 3 areas obviously sum up to the area A of the original triangle. Their areas are AB * r/2, AC * r/2 and BC * r/2, and solving this for r gives: r = 2 * A / (AB + AC + BC) = 2 * A / P.

[jimschneider]

@7:00 - yet another "well-known fact" I'd never heard of :).

[hellohabibi]

That twin drawing animation was really cool!!!

[matheusjahnke]

7:00 I didn't know that... but I think I arrived at that
Call the center of the circle O
Draw OA, OB, OC
The triangle ABC can be almost partitioned[ignoring the dividing lines we made which have 0 area] into OAB, OBC, OAC... in particular, their combined area must be the same

But the triangles we created have a base equal to the side of the original triangle, and height equal to the radius of the circle....
area(OAB ) = r ⋅ length(AB) / 2
area(OBC) = r ⋅ length(BC) / 2
area(OAC) = r ⋅ length(AC) / 2
A=r ⋅ length(AB) / 2 + r ⋅ length(BC) / 2 + r ⋅ length(AC) / 2 = (r/2) ⋅ (length(AB) + length(BC)+length(AC)) = (r/2) p
Solving for r:
r=A/(2p)

[roberttelarket]

I didn't know you had an identical twin?

[matheusjahnke]

1:40 You could argue like you did... but you could also argue that, since AB is parallel to DE(because, as he said before, the outer figure is a square), then CAB ≅ DEC because they are two parallel lines being crossed by a third one;

[philstubblefield]

I *_love_* these geometry problems! I've missed them lately.

[ingiford]

My HS teacher would have marked your similar triangles wrong. She was a stickler that the ordering had to be the same. First letter of each had to have the same angle, second letter, etc so ABC ~ EDC and not ABC ~ CDE

[edcoad]

I set x =-2x + 2 as those are the equations of the 2 lines, got x=y=2/3 which allows lengths via pythag and the r = 2A/p.

[jmcsquared]

Sangaku problems are some of my favorites. They are surprisingly difficult.

[stephenhamer]

Set up a right handed co-ord sys with A as origin and x-axis, AB. Then using parallel projections, we can show C has co-ords (2/3, 2/3). Side lengths of triangle ABC then follow from Pythag. The area of triangle ABC = 1/2 - Area triangle BEC = 1/2 - 1/6 = 1/3, then (proceeding like Penn) we're home.
NOTES:
The r = 2A/P result is easily obtained by cutting triangle ABC into three triangles of height r
The alternate angles theorem (Euclid, book 1) is often a handy piece of kit if you want to prove two triangles are similar

[Bodyknock]

2:11 Just a quick alternative variant step, you could notice that since DE is parallel to AB that angles EDB and DBA are congruent (i.e. interior angles along the transversal of parallel lines.) Like the same reasoning for angles EAB and EDB being congruent.

[Fred-yqfs]

Got the result too.
I knew there was a formula for r.
I had to look it up, but then the calculations were easy.
I found the coordinates of point C (origin A) by intersecting the 2 lines AE and BD: xc=yc=2/3
I named H the point for height HC in triangle ABC.
xH =xC = 2/3
From there, I derived the triangle lengths, the perimeter, the area, and done!

[painfullystupid]

Technically you can put it on a graph map the lines using the coordinate system find the point of intersections for the vertices and the y coords of the incenter using the formula which would be the radius

[ELS]

Thanks, I'd love to see more geometry problems.

[cicik]

triangles above and below are proportional - above is 2 times smaller,
so you can figure out the sides of a triangle and then there is a formula how to get radius of the circle inside the triangle by its sides

[bsmith]

Nice problem. I decided to solve along with the video, but by generalizing by letting DE=x (0<x<1). More algebra involved of course, but the same methodology.
I got BC=sqrt[1+x^2]/(x+1), then AC=sqrt[2]/(x+1), then the altitude of ABC =1/(1+x). Plugging these into the inradius formula I get a final result of radius = 1/(1+sqrt[2]+x+sqrt[1+x^2])
This does get the same answer of 2/(3+2*sqrt[2]+sqrt[5]) when evaluated at x=1/2.

[williamperez-hernandez]

I thought he was going to use Heron's formula to obtain the area of the triangle. Using the similarity of triangles is much easier.

[lammeraalapps]

Nice problem. I took it for granted that your answer was correct and tried to draw it to scale. Unfortunately, I saw that your computed value of r was too large. I figured out that the right answer is 2/(3+sqrt(5)+2*sqrt(2)), approximately 0.248. In case you do not like radicals in the denominator, the expression just mentioned is equal to I saw that I am not the only one who gave a correct answer in a comment. By the way, I wrote two books about similar geometry problems and published them with Amazon. Best wishes, Leen Ammeraal

[ciscoortega]

My approach was more angles/trig-based, and I think it's more elementary. We only need the fact that the incircle is found at the intersection of the three angle bisectors of the triangle. (I promise the solution is really simple, but describing the diagram in text makes it more of a mouthful than necessary hahaha)

Draw the angle bisector at A and the angle bisector at B, and let their point of intersection be denoted by F. Then, the radius of the in-circle is the altitude of the triangle AFB.
Project point F onto the line AB, and give its "shadow" the name G. This splits triangle AFB into the two right triangles AFG and BFG.
Let t = |AG| and so |BG| = 1 - t be the lengths of the two segments that AB was split into. Let h = |FG| be the altitute of the triangle.

Using trig, h = t*tan(GAF) and h = (1-t)*tan(GBF). Simple algebra then gives us that h = tan(GAF)tan(GBF) / (tan(GAF) + tan(GBF))

Both of tan(GAF) and tan(GBF) can be solved for using the half-angle formula (or double-angle formula, I guess) for tangents, since by SOHCAHTOA we know that tan(CAB) = 1 and tan(CBA) = 2. Specifically, the formula tells us that tan(GAF) = (-1 + sqrt(2)) and tan(GBF) = (-1 + sqrt(5))/2.

Plugging these values in gives us the same result as the one you got in the video, after some simplifications! :)