What is the radius of 🔴? -- A fun geometry puzzle.

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[-JiminP-]

For a triangle with side length a, b, c and area A, the radius r of the incircle of the triangle is 2A/(a+b+c) (hint: draw three line segments from the incenter to the each vertex of the triangle).
Since PC bisects RCQ = ACB, AP:PB is equal to AC:CB (by the angle bisector theorem), and thus BP is 5/9 of BA. Since BPQ is similar to BAC, the radius of the incircle of BPQ is 5/9 of the radius of the incircle of BAC, which has the radius 3*4/(3+4+5) = 1. Therefore the answer is 5/9.

[goodplacetostop]

10:28 So what gift did you get, if you celebrate Christmas?

[quantadotonium]

Hey, @ 3:52 is it not supposed to be : AP/AR = AB/AC ? ie, the denominator BC should be AC?

[Qermaq]

Funny, I solved this backwards. Well-known-fact that the incircle of a 3-4-5 RT has radius 1. Then I essentially built the rhombus and the little 3-4-5 from that. I quickly found that by scaling the top 3-4-5 by a factor of 5 made everything into integers. So I wound up with the top triangle at 15-20-25 with incircle r=5, the smaller triangle 12-16-20, and the rhombus has sides 20 and acute angle atan(3/4). The big triangle is thus 27-36-45, so I have to scale everything down by a factor of 9, and the inradius becomes 5/9.

[wannabeactuary]

At 7:40 Area of triangle is 0.5(5/3)(20/9) = 50/27
But the area of the triangle - is of three triangles treating each side as the base going to the centre of circle as the apex. So
0.5r( 25/9 + 20/9 + 5/3) = 0.5r (60/9) = 10r /3
Hence r = 15/27 = 5/9

[felixkolenda]

3:15 | Why is AP/AR=AB/BC?
Shouldn't it be AP/AR=AB/AC?
You wrote down the numbers correct but the formula is wrong, I think.

[davidpraesent]

If someone isn't sure about why the green and blue cirecle arcs intersect (and doesn't know the mentioned "exterior tangent result"):

Since the circle touches all three sides of the triangle, it is the incircle. This means the 3 lines connecting the circle corners to the circle center bisect the corresponding angles. Therefore we get two pairs of congruent triangles.

[marceeaax]

3:22 Why did you put BC as the denominator of the fraction?

[manucitomx]

Merry Christmas and thank you for a truly fun problem.

[MrBlueSkyMrNight]

@3:18  A mistake was made.  He meant to say and write:   AB over AC

[kevinmorgan]

Start with top small triangle with sides 3k, 4k, 5k with inscribed circle of rad r. You can quickly find k = r. Now add the bottom bit with rhombus. Then BC = 5r + 4r = 5 and answer follows immediately.

[DangRenBo]

@5:00 It might be easier to just multiply by the denominators on the left side of the equations instead of cross muktiplying, then set the right sides equal yo each other since both left sides would be 3-x.

[lemon-bcrv]

Merry christmas professor penn and thanks for the great video

[emersonschmidt]

It's Poncelet's theorem time... : AR/PR=CR/BC ; PR=L => (4-L)/L=4/5 then L=20/9. PR^2=AR^2+PA^2 or PA^2+(4-20/9)^2=(20/9)^2 -> PA=12/9 -> PB=3-PA=3-12/9; By Poncelet's theorem in a right triangle, PB+PQ=BQ+2*r then (3-12/9)+L=(5-L)+2*r therefore r=5/9

[Tiqerboy]

For the final part, instead of constructing those two arcs, I preferred to draw the radius perpendicular to each side of the triangle since each side is tangent to the inscribed circle. From this we can bisect each angle of the triangle and end up with 3 pairs of congruent triangles By congruent sides and simple arithmetic one gets the equation to solve for r as he has written.

[stephensu]

for finding the small r, i was thinking about what if we get another r line to the 25/9 side then get a line from the centre of circle to the angle between 20/9 and 25/9 and then this line cut the angle in half (because we have all the sides we knew that angle) then using tan of that half of angle is the ratio between r and (20/9) minus r, then you got the answer of r

[shahinjahanlu]

Marry Christmas dear professor. Have nice times with your family, be safe and life on your side

[charlesbromberick]

Nice job, Mike - I like problems like this.

[erichiseli]

Now call me complicated. Once we zoom into the triangle BPQ, I noted the square with length r. OK. As the center of the inscribed circle is found by dividing the angles in B and Q, that revealed me 4 more triangles, two pairs to be exact.

From there I decided to make an equation to find the sum of the area of those parts, equalling the area of the triangle BPQ:

r^2 + (20/9 -r)*r + (5/3 -r)*r = 5/3*20/9*1/2 which equals to 50/27
By simplifying, we find
27r^2 -105r + 50 = 0

solving by the quadratic formula, I found 2 solutions: 10/3 and 5/9. 10/3 is bigger than BA so it can be discarded as invalid, so we find the final answer 5/9.

[udic]

7:21 why not use the exterier tangent theorem?!
(You are mentioning it later to explain why the 2 circles meet at the same point)
It does exactly what you need without creating more circles.