Congruent AND Concurrent

Geometry Level 4

A B C \triangle ABC is a 13 13 - 14 14 - 15 15 triangle in which three congruent circles are each tangent to two sides of the triangle and are concurrent at Z Z . What is their radius? Express it as p q \dfrac{p}{q} , where p p and q q are coprime positive integers and submit p + q p+q .


The answer is 357.

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1 solution

David Vreken
Apr 9, 2021

Let D D , E E , and F F be the centers of the three congruent circles with radius r r , and draw in and label the diagram as follows:

By Heron's formula, the area of A B C \triangle ABC is A A B C = s ( s a ) ( s b ) ( s c ) = 21 ( 21 13 ) ( 21 14 ) ( 21 15 ) = 84 A_{\triangle ABC} = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{21(21 - 13)(21 - 14)(21 - 15)} = 84 .

Since each circle is tangent to the sides of A B C \triangle ABC , the centers D D , E E , and F F are on the angle bisectors of A A , B B , and C C .

A 13 13 - 14 14 - 15 15 triangle is a 5 5 - 12 12 - 13 13 triangle combined with a 9 9 - 12 12 - 15 15 , so A G = D G cot D A G = r cot ( 1 2 C A B ) = r cot ( 1 2 cos 1 ( 5 13 ) ) = 3 2 r AG = DG \cot \angle DAG = r \cot (\frac{1}{2} \angle CAB) = r \cot (\frac{1}{2} \cos^{-1} (\frac{5}{13})) = \frac{3}{2}r and H B = E H cot E B H = r cot ( 1 2 A B C ) = r cot ( 1 2 cos 1 ( 9 15 ) ) = 2 r HB = EH \cot \angle EBH = r \cot (\frac{1}{2} \angle ABC) = r \cot (\frac{1}{2} \cos^{-1} (\frac{9}{15})) = 2r , so that D E = G H = A B A G H B = 14 3 2 r 2 r = 14 7 2 r DE = GH = AB - AG - HB = 14 - \frac{3}{2}r - 2r = 14 - \frac{7}{2}r .

Since the sides of D E F \triangle DEF are parallel to the sides of A B C \triangle ABC , the two triangles are similar, and have a side ratio of D E A B = 14 7 2 r 14 = 1 1 4 r \cfrac{DE}{AB} = \cfrac{14 - \frac{7}{2}r}{14} = 1 - \frac{1}{4}r .

Since Z D = Z E = Z F = r ZD = ZE = ZF = r , Z Z is the circumcenter of D E F \triangle DEF , so 2 r = D E E F D F 2 A D E F = ( 1 1 4 r ) A B ( 1 1 4 r ) B C ( 1 1 4 r ) A C ( 1 1 4 r ) 2 A A B C = 14 15 13 ( 1 1 4 r ) 84 2r = \cfrac{DE \cdot EF \cdot DF}{2 A_{\triangle DEF}} = \cfrac{(1 - \frac{1}{4}r)AB \cdot (1 - \frac{1}{4}r)BC \cdot (1 - \frac{1}{4}r)AC}{(1 - \frac{1}{4}r)^2 A_{\triangle ABC}} = \cfrac{14 \cdot 15 \cdot 13 \cdot (1 - \frac{1}{4}r)}{84} , and 2 r = 14 15 13 ( 1 1 4 r ) 84 2r = \cfrac{14 \cdot 15 \cdot 13 \cdot (1 - \frac{1}{4}r)}{84} solves to r = 260 97 r = \cfrac{260}{97} .

Therefore, p = 260 p = 260 , q = 97 q = 97 , and p + q = 357 p + q = \boxed{357} .

Very nicely done!

Fletcher Mattox - 2 months ago

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Thank you so much!

David Vreken - 2 months ago

I think, p+q=260+81=341

Arafat Anik - 3 weeks ago

Who knows? You may have found a second solution. Hard to say without any evidence. Have you any?

Fletcher Mattox - 3 weeks ago

(14)(15)(13)(1-0.25r)/84=2r =>(2730)(1-0.25r)/84=2r =>(2730)(1-0.25r)=168r =>2730-682.5r=168r =>2730=850.5r

r=260/81

therefore, p=260, q=81 and p+q=260+81=341

Arafat Anik - 2 weeks, 5 days ago

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How do you justify your initial formula? I do not recognize it.

Fletcher Mattox - 2 weeks, 5 days ago

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