At least there's a diagram

Geometry Level 3

In triangle A B C ABC we have A B = 36 AB=36 , B C = 48 BC=48 , C A = 60 CA=60 . The incircle of A B C ABC is centered at I I and touches A B AB , A C AC , B C BC at M M , N N , D D , respectively. Ray A I AI meets B C BC at K K . The radical axis of the circumcircles of triangles M A N MAN and K I D KID intersects lines A B AB and A C AC at L 1 L_1 and L 2 L_2 , respectively. If L 1 L 2 = x L_1L_2 = x , compute x 2 x^2 .


The answer is 720.

Sharky Kesa by Sharky Kesa , 19, United Kingdom

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4 solutions

Inradius = 12 12 ; A M = 24 ; A I = 720 ; △ A M I ∼ △ A I L 1 ; AM = 24 ; AI = \sqrt {720} ; \triangle AMI \sim \triangle AIL_1 = > I L 1 A I = I M A M ⇒ I L 1 = 12 24 × 720 ⇒ I L 1 = 720 2 ⇒ L 1 L 2 = 720 = > x 2 = 720 => \dfrac{IL_1}{AI}=\dfrac{IM}{AM} \Rightarrow IL_1 = \dfrac{12}{24} \times \sqrt {720} \Rightarrow IL_1 =\dfrac{\sqrt {720}}{2} \Rightarrow L_1L_2 = \sqrt {720} => x^{2} = 720

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Ahmad Saad
Aug 23, 2017

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Marta Reece
Aug 21, 2017

Shrink everything by a factor of 12, and you have a right triangle ABC with sides 3,4,and 5.

∠ B A C = arctan ⁡ 4 3 = 53.1 3 ∘ \angle BAC=\arctan\frac43=53.13^\circ

∠ B A I = ∠ B A C 2 = 26.5 7 ∘ \angle BAI=\dfrac{\angle BAC}2=26.57^\circ

Radius of the incircle = M I = 1 =MI=1 .

A I = 1 sin ⁡ 26.5 7 ∘ = 2.236 AI=\dfrac1{\sin26.57^\circ}=2.236

I L 1 = A I × tan ⁡ 26 , 5 7 ∘ = 1.118 IL_1=AI\times\tan26,57^\circ=1.118

L 1 L 2 = 2 × L 1 = 2.236 L_1L_2=2\times L_1=2.236

Multiply everything by 12 to undo the shrinking at the start

Actual L 1 L 2 = 2.236 × 12 = 26.83 L_1L_2=2.236\times12=26.83

Square the result to get final answer = 26.8 3 2 = 720 =26.83^2=\boxed{720}

(The answer comes out nicely since I, of course, carry the full precision available rather than just the few decimal places reported.)

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How do you know that the answer is EXACTLY 720?

Pi Han Goh - 3 years, 9 months ago

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That came off the calculator. As I said, all I did was carry the available precision. I could have struggled through with the exact figures and all the square roots etc. I did not. So technically speaking I do not know that the figure is absolutely accurate. Whether I know this or not, however, it is the answer I am getting so I give it as the answer. As it obviously came from rounded figures, there is no claim of accuracy involved in doing so.

Marta Reece - 3 years, 9 months ago

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You need not shrink or expand or do anything with the figure. The given data is perfect which gives everything in the terms of whole numbers.

Inradius = 12 12 ; A M = 24 ; A I = 720 ; △ A M I ∼ △ A I L 1 ; AM = 24 ; AI = \sqrt {720} ; \triangle AMI \sim \triangle AIL_1 = > I L 1 A I = I M A M ⇒ I L 1 = 12 24 × 720 ⇒ I L 1 = 720 2 ⇒ L 1 L 2 = 720 = > x 2 = 720 => \dfrac{IL_1}{AI}=\dfrac{IM}{AM} \Rightarrow IL_1 = \dfrac{12}{24} \times \sqrt {720} \Rightarrow IL_1 =\dfrac{\sqrt {720}}{2} \Rightarrow L_1L_2 = \sqrt {720} => x^{2} = 720

Vishwash Kumar ΓΞΩ - 3 years, 9 months ago

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@Vishwash Kumar Γξω – I knew that I did not need to shrink or expand. I did it to make it easier to follow, with a familiar triangle. I see now that it would have worked better the other way.

Marta Reece - 3 years, 9 months ago

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