Maths Olympiad Prep

Track / Stage 5 / 288 of 400 #888 of 1964

Problem 888

AIME late
Geometry Difficulty 5.7 Find the answer

## Task A-3.3.

Let A,B,CA^{\prime}, B^{\prime}, C^{\prime} be the points where the angle bisectors of triangle ABCA B C intersect the opposite sides BC,CA,AB\overline{B C}, \overline{C A}, \overline{A B} respectively, and let SS be the center of the inscribed circle of triangle ABCA B C.

If AS:SA=3:2,BS:SB=4:3|A S|:\left|S A^{\prime}\right|=3: 2,|B S|:\left|S B^{\prime}\right|=4: 3 and if AB=12|A B|=12, determine the lengths of the other sides of the triangle.

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

## Solution.

Let a,ba, b and cc be the lengths of the sides BC,AC\overline{B C}, \overline{A C} and AB\overline{A B}, respectively.

!

Using the Angle Bisector Theorem for triangle ABAA B A^{\prime}, we get

AB:AB=AS:AS=3:2|A B|:\left|A^{\prime} B\right|=|A S|:\left|A^{\prime} S\right|=3: 2, which implies AB=2c3\left|A^{\prime} B\right|=\frac{2 c}{3}.

Similarly, for triangle ABBA B B^{\prime}, we get AB:AB=BS:BS=4:3|A B|:\left|A B^{\prime}\right|=|B S|:\left|B^{\prime} S\right|=4: 3,

which implies AB=3c4\left|A B^{\prime}\right|=\frac{3 c}{4}.

For triangle ACAA C A^{\prime}, we get AC:AC=AS:AS=3:2|A C|:\left|A^{\prime} C\right|=|A S|:\left|A^{\prime} S\right|=3: 2,

which implies b:(a2c3)=3:2b:\left(a-\frac{2 c}{3}\right)=3: 2, hence 3a2c=2b3 a-2 c=2 b.

Similarly, for triangle BCBB C B^{\prime}, we have BC:BC=BS:BS=4:3|B C|:\left|B^{\prime} C\right|=|B S|:\left|B^{\prime} S\right|=4: 3, which implies a:(b3c4)=4:3a:\left(b-\frac{3 c}{4}\right)=4: 3, or 4b3c=3a4 b-3 c=3 a.

Considering c=12c=12, solving the system

3a2c=2b4b3c=3a \begin{aligned} & 3 a-2 c=2 b \\ & 4 b-3 c=3 a \end{aligned}

we get a=28,b=30a=28, b=30.

(2 points)

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.