# 2005 AIME I Problems/Problem 15

(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

## Problem

Triangle $ABC$ has $BC=20.$ The incircle of the triangle evenly trisects the median $AD.$ If the area of the triangle is $m \sqrt{n}$ where $m$ and $n$ are integers and $n$ is not divisible by the square of a prime, find $m+n.$

## Solution 1 $[asy] size(300); pointpen=black;pathpen=black+linewidth(0.65); pen s = fontsize(10); pair A=(0,0),B=(26,0),C=IP(circle(A,10),circle(B,20)),D=(B+C)/2,I=incenter(A,B,C); path cir = incircle(A,B,C); pair E1=IP(cir,B--C),F=IP(cir,A--C),G=IP(cir,A--B),P=IP(A--D,cir),Q=OP(A--D,cir); D(MP("A",A,s)--MP("B",B,s)--MP("C",C,N,s)--cycle); D(cir); D(A--MP("D",D,NE,s)); D(MP("E",E1,NE,s)); D(MP("F",F,NW,s)); D(MP("G",G,s)); D(MP("P",P,SW,s)); D(MP("Q",Q,SE,s)); MP("10",(B+D)/2,NE); MP("10",(C+D)/2,NE); [/asy]$

Let $E$, $F$ and $G$ be the points of tangency of the incircle with $BC$, $AC$ and $AB$, respectively. Without loss of generality, let $AC < AB$, so that $E$ is between $D$ and $C$. Let the length of the median be $3m$. Then by two applications of the Power of a Point Theorem, $DE^2 = 2m \cdot m = AF^2$, so $DE = AF$. Now, $CE$ and $CF$ are two tangents to a circle from the same point, so by the Two Tangent Theorem $CE = CF = c$ and thus $AC = AF + CF = DE + CE = CD = 10$. Then $DE = AF = AG = 10 - c$ so $BG = BE = BD + DE = 20 - c$ and thus $AB = AG + BG = 30 - 2c$.

Now, by Stewart's Theorem in triangle $\triangle ABC$ with cevian $\overline{AD}$, we have $$(3m)^2\cdot 20 + 20\cdot10\cdot10 = 10^2\cdot10 + (30 - 2c)^2\cdot 10.$$

Our earlier result from Power of a Point was that $2m^2 = (10 - c)^2$, so we combine these two results to solve for $c$ and we get $$9(10 - c)^2 + 200 = 100 + (30 - 2c)^2 \quad \Longrightarrow \quad c^2 - 12c + 20 = 0.$$

Thus $c = 2$ or $= 10$. We discard the value $c = 10$ as extraneous (it gives us a line) and are left with $c = 2$, so our triangle has area $\sqrt{28 \cdot 18 \cdot 8 \cdot 2} = 24\sqrt{14}$ and so the answer is $24 + 14 = \boxed{038}$.

## Solution 2

WLOG let E be be between C & D (as in solution 1). Assume $AD = 3m$. We use power of a point to get that $AG = DE = \sqrt{2}m$ and $AB = AG + GB = AG + BE = 10+2\sqrt{2} m$

Since now we have $AC = 10$, $BC = 20, AB = 10+2\sqrt{2} m$ in triangle $\triangle ABC$ and cevian $AD = 3m$. Now, we can apply Stewart's Theorem. $$2000 + 180 m^2 = 10(10+2\sqrt{2}m)^{2} + 1000$$ $$1000 + 180 m^2 = 1000 + 400\sqrt{2}m + 80 m^{2}$$ $$100 m^2 = 400\sqrt{2}m$$ $m = 4\sqrt{2}$ or $m = 0$ if $m = 0$, we get a degenerate triangle, so $m = 4\sqrt{2}$, and thus $AB = 26$. You can now use Heron's Formula to finish. The answer is $24 \sqrt{14}$, or $\boxed{038}$.

-Alexlikemath

## Solution 3

Let $E, F$, and $G$ be the point of tangency (as stated in Solution 1). We can now let $AD$ be $3m$. By using Power of a Point Theorem on A to the incircle, you get that $AG^2 = 2m^2$. We can use it again on point D to the incircle to get the equation $(10 - CE)^2 = 2m^2$. Setting the two equations equal to each other gives $(10 - CE)^2 = AG^2$, and it can be further simplified to be $10 - CE = AG$

Let lengths $AC$ and $AB$ be called $b$ and $c$, respectively. We can write $AG$ as $\frac{b + c - 20}{2}$ and $CE$ as $\frac{b + 20 - c}{2}$. Plugging these into the equation, you get: $$10 - \frac{b + 20 - c}{2} = \frac{b + c - 20}{2}$$ $$b + c - 20 + b + 20 - c = 20 \rightarrow b = 10$$

Additionally, by Median of a triangle formula, you get that $3m = \frac{\sqrt{2c^2 - 200}}{2}$

Refer back to the fact that $AG^2 = 2m^2$. We can now plug in our variables. $$\left(\frac{b + c - 20}{2}\right)^2 = 2m^2 \rightarrow (c - 10)^2 = 8m^2$$ $$c^2 - 20c + 100 = 8 \cdot \frac{2c^2 - 200}{36}$$ $$9c^2 - 180c + 900 = 4c^2 - 400$$ $$5c^2 - 180c + 1300 = 0$$ $$c^2 - 36c + 260 = 0$$

Solving, you get that $c = 26$ or $10$, but the latter will result in a degenerate triangle, so $c = 26$. Finally, you can use Heron's Formula to get that the area is $24\sqrt{14}$, giving an answer of $\boxed{038}$

~sky2025

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 