Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2005 AIME I Problems/Problem 15

Revision as of 06:59, 12 January 2021 by Phoenixfire (talk | contribs) (→‎Solution 2)

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

Use Power of a point similar to the first solution to find that $AB = 30$ and that the side $AC = 2 \cdot x \sqrt{2}$, where $x$ is one third of the median's length. Then use systems of law of cosines, creating two triangles, with $10-10-3x$ with angle $\theta$, and $10-20-2 \cdot x \sqrt{2}$ with the same angle. Solving the system yields $x = 4 \sqrt{2}$. Solving using Heron's Formula gets the answer $24 \sqrt{14}$, or $\boxed{038}$.

Note:

This solution mentions "similar to the first solution find that $AB=30$" but the first solution never uses this value. In fact, it says $AB=30-2c$ where it gets $c=2$ or $c=10$ and further mentions that $c = 10$ as extraneous. Thus it gets $c=2$ and thus $AB=26$. Therefore absurd.

Solution 3

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

See also

2005 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 14 		Followed by
Last Question
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2005_AIME_I_Problems/Problem_15&oldid=141948"