Difference between revisions of "2005 AIME I Problems/Problem 15"

Revision as of 18:21, 16 February 2021

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:

$10 - \frac{b + 20 - c}{2} = \frac{b + c - 20}{2}$ $b + c - 20 + b + 20 - c = 20$ $b = 10$

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.

$\left(\frac{b + c - 20}{2}\right)^2 = 2m^2$ $(c - 10)^2 = 8m^2$ $c^2 - 20c + 100 = 8 \cdot \frac{2c^2 - 200}{36}$ 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}$$ (Error compiling LaTeX. Unknown error_msg)

@@ Line 37: / Line 37: @@
 -Alexlikemath
+== Solution 3 ==
+Let <cmath>E, F</cmath>, and <cmath>G</cmath> be the point of tangency (as stated in Solution 1). We can now let <cmath>AD</cmath> be <cmath>3m</cmath>. By using [[Power of a Point Theorem]] on A to the incircle, you get that <cmath>AG^2 = 2m^2</cmath>. We can use it again on point D to the incircle to get the equation <cmath>(10 - CE)^2 = 2m^2</cmath>. Setting the two equations equal to each other gives <cmath>(10 - CE)^2 = AG^2</cmath>, and it can be further simplified to be <cmath>10 - CE = AG</cmath>
+Let lengths <cmath>AC</cmath> and <cmath>AB</cmath> be called <cmath>b</cmath> and <cmath>c</cmath>, respectively. We can write <cmath>AG</cmath> as <cmath>\frac{b + c - 20}{2}</cmath> and <cmath>CE</cmath> as <cmath>\frac{b + 20 - c}{2}</cmath>. Plugging these into the equation, you get:
+<cmath>10 - \frac{b + 20 - c}{2} = \frac{b + c - 20}{2}</cmath>
+<cmath>b + c - 20 + b + 20 - c = 20</cmath>
+<cmath>b = 10</cmath>
+Additionally, by [[Median of a triangle]] formula, you get that <cmath>3m = \frac{\sqrt{2c^2 - 200}}{2}</cmath>
+Refer back to the fact that <cmath>AG^2 = 2m^2</cmath>. We can now plug in our variables.
+<cmath>\left(\frac{b + c - 20}{2}\right)^2 = 2m^2</cmath>
+<cmath>(c - 10)^2 = 8m^2</cmath>
+<cmath>c^2 - 20c + 100 = 8 \cdot \frac{2c^2 - 200}{36}
+</cmath>9c^2 - 180c + 900 = 4c^2 - 400<cmath>
+</cmath>5c^2 - 180c + 1300 = 0<cmath>
+</cmath>c^2 - 36c + 260 = 0<cmath>
+Solving, you get that </cmath>c = 26<cmath> or </cmath>10<cmath>, 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 </cmath>24\sqrt{14}<cmath>, giving an answer of </cmath>\boxed{038}<math></math>
 == 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

Page

Toolbox

Search