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

Latest revision as of 22:15, 6 January 2024

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

@@ Line 1: / Line 1: @@
 == Problem ==
-Triangle <math> ABC </math> has <math> BC=20. </math> The incircle of the triangle evenly trisects the median <math> AD. </math> If the area of the triangle is <math> m \sqrt{n} </math> where <math> m </math> and <math> n </math> are integers and <math> n </math> is not divisible by the square of a prime, find <math> m+n. </math>
+Triangle <math> ABC </math> has <math> BC=20. </math> The [[incircle]] of the triangle evenly [[trisect]]s the [[median of a triangle | median]] <math> AD. </math> If the area of the triangle is <math> m \sqrt{n} </math> where <math> m </math> and <math> n </math> are integers and <math> n </math> is not [[divisor | divisible]] by the [[perfect square | square]] of a prime, find <math> m+n. </math>
-== Solution ==
+== Solution 1==
+<center><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></center><!-- Asymptote replacement for Image:2005_I_AIME-15.png by azjps -->
+Let <math>E</math>, <math>F</math> and <math>G</math> be the points of tangency of the incircle with <math>BC</math>, <math>AC</math> and <math>AB</math>, respectively.  Without loss of generality, let <math>AC < AB</math>, so that <math>E</math> is between <math>D</math> and <math>C</math>.  Let the length of the median be <math>3m</math>.  Then by two applications of the [[Power of a Point Theorem]], <math>DE^2 = 2m \cdot m = AF^2</math>, so <math>DE = AF</math>.  Now, <math>CE</math> and <math>CF</math> are two tangents to a circle from the same point, so by the Two Tangent Theorem <math>CE = CF = c</math> and thus <math>AC = AF + CF = DE + CE = CD = 10</math>.  Then <math>DE = AF = AG = 10 - c</math> so <math>BG = BE = BD + DE = 20 - c</math> and thus <math>AB = AG + BG = 30 - 2c</math>.
+Now, by [[Stewart's Theorem]] in triangle <math>\triangle ABC</math> with [[cevian]] <math>\overline{AD}</math>, we have
+<cmath>(3m)^2\cdot 20 + 20\cdot10\cdot10 = 10^2\cdot10 + (30 - 2c)^2\cdot 10.</cmath>
+Our earlier result from Power of a Point was that <math>2m^2 = (10 - c)^2</math>, so we combine these two results to solve for <math>c</math> and we get
+<cmath>9(10 - c)^2 + 200 = 100 + (30 - 2c)^2 \quad \Longrightarrow \quad c^2 - 12c + 20 = 0.</cmath>
+Thus <math>c = 2</math> or <math> = 10</math>.  We discard the value <math>c = 10</math> as extraneous (it gives us a line) and are left with <math>c = 2</math>, so our triangle has area <math>\sqrt{28 \cdot 18 \cdot 8 \cdot 2} = 24\sqrt{14}</math> and so the answer is <math>24 + 14 = \boxed{038}</math>.
+== Solution 2 ==
+WLOG let E be be between C & D (as in solution 1). Assume <math>AD = 3m</math>. We use power of a point to get that
+<math>AG = DE = \sqrt{2}m </math> and  <math>AB = AG + GB = AG + BE = 10+2\sqrt{2} m</math>
+Since now we have <math>AC = 10</math>, <math>BC = 20, AB = 10+2\sqrt{2} m </math> in triangle <math>\triangle ABC</math> and [[cevian]] <math>AD = 3m</math>.  Now, we can apply [[Stewart's Theorem]].
+<cmath>2000 + 180 m^2 = 10(10+2\sqrt{2}m)^{2} + 1000</cmath>
+<cmath>1000 + 180 m^2 = 1000 + 400\sqrt{2}m + 80 m^{2}</cmath>
+<cmath>100 m^2 = 400\sqrt{2}m</cmath>
+<math>m = 4\sqrt{2}</math> or <math>m = 0</math> if <math>m = 0</math>, we get a degenerate triangle, so <math>m = 4\sqrt{2}</math>, and thus <math>AB = 26</math>. You can now use [[Heron's Formula]] to finish. The answer is <math>24 \sqrt{14}</math>, or <math>\boxed{038}</math>.
+-Alexlikemath
+== Solution 3 ==
+Let <math>E, F</math>, and <math>G</math> be the point of tangency (as stated in Solution 1). We can now let <math>AD</math> be <math>3m</math>. By using [[Power of a Point Theorem]] on A to the incircle, you get that <math>AG^2 = 2m^2</math>. We can use it again on point D to the incircle to get the equation <math>(10 - CE)^2 = 2m^2</math>. Setting the two equations equal to each other gives <math>(10 - CE)^2 = AG^2</math>, and it can be further simplified to be <math>10 - CE = AG</math>
+Let lengths <math>AC</math> and <math>AB</math> be called <math>b</math> and <math>c</math>, respectively. We can write <math>AG</math> as <math>\frac{b + c - 20}{2}</math> and <math>CE</math> as <math>\frac{b + 20 - c}{2}</math>. 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 \rightarrow b = 10</cmath>
+Additionally, by [[Median of a triangle]] formula, you get that <math>3m = \frac{\sqrt{2c^2 - 200}}{2}</math>
+Refer back to the fact that <math>AG^2 = 2m^2</math>. We can now plug in our variables.
+<cmath>\left(\frac{b + c - 20}{2}\right)^2 = 2m^2 \rightarrow (c - 10)^2 = 8m^2</cmath>
+<cmath>c^2 - 20c + 100 = 8 \cdot \frac{2c^2 - 200}{36}</cmath>
+<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 <math>c = 26</math> or <math>10</math>, but the latter will result in a degenerate triangle, so <math>c = 26</math>.
+Finally, you can use [[Heron's Formula]] to get that the area is <math>24\sqrt{14}</math>, giving an answer of <math>\boxed{038}</math>
+~sky2025
 == See also ==
-* [[2005 AIME I Problems]]
+{{AIME box|year=2005|n=I|num-b=14|after=Last Question}}
+[[Category:Intermediate Geometry Problems]]
+{{MAA Notice}}

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