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

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

(→‎Solution 2)
(→‎See also)
 
(7 intermediate revisions by 4 users not shown)
Line 25: Line 25:
  
 
== Solution 2 ==
 
== Solution 2 ==
−
 
−
Use Power of a point similar to the first solution to find that <math>AB = 30</math> and that the side <math>AC = 2 \cdot x \sqrt{2}</math>, where <math>x</math> is one third of the median's length. Then use systems of law of cosines, creating two triangles, with <math>10-10-3x</math> with angle <math>\theta</math>, and <math>10-20-2 \cdot x \sqrt{2}</math> with the same angle. Solving the system yields <math>x = 4 \sqrt{2}</math>. Solving using Heron's Formula gets the answer <math>24 \sqrt{14}</math>, or <math>\boxed{038}</math>.
 
−
 
−
 
−
 
−
Note:
 
−
 
−
This solution mentions "similar to the first solution find that <math>AB=30</math>" but the first solution never uses this value. In fact, it says <math>AB=30-2c</math> where it gets <math>c=2</math> or <math>c=10</math> and further mentions that <math>c = 10</math> as extraneous (it gives us a line). Thus it gets <math>c=2</math> and thus <math>AB=26</math>. Therefore absurd.
 
−
 
−
== Solution 3 ==
 
 
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  
 
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>  
 
<math>AG = DE = \sqrt{2}m </math> and  <math>AB = AG + GB = AG + BE = 10+2\sqrt{2} m</math>  
Line 47: Line 37:
  
 
-Alexlikemath
 
-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 ==
 
== See also ==

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

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=210147"