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 "1962 AHSME Problems/Problem 39"

(→‎Solution)
(→‎Solution)
 
(4 intermediate revisions by 2 users not shown)
Line 1: Line 1:
 
==Problem==
 
==Problem==
−
Two medians of a triangle with unequal sides are <math>3</math> inches and <math>6</math> inches. Its area is <math>3 \sqrt{15}</math> square inches. The length of the third median in inches, is:  
+
Two medians of a triangle with unequal sides are <math>3</math> inches and <math>6</math> inches. Its area is <math>3 \sqrt{15}</math> square inches. The length of the third median in inches, is:
 +
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{6}\qquad\textbf{(D)}\ 6\sqrt{3}\qquad\textbf{(E)}\ 6\sqrt{6} </math>
 +
==Solution==
 +
By the area formula:
 +
<cmath>A = \frac43\sqrt{s(s-m_1)(s-m_2)(s-m_3)}</cmath>
 +
Where <math>s = \frac{m_1+m_2+m_3}{2}</math>.
 +
Plugging in the numbers:
 +
<cmath>3\sqrt{15} = \frac43\sqrt{\frac{9+m}2\cdot\frac{9-m}2\cdot\frac{m+3}2\cdot\frac{m-3}2}</cmath>
 +
Simplifying and squaring both sides:
 +
<cmath>1215 = (9-m)(9+m)(m+3)(m-3)</cmath>
 +
Now, we can just plug in the answer choices and find that <math>\boxed{3\sqrt6}</math> works.
 +
 
 +
==Solution 2==
 +
 
 +
We connect all the medians of the triangle. We know that the ratio of the vertice to orthocenter / median is in a <math>2:3</math> ratio. For convenience, call the orthocenter <math>O</math>, and label the triangle <math>\triangle ABC</math> such that the median from <math>A</math> to <math>BC</math> is 3 and the median from <math>B</math> to <math>AC</math> is 6. Then, <math>BO</math> is 4 and <math>AO</math> is 2. Let us call the portion of the third median that goes from <math>O</math> to <math>AB</math> have length <math>x</math>, and and <math>OC</math> have length <math>2x</math>. Note that the median from <math>O</math> to <math>AB</math> in <math>\triangle AOB</math> is equal to <math>x</math>.
 +
 
 +
Note that the medians split the triangle into 6 triangles of equal area, so <math>\triangle AOB</math> has area equal to <math>\frac{1}{3}</math> of <math>\triangle ABC = \sqrt{15}</math>. Let <math>AB=2s</math>. Using Herons*, we get:
 +
 
 +
<cmath>15=(3+s)(s-1)(s+1)(3-2)</cmath>
 +
<cmath>=(9-s^2)(s^2-1)</cmath>
  
−
<math> \textbf{(A)}\ 4\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{6}\qquad\textbf{(D)}\ 6\sqrt{3}\qquad\textbf{(E)}\ 6\sqrt{6} </math>
+
We can see that <math>s^2=4</math>, meaning that <math>s</math> is 2 and <math>AB=4</math>. We can then use Steward's* to find the length of the median from <math>O</math>, since we know the median cuts <math>AB</math> into segments each of length <math>2</math>. We get:
 +
 
 +
<cmath>(2^2)(4)+4x^2=(4^2)(2)+(2^2)(2)</cmath>
 +
<cmath>16+4x^2=32+8</cmath>
 +
<cmath>4x^2=24</cmath>
 +
<cmath>x^2=6</cmath>
 +
<cmath>x=\sqrt{6}</cmath>
 +
 
 +
Since the length of the actual median from <math>C</math> is equal to <math>3</math>, we have that the answer is <math>\boxed{3\sqrt{6}}</math>.
  
−
==Solution==
+
*If anyone has a better method of either finding <math>AB</math> or the median of <math>AOB</math> from <math>O</math>, please feel free to edit
−
{{solution}}
+
~williamxiao

Latest revision as of 12:48, 19 July 2023

Problem

Two medians of a triangle with unequal sides are $3$ inches and $6$ inches. Its area is $3 \sqrt{15}$ square inches. The length of the third median in inches, is: $\textbf{(A)}\ 4\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 3\sqrt{6}\qquad\textbf{(D)}\ 6\sqrt{3}\qquad\textbf{(E)}\ 6\sqrt{6}$

Solution

By the area formula: \[A = \frac43\sqrt{s(s-m_1)(s-m_2)(s-m_3)}\] Where $s = \frac{m_1+m_2+m_3}{2}$. Plugging in the numbers: \[3\sqrt{15} = \frac43\sqrt{\frac{9+m}2\cdot\frac{9-m}2\cdot\frac{m+3}2\cdot\frac{m-3}2}\] Simplifying and squaring both sides: \[1215 = (9-m)(9+m)(m+3)(m-3)\] Now, we can just plug in the answer choices and find that $\boxed{3\sqrt6}$ works.

Solution 2

We connect all the medians of the triangle. We know that the ratio of the vertice to orthocenter / median is in a $2:3$ ratio. For convenience, call the orthocenter $O$, and label the triangle $\triangle ABC$ such that the median from $A$ to $BC$ is 3 and the median from $B$ to $AC$ is 6. Then, $BO$ is 4 and $AO$ is 2. Let us call the portion of the third median that goes from $O$ to $AB$ have length $x$, and and $OC$ have length $2x$. Note that the median from $O$ to $AB$ in $\triangle AOB$ is equal to $x$.

Note that the medians split the triangle into 6 triangles of equal area, so $\triangle AOB$ has area equal to $\frac{1}{3}$ of $\triangle ABC = \sqrt{15}$. Let $AB=2s$. Using Herons*, we get:

\[15=(3+s)(s-1)(s+1)(3-2)\] \[=(9-s^2)(s^2-1)\]

We can see that $s^2=4$, meaning that $s$ is 2 and $AB=4$. We can then use Steward's* to find the length of the median from $O$, since we know the median cuts $AB$ into segments each of length $2$. We get:

\[(2^2)(4)+4x^2=(4^2)(2)+(2^2)(2)\] \[16+4x^2=32+8\] \[4x^2=24\] \[x^2=6\] \[x=\sqrt{6}\]

Since the length of the actual median from $C$ is equal to $3$, we have that the answer is $\boxed{3\sqrt{6}}$.

  • If anyone has a better method of either finding $AB$ or the median of $AOB$ from $O$, please feel free to edit

~williamxiao

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1962_AHSME_Problems/Problem_39&oldid=195889"