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Difference between revisions of "1962 AHSME Problems/Problem 39"

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Now, we can just plug in the answer choices and find that <math>\boxed{3\sqrt6}</math> works.
 
Now, we can just plug in the answer choices and find that <math>\boxed{3\sqrt6}</math> works.
  
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Solution 2
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==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>.  
 
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>.  

Latest revision as of 11:48, 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

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