Difference between revisions of "1962 AHSME Problems/Problem 39"
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==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> | + | 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>. | ||
− | + | *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 | |
− | + | ~williamxiao |
Latest revision as of 11:48, 19 July 2023
Problem
Two medians of a triangle with unequal sides are inches and inches. Its area is square inches. The length of the third median in inches, is:
Solution
By the area formula: Where . Plugging in the numbers: Simplifying and squaring both sides: Now, we can just plug in the answer choices and find that 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 ratio. For convenience, call the orthocenter , and label the triangle such that the median from to is 3 and the median from to is 6. Then, is 4 and is 2. Let us call the portion of the third median that goes from to have length , and and have length . Note that the median from to in is equal to .
Note that the medians split the triangle into 6 triangles of equal area, so has area equal to of . Let . Using Herons*, we get:
We can see that , meaning that is 2 and . We can then use Steward's* to find the length of the median from , since we know the median cuts into segments each of length . We get:
Since the length of the actual median from is equal to , we have that the answer is .
- If anyone has a better method of either finding or the median of from , please feel free to edit
~williamxiao