1962 AHSME Problems/Problem 39
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:
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.
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