2006 AMC 12B Problems/Problem 23
Problem
Isosceles has a right angle at . Point is inside , such that , , and . Legs and have length , where and are positive integers. What is ?
Solution
Using the Law of Cosines on , we have:
Using the Law of Cosines on , we have:
Now we use .
Note that we know that we want the solution with since we know that . Thus, .
Solution 2
Rotate triangle 90 degrees counterclockwise about so that the image of rests on . Now let the image of be . Note that , meaning triangle is right isosceles, and . Then . Now because and , we observe that , by the Pythagorean Theorem on . Now we have that . So we take the cosine of the second equality, using that fact that , to get . Finally, we use the fact that and use the Law of Cosines on triangle to arrive at the value of .
Or notice that since and , we have , and Law of Cosines on triangle gives the value of .
Solution 3 (coordinate bash)
Let point have coordinates and have coordinates Then, has and has .
By distance formula, we have
Expanding and gives and respectively. Then, using equation , we have that and
Then, solving for and gives and Plugging these values of and into equation yields
We multiply both sides by and expand, yielding the equation Simplifying gives the equation Solving this quadratic gives Now, if this were the actual test, we stop here, noting that the question tells us and are positive, so must be , and our solution is .
However, here is why cannot be :
If , using as an approximation for , , and is slightly greater than . Also note that this implies that .
Note that at least one of , and must be obtuse, since they sum to . Then, note the well known fact that if is the largest angle in , must be the largest side. However, combined with the first fact, implies that either is the largest side of , is the largest side of , or is the largest side of . By our approximations, this cannot possibly be true, even if we are generous with our margin of error, so cannot equal , and .
The answer is
See also
2006 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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All AMC 12 Problems and Solutions |
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