2019 AMC 12A Problems/Problem 25
Contents
Problem
Let be a triangle whose angle measures are exactly , , and . For each positive integer , define to be the foot of the altitude from to line . Likewise, define to be the foot of the altitude from to line , and to be the foot of the altitude from to line . What is the least positive integer for which is obtuse?
Solution 1
For all nonnegative integers , let , , and .
Note that quadrilateral is cyclic since ; thus, . By a similar argument, . Thus, . By a similar argument, and .
Therefore, for any positive integer , we have (identical recurrence relations can be derived for and ). To find an explicit form for this recurrence, we guess that the constant term is related exponentially to (and the coefficient of is ). Hence, we let . We will solve for , , and by iterating the recurrence to obtain , , and . Letting respectively, we have
Subtracting from , we have , and subtracting from gives . Dividing these two equations gives , so . Substituting back, we get and .
We will now prove that for all positive integers , via induction. Clearly the base case of holds, so it is left to prove that assuming our inductive hypothesis holds for . Using the recurrence relation, we have
Our induction is complete, so for all positive integers , . Identical equalities hold for and .
The problem asks for the smallest such that either , , or is greater than . WLOG, let , , and . Thus, for all , , and . Solving for the smallest possible value of in each sequence, we find that gives . Therefore, the answer is .
Solution 2
We start from Solution 1 until we reach the recurrence relation Iterate this again, to get Subtract the two, getting This recurrence has characteristic equation Now, write We obtain similar recursions for that can be easily solved by getting from the original recursive formula and then using those two values to solve for and Then proceed with the last paragraph of Solution 1.
Solution 3 (If you're short on time)
We note that the problem seems quite complicated, but since it is an AMC 12, the difference between the largest angle of and (we call this quantity S) most likely reduces to a simpler problem like some repeating sequence. The only obvious sequence (for the answer choices) is a geometric sequence with an integer common ratio. From here, we get that and we need . The first power of two greater than is thus our answer is .
~Dhillonr25
Video Solution by Richard Rusczyk
https://artofproblemsolving.com/videos/amc/2019amc12a/497
~ dolphin7
See Also
2019 AMC 12A (Problems • Answer Key • Resources) | |
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