1992 AHSME Problems/Problem 25
Contents
Problem
In , and . If perpendiculars constructed to at and to at meet at , then
Solution 1 (Extending Line Segments)
We begin by drawing a diagram. We extend and to meet at This gives us a couple right triangles in and We see that . Hence, and are 30-60-90 triangles.
Using the side ratios of 30-60-90 triangles, we have . This tells us that . Also, .
Because , we have Solving the equation, we have Hence, .
Solution 2 (Cyclic Quadrilaterals, Right Triangles, LoC)
Since is cyclic. Using Ptolemy's Theorem gets
Right triangles and obtain and respectively.
Seeing squares in and , we square and get
We don't like that term, but fortunately LoC exists: . Solving for and plugging it into , and using and from the first two equations, gets
Solve for .
~PureSwag
See also
1992 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Problem 26 | |
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