Difference between revisions of "1992 AHSME Problems/Problem 25"
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\text{(E) } \frac{10}{\sqrt{3}}</math> | \text{(E) } \frac{10}{\sqrt{3}}</math> | ||
− | == Solution == | + | == Solution 1 (Extending Line Segments) == |
We begin by drawing a diagram. | We begin by drawing a diagram. | ||
<asy> | <asy> | ||
Line 67: | Line 67: | ||
\end{align*}</cmath> | \end{align*}</cmath> | ||
Hence, <math>CD=\boxed{\textbf{E}}</math>. | Hence, <math>CD=\boxed{\textbf{E}}</math>. | ||
+ | |||
+ | == Solution 2 (Cyclic Quadrilaterals, Right Triangles, LoC) == | ||
+ | |||
+ | Since <math>\angle{A}+\angle{C} = 180^{\circ}, ABCD</math> is cyclic. Using Ptolemy's Theorem gets <math>\text{(1)} 4AD + 3CD = \sqrt{37}BD.</math> | ||
+ | |||
+ | Right triangles <math>\Delta ABD</math> and <math>\Delta CBD</math> obtain <math>\text{(2)} 9+AD^2=BD^2</math> and <math>\text{(3)} 16+CD^2=BD^2,</math> respectively. | ||
+ | |||
+ | Seeing squares in <math>\text{(2)}</math> and <math>\text{(3)}</math>, we square <math>\text{(1)}</math> and get <math>\text{(4)} 16AD^2+9CD^2 + 24AD\cdot CD = 37BD^2.</math> | ||
+ | |||
+ | We don't like that <math>AD\cdot CD</math> term, but fortunately LoC exists: <math>37 = AD^2 + CD^2 - 2*AD*CD*\cos(60^{\circ})</math>. Solving for <math>AD\cdot CD</math> and plugging it into <math>\text{(4)}</math>, and using <math>AD^2 = 7 + CD^2</math> and <math>BD^2 = 16 + CD^2</math> from the first two equations, gets <math>40(CD^2+7)+33CD^2 - 24\cdot 37 = 37(16+CD^2).</math> | ||
+ | |||
+ | Solve for <math>CD = \boxed{\textbf{E}}</math>. | ||
+ | |||
+ | ~PureSwag | ||
== See also == | == See also == |
Revision as of 15:22, 24 October 2022
Contents
[hide]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 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.