Difference between revisions of "1992 AHSME Problems/Problem 25"
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== Solution == | == Solution == | ||
− | <math>\ | + | We begin by drawing a diagram. |
+ | <asy> | ||
+ | import olympiad; import cse5; import geometry; size(150); | ||
+ | defaultpen(fontsize(10pt)); | ||
+ | defaultpen(0.8); | ||
+ | dotfactor = 4; | ||
+ | pair A = origin; | ||
+ | pair C = A+dir(55); | ||
+ | pair D = A+dir(0); | ||
+ | pair B = extension(A,A+dir(90),C,C+dir(-155)); | ||
+ | label("$A$",A,S); | ||
+ | label("$C$",C,NE); | ||
+ | label("$D$",D,SE); | ||
+ | label("$B$",B,NW); | ||
+ | label("$4$",B--C,NW); | ||
+ | label("$3$",A--B,W); | ||
+ | draw(A--C--D--cycle); | ||
+ | draw(A--B--C); | ||
+ | draw(rightanglemark(B,C,D,2)); | ||
+ | draw(rightanglemark(B,A,D,2)); | ||
+ | </asy> | ||
+ | We extend <math>CB</math> and <math>DA</math> to meet at <math>E.</math> This gives us a couple right triangles in <math>CED</math> and <math>BEA.</math> | ||
+ | <asy> | ||
+ | import olympiad; import cse5; import geometry; size(250); | ||
+ | defaultpen(fontsize(10pt)); | ||
+ | defaultpen(0.8); | ||
+ | dotfactor = 4; | ||
+ | pair A = origin; | ||
+ | pair C = A+dir(55); | ||
+ | pair D = A+dir(0); | ||
+ | pair B = extension(A,A+dir(90),C,C+dir(-155)); | ||
+ | pair E = extension(A,A+2*dir(180),B,B+2*dir(-155)); | ||
+ | label("$A$",A,S); | ||
+ | label("$C$",C,NE); | ||
+ | label("$D$",D,SE); | ||
+ | label("$B$",B,NW); | ||
+ | label("$4$",B--C,NW); | ||
+ | label("$3$",A--B,W); | ||
+ | label("$E$",E,SW); | ||
+ | draw(A--C--D--cycle); | ||
+ | draw(A--B--C); | ||
+ | draw(rightanglemark(B,C,D,2)); | ||
+ | draw(rightanglemark(B,A,D,2)); | ||
+ | draw(A--E--B,dashed); | ||
+ | </asy> | ||
+ | We see that <math>\angle E = 30^\circ</math>. Hence, <math>\triangle BEA</math> and <math>\triangle DEC</math> are 30-60-90 triangles. | ||
+ | |||
+ | Using the side ratios of 30-60-90 triangles, we have <math>BE=2BA=6</math>. This tells us that <math>CE=BC+BE=4+6=10</math>. Also, <math>EA=3\sqrt{3}</math>. | ||
+ | |||
+ | Because <math>\triangle DEC\sim\triangle BEA</math>, we have <cmath>\frac{10}{3\sqrt{3}}=\frac{CD}{3}.</cmath> | ||
+ | Solving the equation, we have | ||
+ | <cmath>\begin{align*} | ||
+ | \frac{CD}3&=\frac{10}{3\sqrt{3}}\ | ||
+ | CD&=3\cdot\frac{10}{3\sqrt{3}}\ | ||
+ | CD&=\frac{10}{\sqrt{3}}\ | ||
+ | \end{align*}</cmath> | ||
+ | Hence, <math>CD=\boxed{\textbf{E}}</math>. | ||
== See also == | == See also == |
Revision as of 13:00, 10 April 2015
Problem
In , and . If perpendiculars constructed to at and to at meet at , then
Solution
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, .
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.