1992 AHSME Problems/Problem 25
Problem
In 
, 
and 
. If perpendiculars constructed to 
at 
and to 
at 
meet at 
, then 
Solution
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]](http://latex.artofproblemsolving.com/f/a/0/fa08c2e7dcbccff97c2e9fba0f4079fe9dccae1c.png)
We extend 
and 
to meet at 
This gives us a couple right triangles in 
and 
![[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]](http://latex.artofproblemsolving.com/d/8/1/d8199b11a714357a05f6f16681656d568c310285.png)
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 ![\[\frac{10}{3\sqrt{3}}=\frac{CD}{3}.\]](http://latex.artofproblemsolving.com/9/4/9/9491382a13324d214d9345ee434f91c5ee2b57f1.png)
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 | ||
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