1993 AHSME Problems/Problem 23

Problem

$[asy] draw(circle((0,0),10),black+linewidth(.75)); draw((-10,0)--(10,0),black+linewidth(.75)); draw((-10,0)--(9,sqrt(19)),black+linewidth(.75)); draw((-10,0)--(9,-sqrt(19)),black+linewidth(.75)); draw((2,0)--(9,sqrt(19)),black+linewidth(.75)); draw((2,0)--(9,-sqrt(19)),black+linewidth(.75)); MP("X",(2,0),N);MP("A",(-10,0),W);MP("D",(10,0),E);MP("B",(9,sqrt(19)),E);MP("C",(9,-sqrt(19)),E); [/asy]$

Points $A,B,C$ and $D$ are on a circle of diameter $1$ , and $X$ is on diameter $\overline{AD}.$

If $BX=CX$ and $3\angle{BAC}=\angle{BXC}=36^\circ$ , then $AX=$

$\text{(A) } cos(6^\circ)cos(12^\circ)sec(18^\circ)\quad\\ \text{(B) } cos(6^\circ)sin(12^\circ)csc(18^\circ)\quad\\ \text{(C) } cos(6^\circ)sin(12^\circ)sec(18^\circ)\quad\\ \text{(D) } sin(6^\circ)sin(12^\circ)csc(18^\circ)\quad\\ \text{(E) } sin(6^\circ)sin(12^\circ)sec(18^\circ)$

Solution

We have all the angles we need, but most obviously, we see that right angle in triangle $ABD$ .

Note also that angle $BAD$ is 6 degrees, so length $AB = cos(6)$ because the diameter, $AD$ , is 1.

Now, we can concentrate on triangle $ABX$ (after all, now we can decipher all angles easily and use Law of Sines).

We get:

$\frac{AB}{\sin(\angle{AXB})} =\frac{AX}{\sin(\angle{ABX})}$

That's equal to

$\frac{\cos(6)}{\sin(180-18)} =\frac{AX}{\sin(12)}$

Therefore, our answer is equal to: $\fbox{B}$

Note that $\sin(162) = \sin(18)$ , and don't accidentally put $\fbox{C}$ because you thought $\frac{1}{\sin}$ was $\sec$ !

1993 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 22	Followed by Problem 24
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
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1993 AHSME Problems/Problem 23

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