1979 AHSME Problems/Problem 12
Problem 12
![[asy] size(200); pair A=(-2,0),B,C=(-1,0),D=(1,0),EE,O=(0,0); draw(arc(O,1, 0, 180)); EE=midpoint(arc(O,1, 0, 90)); draw(A--EE); draw(A--D); B=intersectionpoint(arc(O,1, 180, 0),EE--A); draw(O--EE); label("$A$",A,W); label("$B$",B,NW); label("$C$",C,S);label("$D$",D,E);label("$E$",EE,NE);label("$O$",O,S);label("$45^\circ$",(0.25,0.1),fontsize(10pt)); //Credit to TheMaskedMagician for the diagram[/asy]](http://latex.artofproblemsolving.com/5/3/2/5322d08b6af2119cb1176c4c5d29209cabc0db2c.png)
In the adjoining figure, 
is the diameter of a semi-circle with center 
. Point 
lies on the extension of 
past 
;
point 
lies on the semi-circle, and 
is the point of intersection (distinct from ) of line segment 
with the semi-circle.
If length 
equals length 
, and the measure of 
is 
, then the
measure of 
is
Solution
Solution by e_power_pi_times_i
Because 
, triangles 
and 
are isosceles. Denote $\measuredangleBAO = \measuredangleAOB = \theta$ (Error compiling LaTeX. Unknown error_msg). Then $\measuredangleABO = 180^\circ-2\theta$ (Error compiling LaTeX. Unknown error_msg), and $\measuredangleEBO = \measuredangleOEB = 2\theta$ (Error compiling LaTeX. Unknown error_msg), so $\measuredangleBOE = 180^\circ-4\theta$ (Error compiling LaTeX. Unknown error_msg). Notice that $\measuredangleAOB + \measuredangleBOE + 45^\circ = 180^\circ$ (Error compiling LaTeX. Unknown error_msg). Therefore 
, and 
.
See also
| 1979 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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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