Difference between revisions of "1996 AHSME Problems/Problem 21"
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==Solution== | ==Solution== | ||
+ | Redraw the figure as a concave pentagon <math>ADECB</math>: | ||
+ | |||
+ | <asy> | ||
+ | size(120); | ||
+ | pair B=origin, A=1*dir(70), M=foot(A, B, (3,0)), C=reflect(A, M)*B, E=foot(B, A, C), D=1*dir(20); | ||
+ | dot(A^^B^^C^^D^^E); | ||
+ | draw(A--D--E--C--B--A); | ||
+ | markscalefactor=0.005; | ||
+ | draw(rightanglemark(D, E, C)); | ||
+ | dot(A^^B^^C^^D^^E); | ||
+ | pair point=midpoint(A--M); | ||
+ | label("$A$", A, dir(point--A)); | ||
+ | label("$B$", B, dir(point--B)); | ||
+ | label("$C$", C, dir(point--C)); | ||
+ | label("$D$", D, dir(point--D)); | ||
+ | label("$E$", E, dir(point--E)); | ||
+ | </asy> | ||
+ | |||
+ | The angles of the pentagon will still sum to <math>180^\circ \cdot 3 = 540^\circ</math>, regardless of whether the pentagon is concave or not. As a quick proof, note that the nine angles of three original triangles <math>\triangle AEB</math>, <math>\triangle CBE</math>, and <math>\triangle DEA</math> all make up the angles of the pentagon without overlap. | ||
+ | |||
+ | Since reflex <math>\angle E = 270^\circ</math>, we have: | ||
+ | |||
+ | <math>\angle C + \angle B + \angle A + \angle D = 540^\circ - 270^\circ = 270^\circ</math>. | ||
+ | |||
+ | From isosceles <math>\triangle ABC</math>, we get <math>\angle B = \angle C</math>, so: | ||
+ | |||
+ | <math>2\angle C + \angle D + \angle A = 270^\circ</math> | ||
+ | |||
+ | From isosceles <math>/triangle ABD</math>, we get <math>\angle A = \angle D</math>, so: | ||
+ | |||
+ | <math>2\angle C + 2\angle D = 270^\circ</math> | ||
+ | |||
+ | $\angle C + \angle D = 135^\circ | ||
==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=20|num-a=22}} | {{AHSME box|year=1996|num-b=20|num-a=22}} |
Revision as of 14:00, 20 August 2011
Problem
Triangles and are isosceles with , and intersects at . If is perpendicular to , then is
Solution
Redraw the figure as a concave pentagon :
The angles of the pentagon will still sum to , regardless of whether the pentagon is concave or not. As a quick proof, note that the nine angles of three original triangles , , and all make up the angles of the pentagon without overlap.
Since reflex , we have:
.
From isosceles , we get , so:
From isosceles , we get , so:
$\angle C + \angle D = 135^\circ
See also
1996 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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 |