Difference between revisions of "1996 AHSME Problems/Problem 21"
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<math> \text{(A)}\ 115^\circ\qquad\text{(B)}\ 120^\circ\qquad\text{(C)}\ 130^\circ\qquad\text{(D)}\ 135^\circ\qquad\text{(E)}\ \text{not uniquely determined} </math> | <math> \text{(A)}\ 115^\circ\qquad\text{(B)}\ 120^\circ\qquad\text{(C)}\ 130^\circ\qquad\text{(D)}\ 135^\circ\qquad\text{(E)}\ \text{not uniquely determined} </math> | ||
− | ==Solution== | + | ==Solution 1== |
Redraw the figure as a concave pentagon <math>ADECB</math>: | Redraw the figure as a concave pentagon <math>ADECB</math>: | ||
Line 56: | Line 56: | ||
<math>\angle C + \angle D = 135^\circ</math>, which is answer <math>\boxed{D}</math> | <math>\angle C + \angle D = 135^\circ</math>, which is answer <math>\boxed{D}</math> | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Let <math>m\angle ABC = x</math>. By the isosceles triangle theorem, we have <math>m\angle ACB = x</math> and <math>m\angle BAD = m\angle BDA</math>. Because the angles of a triangle sum to <math>\pi</math>, we have <math>m\angle BAC = \pi - 2x</math>, then <math>m\angle ABD=\frac{\pi}{2}-(\pi-2x)=2x-\frac{\pi}{2}</math>. Then we have <math>m\angle BAD + m\angle BDA = \pi - m\angle ABD</math>. Substituting, this becomes <math>2m\angle BDA = \pi-\left(\frac{\pi}{2}-(\pi-2x)\right) \to m \angle BDA = \frac{3 \pi}{4} - x </math>. Adding <math>x</math>, which is <math>m\angle ACB</math>, we have <math>m\angle BDA + m\angle ACB = \frac{3 \pi}{4} = 135^\circ \to \boxed{\textbf{D}}</math> | ||
+ | |||
==See also== | ==See also== | ||
{{AHSME box|year=1996|num-b=20|num-a=22}} | {{AHSME box|year=1996|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 20:40, 13 January 2014
Contents
[hide]Problem
Triangles and are isosceles with , and intersects at . If is perpendicular to , then is
Solution 1
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:
, which is answer
Solution 2
Let . By the isosceles triangle theorem, we have and . Because the angles of a triangle sum to , we have , then . Then we have . Substituting, this becomes . Adding , which is , we have
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 |
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