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Difference between revisions of "1996 AHSME Problems/Problem 21"

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==Problem==
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Triangles <math>ABC</math> and <math>ABD</math> are isosceles with <math>AB=AC=BD</math>, and <math>BD</math> intersects <math>AC</math> at <math>E</math>. If <math>BD</math> is perpendicular to <math>AC</math>, then <math> \angle C+\angle D </math> is
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<asy>
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size(120);
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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);
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dot(A^^B^^C^^D^^E);
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draw(A--D--B--A--C--B);
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markscalefactor=0.005;
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draw(rightanglemark(A, E, B));
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dot(A^^B^^C^^D^^E);
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pair point=midpoint(A--M);
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label("$A$", A, dir(point--A));
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label("$B$", B, dir(point--B));
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label("$C$", C, dir(point--C));
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label("$D$", D, dir(point--D));
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label("$E$", E, dir(point--E));
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</asy>
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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>
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==Solution 1==
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Redraw the figure as a concave pentagon <math>ADECB</math>:
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<asy>
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size(120);
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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);
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dot(A^^B^^C^^D^^E);
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draw(A--D--E--C--B--A);
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markscalefactor=0.005;
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draw(rightanglemark(D, E, C));
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dot(A^^B^^C^^D^^E);
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pair point=midpoint(A--M);
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label("$A$", A, dir(point--A));
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label("$B$", B, dir(point--B));
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label("$C$", C, dir(point--C));
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label("$D$", D, dir(point--D));
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label("$E$", E, dir(point--E));
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</asy>
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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.
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Since reflex <math>\angle E = 270^\circ</math>, we have:
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<math>\angle C + \angle B + \angle A + \angle D = 540^\circ - 270^\circ = 270^\circ</math>.
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From isosceles <math>\triangle ABC</math>, we get <math>\angle B = \angle C</math>, so:
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<math>2\angle C + \angle D + \angle A = 270^\circ</math>
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From isosceles <math>\triangle ABD</math>, we get <math>\angle A = \angle D</math>, so:
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<math>2\angle C + 2\angle D = 270^\circ</math>
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<math>\angle C + \angle D = 135^\circ</math>, which is answer <math>\boxed{D}</math>
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==Solution 2==
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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>
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==See also==
 
==See also==
 
{{AHSME box|year=1996|num-b=20|num-a=22}}
 
{{AHSME box|year=1996|num-b=20|num-a=22}}
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{{MAA Notice}}

Latest revision as of 20:40, 13 January 2014

Problem

Triangles $ABC$ and $ABD$ are isosceles with $AB=AC=BD$, and $BD$ intersects $AC$ at $E$. If $BD$ is perpendicular to $AC$, then $\angle C+\angle D$ is

[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--B--A--C--B); markscalefactor=0.005; draw(rightanglemark(A, E, B)); 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]

$\text{(A)}\ 115^\circ\qquad\text{(B)}\ 120^\circ\qquad\text{(C)}\ 130^\circ\qquad\text{(D)}\ 135^\circ\qquad\text{(E)}\ \text{not uniquely determined}$

Solution 1

Redraw the figure as a concave pentagon $ADECB$:

[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 $180^\circ \cdot 3 = 540^\circ$, regardless of whether the pentagon is concave or not. As a quick proof, note that the nine angles of three original triangles $\triangle AEB$, $\triangle CBE$, and $\triangle DEA$ all make up the angles of the pentagon without overlap.

Since reflex $\angle E = 270^\circ$, we have:

$\angle C + \angle B + \angle A + \angle D = 540^\circ - 270^\circ = 270^\circ$.

From isosceles $\triangle ABC$, we get $\angle B = \angle C$, so:

$2\angle C + \angle D + \angle A = 270^\circ$

From isosceles $\triangle ABD$, we get $\angle A = \angle D$, so:

$2\angle C + 2\angle D = 270^\circ$

$\angle C + \angle D = 135^\circ$, which is answer $\boxed{D}$

Solution 2

Let $m\angle ABC = x$. By the isosceles triangle theorem, we have $m\angle ACB = x$ and $m\angle BAD = m\angle BDA$. Because the angles of a triangle sum to $\pi$, we have $m\angle BAC = \pi - 2x$, then $m\angle ABD=\frac{\pi}{2}-(\pi-2x)=2x-\frac{\pi}{2}$. Then we have $m\angle BAD + m\angle BDA = \pi - m\angle ABD$. Substituting, this becomes $2m\angle BDA = \pi-\left(\frac{\pi}{2}-(\pi-2x)\right) \to m \angle BDA = \frac{3 \pi}{4} - x$. Adding $x$, which is $m\angle ACB$, we have $m\angle BDA + m\angle ACB = \frac{3 \pi}{4} = 135^\circ \to \boxed{\textbf{D}}$


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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