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>
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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.
 +
 +
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:
 +
 +
<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:
 +
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<math>2\angle C + 2\angle D = 270^\circ</math>
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$\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 $ABC$ 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

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

See also

1996 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 20
Followed by
Problem 22
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