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Difference between revisions of "1975 AHSME Problems/Problem 24"

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==Problem==
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In triangle <math>ABC</math>, <math>\angle C = \theta</math> and <math>\angle B = 2\theta</math>, where <math>0^{\circ} < \theta < 60^{\circ}</math>. The circle with center <math>A</math> and radius <math>AB</math> intersects <math>AC</math> at <math>D</math> and intersects <math>BC</math>, extended if necessary, at <math>B</math> and at <math>E</math> (<math>E</math> may coincide with <math>B</math>). Then <math>EC = AD</math>
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<math>
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\textbf{(A)}\ \text{for no values of}\ \theta \qquad
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\textbf{(B)}\ \text{only if}\ \theta = 45^{\circ} \\
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\textbf{(C)}\ \text{only if}\ 0^{\circ} < \theta \leq 45^{\circ} \qquad
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\textbf{(D)}\ \text{only if}\ 45^{\circ} \leq \theta \leq 60^{\circ} \\
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\textbf{(E)}\ \text{for all}\ \theta\ \text{such that}\ 0^{\circ} < \theta < 60^{\circ}
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</math>
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==Solution==
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Since <math>AD = AE</math>, we know <math>EC = AD</math> if and only if triangle <math>ACE</math> is isosceles and <math>\angle ACE = \angle CAE</math>. Letting <math>\angle ACE = \theta</math>, we want to find when <math>\angle CAE = \theta</math>. We know <math>\angle ABC = \angle AEB = 2\theta</math>, so <math>\angle EAB = 180-4\theta</math>. We also know <math>\angle CAB = 180-3\theta</math>, and since <math>\angle CAE = \angle CAB - \angle EAB</math>, <math>\angle CAE = \theta</math>. Since we now know that <math>\angle CAE = \angle ACE</math> regardless of <math>\theta</math>, we have <math>0^{\circ} < \theta < 60^{\circ}</math>, or <math>\boxed{E}</math>.
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==See Also==
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{{AHSME box|year=1975|num-b=23|num-a=25}}
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{{MAA Notice}}

Latest revision as of 19:49, 28 March 2021

Problem

In triangle $ABC$, $\angle C = \theta$ and $\angle B = 2\theta$, where $0^{\circ} < \theta < 60^{\circ}$. The circle with center $A$ and radius $AB$ intersects $AC$ at $D$ and intersects $BC$, extended if necessary, at $B$ and at $E$ ($E$ may coincide with $B$). Then $EC = AD$

$\textbf{(A)}\ \text{for no values of}\ \theta \qquad \textbf{(B)}\ \text{only if}\ \theta = 45^{\circ} \\ \textbf{(C)}\ \text{only if}\ 0^{\circ} < \theta \leq 45^{\circ} \qquad \textbf{(D)}\ \text{only if}\ 45^{\circ} \leq \theta \leq 60^{\circ} \\ \textbf{(E)}\ \text{for all}\ \theta\ \text{such that}\ 0^{\circ} < \theta < 60^{\circ}$

Solution

Since $AD = AE$, we know $EC = AD$ if and only if triangle $ACE$ is isosceles and $\angle ACE = \angle CAE$. Letting $\angle ACE = \theta$, we want to find when $\angle CAE = \theta$. We know $\angle ABC = \angle AEB = 2\theta$, so $\angle EAB = 180-4\theta$. We also know $\angle CAB = 180-3\theta$, and since $\angle CAE = \angle CAB - \angle EAB$, $\angle CAE = \theta$. Since we now know that $\angle CAE = \angle ACE$ regardless of $\theta$, we have $0^{\circ} < \theta < 60^{\circ}$, or $\boxed{E}$.

See Also

1975 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
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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