Difference between revisions of "1975 AHSME Problems/Problem 24"
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− | + | ==Problem== | |
+ | 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> | ||
+ | |||
+ | <math> | ||
+ | \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} | ||
+ | </math> | ||
+ | |||
+ | ==Solution== | ||
+ | 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>. | ||
+ | |||
+ | ==See Also== | ||
+ | {{AHSME box|year=1975|num-b=23|num-a=25}} | ||
+ | {{MAA Notice}} |
Latest revision as of 20:49, 28 March 2021
Problem
In triangle ,
and
, where
. The circle with center
and radius
intersects
at
and intersects
, extended if necessary, at
and at
(
may coincide with
). Then
Solution
Since , we know
if and only if triangle
is isosceles and
. Letting
, we want to find when
. We know
, so
. We also know
, and since
,
. Since we now know that
regardless of
, we have
, or
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.