Difference between revisions of "1966 AHSME Problems/Problem 31"
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== Problem == | == Problem == | ||
+ | <asy> | ||
+ | draw(circle((0,0),10),black+linewidth(1)); | ||
+ | draw(circle((-1.25,2.5),4.5),black+linewidth(1)); | ||
+ | dot((0,0)); | ||
+ | dot((-1.25,2.5)); | ||
+ | draw((-sqrt(96),-2)--(-2,sqrt(96)),black+linewidth(.5)); | ||
+ | draw((-2,sqrt(96))--(sqrt(96),-2),black+linewidth(.5)); | ||
+ | draw((-sqrt(96),-2)--(sqrt(96)-2.5,7),black+linewidth(.5)); | ||
+ | draw((-sqrt(96),-2)--(sqrt(96),-2),black+linewidth(.5)); | ||
+ | MP("O'", (0,0), W); | ||
+ | MP("O", (-2,2), W); | ||
+ | MP("A", (-10,-2), W); | ||
+ | MP("B", (10,-2), E); | ||
+ | MP("C", (-2,sqrt(96)), N); | ||
+ | MP("D", (sqrt(96)-2.5,7), NE); | ||
+ | </asy> | ||
Triangle <math>ABC</math> is inscribed in a circle with center <math>O'</math>. A circle with center <math>O</math> is inscribed in triangle <math>ABC</math>. <math>AO</math> is drawn, and extended to intersect the larger circle in <math>D</math>. Then we must have: | Triangle <math>ABC</math> is inscribed in a circle with center <math>O'</math>. A circle with center <math>O</math> is inscribed in triangle <math>ABC</math>. <math>AO</math> is drawn, and extended to intersect the larger circle in <math>D</math>. Then we must have: | ||
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== Solution == | == Solution == | ||
+ | We will prove that <math>\triangle DOB</math> and <math>\triangle COD</math> is isosceles, meaning that <math>CD=OD=BD</math> and hence <math>\fbox{D}</math>. | ||
+ | Let <math>\angle A=2\alpha</math> and <math>\angle B=2\beta</math>. Since the incentre of a triangle is the intersection of its angle bisectors, <math>\angle OAB=\alpha</math> and <math>\angle ABO=\beta</math>. Hence <math>\angle DOB=\alpha +\beta</math>. Since quadrilateral <math>ABCD</math> is cyclic, <math>\angle CAD=\alpha=\angle CBD</math>. So <math>\angle OBD=\angle OBC+\angle CBD=\alpha + \beta = \angle DOB</math>. This means that <math>\triangle DOB</math> is isosceles, and hence <math>BD=OD</math>. | ||
+ | |||
+ | Now let <math>\angle C=2\gamma</math> which means <math>\angle ACO=COD=\gamma</math>. Since <math>ABCD</math> is cyclic, <math>\angle DAB=\alpha=\angle DCB.</math> Also, <math>\angle DAC\alpha</math> so <math>\angle DOC=\alpha + \gamma</math>. Thus, <math>\angle OCD=\angle OCB+\angle BCD=\gamma + \alpha=\angle DOC</math> which means <math>\triangle COD</math> is isosceles, and hence <math>CD=OD=BD</math>. | ||
+ | |||
+ | Thus our answer is <math>\fbox{D.}</math> | ||
== See also == | == See also == |
Latest revision as of 18:52, 22 April 2024
Problem
Triangle
is inscribed in a circle with center
. A circle with center
is inscribed in triangle
.
is drawn, and extended to intersect the larger circle in
. Then we must have:
Solution
We will prove that and
is isosceles, meaning that
and hence
.
Let and
. Since the incentre of a triangle is the intersection of its angle bisectors,
and
. Hence
. Since quadrilateral
is cyclic,
. So
. This means that
is isosceles, and hence
.
Now let which means
. Since
is cyclic,
Also,
so
. Thus,
which means
is isosceles, and hence
.
Thus our answer is
See also
1966 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 30 |
Followed by Problem 32 | |
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.