Difference between revisions of "1966 AHSME Problems/Problem 6"
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− | + | == Problem == | |
+ | <math>AB</math> is the diameter of a circle centered at <math>O</math>. <math>C</math> is a point on the circle such that angle <math>BOC</math> is <math>60^\circ</math>. If the diameter of the circle is <math>5</math> inches, the length of chord <math>AC</math>, expressed in inches, is: | ||
+ | |||
+ | <math>\text{(A)} \ 3 \qquad \text{(B)} \ \frac {5\sqrt {2}}{2} \qquad \text{(C)} \frac {5\sqrt3}{2} \ \qquad \text{(D)} \ 3\sqrt3 \qquad \text{(E)} \ \text{none of these}</math> | ||
+ | |||
+ | == Solution == | ||
+ | <asy> | ||
+ | draw(unitcircle); | ||
+ | draw((-1,0)--(1,0)--(1/2, sqrt(3)/2)--cycle); | ||
+ | label( "A", (-1,0), W); | ||
+ | label( "B", (1,0), E); | ||
+ | label( "C", (1/2, sqrt(3)/2), N); | ||
+ | </asy> | ||
+ | We see that <math>\angle A</math> is half the measure of <math>\angle BOC</math>, so <math>\angle A = 30^{\circ}</math>. That makes <math>ABC</math> a <math>30-60-90</math> triangle and sidelength <math>\overline{AC}</math> equal to <math>\frac{5\sqrt{3}}{2}</math>. | ||
+ | <math>\fbox{C}</math> | ||
+ | |||
+ | == See also == | ||
+ | {{AHSME box|year=1966|num-b=5|num-a=7}} | ||
+ | |||
+ | [[Category:Introductory Geometry Problems]] | ||
+ | {{MAA Notice}} |
Latest revision as of 12:42, 28 January 2020
Problem
is the diameter of a circle centered at
.
is a point on the circle such that angle
is
. If the diameter of the circle is
inches, the length of chord
, expressed in inches, is:
Solution
We see that
is half the measure of
, so
. That makes
a
triangle and sidelength
equal to
.
See also
1966 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 5 |
Followed by Problem 7 | |
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.