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1966 AHSME Problems/Problem 6: Difference between revisions

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Answer is C
== 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}}
{{MAA Notice}}

Latest revision as of 11:42, 28 January 2020

Problem

$AB$ is the diameter of a circle centered at $O$. $C$ is a point on the circle such that angle $BOC$ is $60^\circ$. If the diameter of the circle is $5$ inches, the length of chord $AC$, expressed in inches, is:

$\text{(A)} \ 3 \qquad \text{(B)} \ \frac {5\sqrt {2}}{2} \qquad \text{(C)} \frac {5\sqrt3}{2} \ \qquad \text{(D)} \ 3\sqrt3 \qquad \text{(E)} \ \text{none of these}$

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 $\angle A$ is half the measure of $\angle BOC$, so $\angle A = 30^{\circ}$. That makes $ABC$ a $30-60-90$ triangle and sidelength $\overline{AC}$ equal to $\frac{5\sqrt{3}}{2}$. $\fbox{C}$

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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