Difference between revisions of "2008 AMC 12B Problems/Problem 9"

Revision as of 23:21, 14 August 2011

Problem 9

Points $A$ and $B$ are on a circle of radius $5$ and $AB = 6$ . Point $C$ is the midpoint of the minor arc $AB$ . What is the length of the line segment $AC$ ?

$\textbf{(A)}\ \sqrt {10} \qquad \textbf{(B)}\ \frac {7}{2} \qquad \textbf{(C)}\ \sqrt {14} \qquad \textbf{(D)}\ \sqrt {15} \qquad \textbf{(E)}\ 4$

Solutions

Solution 1

Let $\alpha$ be the angle that subtends the arc $AB$ . By the law of cosines, $6^2=5^2+5^2-2\cdot 5\cdot 5\cos(\alpha)$ implies $\cos(\alpha) = 7/25$ .

The half-angle formula says that $\cos(\alpha/2) = \frac{\sqrt{1+\cos(\alpha)}}{2} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$ . The law of cosines tells us $AC = \sqrt{5^2+5^2-2*5\cdot 5\cdot \frac{4}{5}} = \sqrt{50-50\frac{4}{5}} = \sqrt{10}$ , which is answer choice $\boxed{\text{A}}$ .

Solution 2

$[asy] defaultpen(fontsize(8)); pair A=(-3,4), B=(3,4), C=(0,5), D=(0,4), O=(0,0); D(Circle(O,5)); D(O--B--A--O--C);D(A--C--B); label("$A$",A,(-1,1));label("$O$",O,(0,-1));label("$B$",B,(1,1));label("$C$",C,(0,1));label("$D$",D,(-1,-1)); [/asy]$

Figure 1

Define $D$ as the midpoint of line segment $\overline{AB}$ , and $O$ the center of the circle. Then $O$ , $C$ , and $D$ are collinear, and since $D$ is the midpoint of $AB$ , $m\angle ODA=90\deg$ and so $OD=\sqrt{5^2-3^2}=4$ . Since $OD=4$ , $CD=5-4=1$ , and so $AC=\sqrt{3^2+1^2}=\sqrt{10} \rightarrow \boxed{\text{A}}$ .

@@ Line 4: / Line 4: @@
 <math>\textbf{(A)}\ \sqrt {10} \qquad \textbf{(B)}\ \frac {7}{2} \qquad \textbf{(C)}\ \sqrt {14} \qquad \textbf{(D)}\ \sqrt {15} \qquad \textbf{(E)}\ 4</math>
-==Solution==
+==Solutions==
-===Trig Solution:===
+===Solution 1===
-Let <math>\alpha</math> be the angle that subtends the arc AB. By the law of cosines,
+Let <math>\alpha</math> be the angle that subtends the arc <math>AB</math>. By the law of cosines,
-<math>6^2=5^2+5^2-2*5*5cos(\alpha)</math>
+<math>6^2=5^2+5^2-2\cdot 5\cdot 5\cos(\alpha)</math> implies <math>\cos(\alpha) = 7/25</math>.
-<math>\alpha = cos^{-1}(7/25)</math>
+The [[Trigonometric_identities#Half_Angle_Identities | half-angle formula]] says that
+<math>\cos(\alpha/2) = \frac{\sqrt{1+\cos(\alpha)}}{2} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}</math>. The law of cosines tells us <math>AC = \sqrt{5^2+5^2-2*5\cdot 5\cdot \frac{4}{5}} = \sqrt{50-50\frac{4}{5}} = \sqrt{10}</math>, which is answer choice <math>\boxed{\text{A}}</math>.
-The half-angle formula says that
+===Solution 2===
-<math>cos(\alpha/2) = \frac{\sqrt{1+cos(\alpha)}}{2} = \sqrt{\frac{32/25}{2}} = \sqrt{\frac{16}{25}} = \frac{4}{5}</math>
+{{asy image|
-<math>AC = \sqrt{5^2+5^2-2*5*5*\frac{4}{5}}</math>
+<asy>
+defaultpen(fontsize(8));
-<math>AC = \sqrt{50-50\frac{4}{5}}</math>
+pair A=(-3,4), B=(3,4), C=(0,5), D=(0,4), O=(0,0);
+D(Circle(O,5));
-<math>AC = \sqrt{10}</math>, which is answer choice A.
+D(O--B--A--O--C);D(A--C--B);
+label("$A$",A,(-1,1));label("$O$",O,(0,-1));label("$B$",B,(1,1));label("$C$",C,(0,1));label("$D$",D,(-1,-1));
-===More Elegant Solution===
+</asy>
-Define D as the midpoint of AB, and R the center of the circle. R, C, and D are collinear, and since D is the midpoint of AB,
+|right|Figure 1
+}}
-<math>m\angle RDA=90\deg</math>
+Define <math>D</math> as the midpoint of line segment <math>\overline{AB}</math>, and <math>O</math> the center of the circle. Then <math>O</math>, <math>C</math>, and <math>D</math> are collinear, and since <math>D</math> is the midpoint of <math>AB</math>, <math>m\angle ODA=90\deg</math> and so <math>OD=\sqrt{5^2-3^2}=4</math>. Since <math>OD=4</math>, <math>CD=5-4=1</math>, and so <math>AC=\sqrt{3^2+1^2}=\sqrt{10} \rightarrow \boxed{\text{A}}</math>.
- and so
-<math>RD=\sqrt{5^2-3^2}=4</math>.
-Since
-<math>RD=4</math>,
- <math>CD=5-4=1</math>,
- and so
-<math>AC=\sqrt{3^2+1^2}=\sqrt{10} \rightarrow A</math>
 ==See Also==
 {{AMC12 box|year=2008|ab=B|num-b=8|num-a=10}}

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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