Difference between revisions of "2010 AMC 10B Problems/Problem 19"
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== Problem == | == Problem == | ||
− | A circle with center <math>O</math> has area <math>156\pi</math>. Triangle <math>ABC</math> is equilateral, <math>\ | + | A circle with center <math>O</math> has area <math>156\pi</math>. Triangle <math>ABC</math> is equilateral, <math>\overline{BC}</math> is a chord on the circle, <math>OA = 4\sqrt{3}</math>, and point <math>O</math> is outside <math>\triangle ABC</math>. What is the side length of <math>\triangle ABC</math>? |
<math>\textbf{(A)}\ 2\sqrt{3} \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 4\sqrt{3} \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18</math> | <math>\textbf{(A)}\ 2\sqrt{3} \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 4\sqrt{3} \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18</math> | ||
− | ==Solution== | + | ==Solution 1== |
The formula for the area of a circle is <math>\pi r^2</math> so the radius of this circle is <math>\sqrt{156}.</math> | The formula for the area of a circle is <math>\pi r^2</math> so the radius of this circle is <math>\sqrt{156}.</math> | ||
Line 48: | Line 48: | ||
s &= \boxed{\textbf{(B)}\ 6} | s &= \boxed{\textbf{(B)}\ 6} | ||
\end{align*} </cmath> | \end{align*} </cmath> | ||
+ | |||
+ | ==Solution 2== | ||
+ | We can use the same diagram as Solution 1 and label the side length of <math>\triangle ABC</math> as <math>s</math>. Using congruent triangles, namely the two triangles <math>\triangle BOA</math> and <math>\triangle COA</math>, we get that <math>\angle BAO = \angle CAO \implies \angle BAO = \frac{360-60}{2} = 150</math>. From this, we can use the [[Law of Cosines]], to get <cmath>s^2 + (4 \sqrt{3})^2 - 2 \times s \times 4 \sqrt{3} \times - \frac{\sqrt{3}}{2} = (2 \sqrt{39})^2</cmath> Simplifying, we get <cmath>s^2 + 12s + 48 = 156 \implies s^2 + 12s - 108 = 0</cmath> We can factor this to get <cmath>(x-6)(x+18)</cmath> Obviously, we want the positive solution to get <math>\boxed{\textbf{(B)}\ 6}</math> | ||
+ | ~hansenhe | ||
+ | |||
+ | ==Video Solution== | ||
+ | https://youtu.be/FQO-0E2zUVI?t=906 | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2010|ab=B|num-b=18|num-a=20}} | {{AMC10 box|year=2010|ab=B|num-b=18|num-a=20}} | ||
+ | {{MAA Notice}} |
Revision as of 23:51, 8 March 2021
Problem
A circle with center has area . Triangle is equilateral, is a chord on the circle, , and point is outside . What is the side length of ?
Solution 1
The formula for the area of a circle is so the radius of this circle is
Because must be in the interior of circle
Let be the unknown value, the sidelength of the triangle, and let be the point on where Since is equilateral, and We are given Use the Pythagorean Theorem and solve for
Solution 2
We can use the same diagram as Solution 1 and label the side length of as . Using congruent triangles, namely the two triangles and , we get that . From this, we can use the Law of Cosines, to get Simplifying, we get We can factor this to get Obviously, we want the positive solution to get ~hansenhe
Video Solution
https://youtu.be/FQO-0E2zUVI?t=906
~IceMatrix
See Also
2010 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 | ||
All AMC 10 Problems and Solutions |
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