Difference between revisions of "2010 AMC 10B Problems/Problem 19"
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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> | ||
− | Because <math>OA=4\sqrt{3} < \sqrt{156}, A</math> must be in the interior of circle <math>O.</math> | + | Because <math>OA=4\sqrt{3}=\sqrt{48} < \sqrt{156}, A</math> must be in the interior of circle <math>O.</math> |
<center><asy> | <center><asy> | ||
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</asy></center> | </asy></center> | ||
− | Let <math>s</math> be the | + | Let <math>s</math> be the side length of the triangle, the unknown value, and let <math>X</math> be the point on <math>BC</math> where <math>OX \perp BC.</math> Since <math>\triangle ABC</math> is equilateral, <math>BX=\frac{s}{2}</math> and <math>AX=\frac{s\sqrt{3}}{2}.</math> We are given <math>AO=4\sqrt{3}.</math> Use the [[Pythagorean Theorem]] and solve for <math>s.</math> |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
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==Solution 2== | ==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^\circ-60^\circ}{2} = 150^\circ</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> Lengths must be non-negative, so the answer is <math>\boxed{\textbf{(B)}\ 6}</math> | |
+ | ~bryan gao | ||
+ | |||
+ | ==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}} | {{MAA Notice}} |
Latest revision as of 18:34, 4 May 2024
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
![[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=3; real r=sqrt(156); pair A=(0,sqrt(48)), B=(-3,sqrt(147)), C=(3,sqrt(147)); pair O=(0,0); pair X=(0,7sqrt(3)); path outer=Circle(O,r); draw(outer); draw(A--B--C--cycle); draw(O--X); draw(O--B); pair[] ps={A,B,C,O,X}; dot(ps); label("$A$",A,SE); label("$B$",B,NW); label("$C$",C,NE); label("$O$",O,S); label("$X$",X,N); label("$s$",A--C,SE); label("$\frac{s}{2}$",B--X,N); label("$\frac{s\sqrt{3}}{2}$",A--X,NE); label("$\sqrt{156}$",O--B,SW); label("$4\sqrt{3}$",A--O,E); [/asy]](http://latex.artofproblemsolving.com/0/3/f/03fe10542785e49de90d325a8bb86d13462ffd6a.png)
Let be the side length of the triangle, the unknown value, 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
Lengths must be non-negative, so the answer is
~bryan gao
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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