Difference between revisions of "2007 AMC 10B Problems/Problem 11"
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By the Pythagorean Theorem <cmath>AD^2 = 3^2 - 1^2</cmath> <cmath>AD = 2\sqrt{2}</cmath> | By the Pythagorean Theorem <cmath>AD^2 = 3^2 - 1^2</cmath> <cmath>AD = 2\sqrt{2}</cmath> | ||
− | <math>\triangle ADC</math> is similar to <math>\triangle ACR</math>, so <cmath>\frac {2\sqrt{2}}{3} = \frac {3}{2r}</cmath> which gives us <cmath>2r = \frac {9}{2\sqrt{2}} = \frac {9\sqrt{2}}{4}</cmath> | + | <math>\triangle ADC</math> is similar to <math>\triangle ACR</math>, so <cmath>\frac {2\sqrt{2}}{3} = \frac {3}{2r}</cmath> which gives us <cmath>2r = \frac {9}{2\sqrt{2}} = \frac {9\sqrt{2}}{4}</cmath> therefore <cmath>r = \frac {9\sqrt{2}}{8}</cmath> |
The area of the circle is therefore <math>\pi r^2 = \left(\frac {9\sqrt{2}}{8}\right)^2 \pi = \boxed{\mathrm{(C) \ } \frac {81}{32} \pi}</math> | The area of the circle is therefore <math>\pi r^2 = \left(\frac {9\sqrt{2}}{8}\right)^2 \pi = \boxed{\mathrm{(C) \ } \frac {81}{32} \pi}</math> |
Revision as of 00:25, 25 December 2013
Problem
A circle passes through the three vertices of an isosceles triangle that has two sides of length and a base of length
. What is the area of this circle?
Solution
Solution 1
Let have vertex
and center
, with foot of altitude from
at
.
![[asy] import olympiad; pair B=(0,0), C=(2,0), A=(1,3), D=(1,0); pair O=circumcenter(A,B,C); draw(A--B--C--A--D); draw(B--O--C); draw(circumcircle(A,B,C)); dot(O); label("\(A\)",A,N); label("\(B\)",B,S); label("\(C\)",C,S); label("\(D\)",D,S); label("\(O\)",O,W); label("\(r\)",(O+A)/2,SE); label("\(r\)",(O+B)/2,N); label("\(h\)",(O+D)/2,SE); label("\(3\)",(A+B)/2,NW); label("\(1\)",(B+D)/2,N); [/asy]](http://latex.artofproblemsolving.com/d/d/6/dd6654d76ebcb5aa78bf644545d9639c8fccd53a.png)
Then by Pythagorean Theorem (with radius , height
) on
Substituting and solving gives . Then the area of the circle is
.
Solution 2
By (or we could use
and Heron's formula),
and the answer is
Alternatively, by the Extended Law of Sines,
Answer follows as above.
Solution 3
Extend segment to
on Circle
.
![[asy] import olympiad; pair B=(0,0), C=(2,0), A=(1,3), D=(1,0), R=(1,-0.35); pair O=circumcenter(A,B,C); draw(A--B--C--A--D--R--C); draw(B--O--C); draw(circumcircle(A,B,C)); dot(O); label("\(A\)",A,N); label("\(B\)",B,S); label("\(C\)",C,S); label("\(D\)",D,S); label("\(O\)",O,W); label("\(R\)",R,S); label("\(r\)",(O+A)/2,SE); label("\(r\)",(O+R)/2,SE); label("\(3\)",(A+C)/2,NE); label("\(1\)",(C+D)/2,N); [/asy]](http://latex.artofproblemsolving.com/7/6/0/7609b3287774915ebf4e80694297138491851a07.png)
By the Pythagorean Theorem
is similar to
, so
which gives us
therefore
The area of the circle is therefore
See also
2007 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.