Difference between revisions of "2007 AMC 10B Problems/Problem 11"
(→Solution) |
|||
Line 46: | Line 46: | ||
2R = \frac {AC}{\sin \angle ABC} = \frac {3}{\frac {2\sqrt {2}}{3}} \Longrightarrow R = \frac {9}{4\sqrt {2}} | 2R = \frac {AC}{\sin \angle ABC} = \frac {3}{\frac {2\sqrt {2}}{3}} \Longrightarrow R = \frac {9}{4\sqrt {2}} | ||
</cmath> | </cmath> | ||
− | Answer follows as above. | + | Answer follows as above. |
+ | |||
+ | === Solution 3 === | ||
+ | |||
+ | Extend segment <math>AD</math> to <math>R</math> on Circle <math>O</math>. | ||
+ | <center><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></center> | ||
+ | |||
+ | By the Pythagorean Theorem <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> so <cmath>r = \frac {9\sqrt{2}}{8}</cmath> | ||
+ | |||
+ | The area of the circle is therefor <math>\pi r^2 = \left(\frac {9\sqrt{2}}{8}\right)^2 \pi = \boxed{\mathrm{(C) \ } \frac {81}{32} \pi}</math> | ||
== See also == | == See also == |
Revision as of 19:53, 23 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 .
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 .
By the Pythagorean Theorem
is similar to , so which gives us so
The area of the circle is therefor
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.