Difference between revisions of "2007 AMC 10B Problems/Problem 11"
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<math>\textbf{(A) } 2\pi \qquad\textbf{(B) } \frac{5}{2}\pi \qquad\textbf{(C) } \frac{81}{32}\pi \qquad\textbf{(D) } 3\pi \qquad\textbf{(E) } \frac{7}{2}\pi</math> | <math>\textbf{(A) } 2\pi \qquad\textbf{(B) } \frac{5}{2}\pi \qquad\textbf{(C) } \frac{81}{32}\pi \qquad\textbf{(D) } 3\pi \qquad\textbf{(E) } \frac{7}{2}\pi</math> | ||
− | == | + | == Solutions == |
=== Solution 1 === | === Solution 1 === | ||
− | Let <math>\triangle ABC</math> have vertex <math>A</math> and center <math>O</math>, with foot of altitude from <math>A</math> at <math>D</math>. | + | Let <math>\triangle ABC</math> have vertex <math>A</math> and center <math>O</math>, with foot of altitude from <math>A</math> intersecting <math>BC</math> at <math>D</math>. |
<center><asy> | <center><asy> | ||
import olympiad; | import olympiad; | ||
Line 27: | Line 27: | ||
label("\(1\)",(B+D)/2,N); | label("\(1\)",(B+D)/2,N); | ||
</asy></center> | </asy></center> | ||
− | Then by Pythagorean Theorem (with radius <math>r</math> | + | Then by the Pythagorean Theorem (with radius <math>r</math> and height <math>OD = h</math>) on <math>\triangle OBD</math> and <math>\triangle ABD</math> |
<cmath> | <cmath> | ||
\begin{align*} h^2 + 1 & = r^2 \\ | \begin{align*} h^2 + 1 & = r^2 \\ | ||
Line 73: | Line 73: | ||
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> | ||
+ | |||
+ | ===Solution 4=== | ||
+ | First, we extend <math>AD</math> to intersect the circle at <math>E.</math> | ||
+ | <center><asy> | ||
+ | import olympiad; | ||
+ | pair B=(0,0), C=(2,0), A=(1,3), D=(1,0), E=(1,-(8^0.5)/8); | ||
+ | pair O=circumcenter(A,B,C); | ||
+ | draw(A--B--C--A--E); | ||
+ | draw(circumcircle(A,B,C)); | ||
+ | dot(O); | ||
+ | dot(D); | ||
+ | dot(B); | ||
+ | dot(C); | ||
+ | dot(A); | ||
+ | dot(E); | ||
+ | label("$A$",A,N); | ||
+ | label("$B$",B,S); label("\(C\)",C,S); | ||
+ | label("$D$",D,NE); | ||
+ | label("$O$",O,W); | ||
+ | label("$E$",E,S); | ||
+ | label("$3$",(A+B)/2,NW); | ||
+ | label("$1$",(B+D)/2,N); | ||
+ | </asy></center> | ||
+ | By the [[Pythagorean Theorem]], we know that | ||
+ | <cmath>1^2+AD^2=3^2\implies AD=2\sqrt{2}.</cmath> | ||
+ | We also know that, from the [[Power of a Point Theorem]], <cmath>AD\cdot DE=BD\cdot DC.</cmath> | ||
+ | We can substitute the values we know to get | ||
+ | <cmath>2\sqrt{2}\cdot DE=1</cmath> | ||
+ | We can simplify this to get that | ||
+ | <cmath>DE=\dfrac{2\sqrt{2}}{8}.</cmath> | ||
+ | We add <math>AD</math> and <math>DE</math> together to get the length of the diameter, and then we can find the area. | ||
+ | <cmath>AE=AD+DE=2\sqrt{2}+\dfrac{2\sqrt{2}}{8}=\dfrac{9\sqrt{2}}{4}.</cmath> | ||
+ | Therefore, the radius is <math>\dfrac{9\sqrt{2}}{8}</math>, so the area is | ||
+ | <cmath>\left(\dfrac{9\sqrt{2}}{8}\right)^2\pi=\boxed{(\text{C})\dfrac{81}{32}\pi.}</cmath> | ||
+ | |||
+ | ===Solution 5=== | ||
+ | Another possible solution is to plot the circle and triangle on a graph with the circle having center <math>(0,0)</math>. | ||
+ | Let the radius of the circle = <math>r</math>. | ||
+ | Let the distance between the origin and the base of triangle = <math>a</math>. | ||
+ | <cmath>1 + a^2 = r^2</cmath> | ||
+ | <cmath> r + a = 2\sqrt{2} </cmath> | ||
+ | <cmath> a = 2\sqrt{2} - r </cmath> | ||
+ | <cmath>9 - 4r\sqrt{2} = 0 </cmath> | ||
+ | <cmath>r = \frac{9\sqrt{2}}{8}</cmath> | ||
+ | <cmath>\pi r^2 = \boxed{(\text{C})\frac{81}{32}\pi}</cmath> | ||
+ | |||
+ | ===Solution 6=== | ||
+ | We can also use LOC to solve this problem. If O is the center of the circle, angle <math>OBC</math> is double angle <math>BAC</math>. From law of cosines, we have that <math>\cos(\angle BAC) = \frac{7}{9}</math>. We now apply law of cosines on triangle <math>OBC</math> to get that <math>4 = r^2+r^2-2r^2\cos(2 \angle BAC)</math> but we know that <math>2 \cos (\angle BAC) = 2 \cdot (\frac{7}{9})^2 - 1 = \frac{17}{81}</math>. Plugging this in and simplifying, we get <math>r = \frac{9}{8} \cdot \sqrt{2}</math> and thus the area of the circle is equal to <math>\frac{81}{32} \pi</math>. | ||
== See also == | == See also == |
Latest revision as of 22:02, 26 July 2021
Contents
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?
Solutions
Solution 1
Let have vertex
and center
, with foot of altitude from
intersecting
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 the Pythagorean Theorem (with radius and height
) on
and
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
Solution 4
First, we extend to intersect the circle at
![[asy] import olympiad; pair B=(0,0), C=(2,0), A=(1,3), D=(1,0), E=(1,-(8^0.5)/8); pair O=circumcenter(A,B,C); draw(A--B--C--A--E); draw(circumcircle(A,B,C)); dot(O); dot(D); dot(B); dot(C); dot(A); dot(E); label("$A$",A,N); label("$B$",B,S); label("\(C\)",C,S); label("$D$",D,NE); label("$O$",O,W); label("$E$",E,S); label("$3$",(A+B)/2,NW); label("$1$",(B+D)/2,N); [/asy]](http://latex.artofproblemsolving.com/3/1/5/315635f0b6afac6eba65f160b4765f8a089b242a.png)
By the Pythagorean Theorem, we know that
We also know that, from the Power of a Point Theorem,
We can substitute the values we know to get
We can simplify this to get that
We add
and
together to get the length of the diameter, and then we can find the area.
Therefore, the radius is
, so the area is
Solution 5
Another possible solution is to plot the circle and triangle on a graph with the circle having center .
Let the radius of the circle =
.
Let the distance between the origin and the base of triangle =
.
Solution 6
We can also use LOC to solve this problem. If O is the center of the circle, angle is double angle
. From law of cosines, we have that
. We now apply law of cosines on triangle
to get that
but we know that
. Plugging this in and simplifying, we get
and thus the area of the circle is equal to
.
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.