Difference between revisions of "2005 AMC 12A Problems/Problem 15"
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== Problem == | == Problem == | ||
Let <math>\overline{AB}</math> be a [[diameter]] of a [[circle]] and <math>C</math> be a point on <math>\overline{AB}</math> with <math>2 \cdot AC = BC</math>. Let <math>D</math> and <math>E</math> be [[point]]s on the circle such that <math>\overline{DC} \perp \overline{AB}</math> and <math>\overline{DE}</math> is a second diameter. What is the [[ratio]] of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math>? | Let <math>\overline{AB}</math> be a [[diameter]] of a [[circle]] and <math>C</math> be a point on <math>\overline{AB}</math> with <math>2 \cdot AC = BC</math>. Let <math>D</math> and <math>E</math> be [[point]]s on the circle such that <math>\overline{DC} \perp \overline{AB}</math> and <math>\overline{DE}</math> is a second diameter. What is the [[ratio]] of the area of <math>\triangle DCE</math> to the area of <math>\triangle ABD</math>? | ||
+ | |||
+ | <asy> | ||
+ | unitsize(2.5cm); | ||
+ | defaultpen(fontsize(10pt)+linewidth(.8pt)); | ||
+ | dotfactor=3; | ||
+ | pair O=(0,0), C=(-1/3.0), B=(1,0), A=(-1,0); | ||
+ | pair D=dir(aCos(C.x)), E=(-D.x,-D.y); | ||
+ | draw(A--B--D--cycle); | ||
+ | draw(D--E--C); | ||
+ | draw(unitcircle,white); | ||
+ | drawline(D,C); | ||
+ | dot(O); | ||
+ | clip(unitcircle); | ||
+ | draw(unitcircle); | ||
+ | label("$E$",E,SSE); | ||
+ | label("$B$",B,E); | ||
+ | label("$A$",A,W); | ||
+ | label("$D$",D,NNW); | ||
+ | label("$C$",C,SW); | ||
+ | draw(rightanglemark(D,C,B,2));</asy> | ||
<math>(\text {A}) \ \frac {1}{6} \qquad (\text {B}) \ \frac {1}{4} \qquad (\text {C})\ \frac {1}{3} \qquad (\text {D}) \ \frac {1}{2} \qquad (\text {E})\ \frac {2}{3}</math> | <math>(\text {A}) \ \frac {1}{6} \qquad (\text {B}) \ \frac {1}{4} \qquad (\text {C})\ \frac {1}{3} \qquad (\text {D}) \ \frac {1}{2} \qquad (\text {E})\ \frac {2}{3}</math> | ||
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== Solution == | == Solution == |
Revision as of 17:21, 24 November 2011
Problem
Let be a diameter of a circle and
be a point on
with
. Let
and
be points on the circle such that
and
is a second diameter. What is the ratio of the area of
to the area of
?
Solution
Solution 1
Notice that the bases of both triangles are diameters of the circle. Hence the ratio of the areas is just the ratio of the heights of the triangles, or (
is the foot of the perpendicular from
to
).
Call the radius . Then
,
. Using the Pythagorean Theorem in
, we get
.
Now we have to find . Notice
, so we can write the proportion:
![$\frac{OF}{OC} = \frac{OC}{OD}$](http://latex.artofproblemsolving.com/0/2/2/02291691270e57eea057ea561416318b8de9fc01.png)
![$\frac{OF}{\frac{1}{3}r} = \frac{\frac{1}{3}r}{r}$](http://latex.artofproblemsolving.com/6/e/1/6e1dcd67ba34e71f9f880510df1fd482941f6e37.png)
![$OF = \frac 19r$](http://latex.artofproblemsolving.com/5/a/4/5a4dcf36a069d28f1265cf8c1b825c9ae46c0cbd.png)
By the Pythagorean Theorem in , we have
.
Our answer is .
Solution 2
Let the centre of the circle be .
Note that .
is midpoint of
.
is midpoint of
Area of
Area of
Area of
Area of
.
See also
2005 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 14 |
Followed by Problem 16 |
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 12 Problems and Solutions |