Difference between revisions of "2005 AMC 10A Problems/Problem 23"
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O=(0,0);A=(15,0);B=(-15,0);C=(5,0);D=(5,14.142135623730950488016887242097);E=(-5,-14.142135623730950488016887242097);F=(0.5555555555555555,1.5713484026367722764463208046774); | O=(0,0);A=(15,0);B=(-15,0);C=(5,0);D=(5,14.142135623730950488016887242097);E=(-5,-14.142135623730950488016887242097);F=(0.5555555555555555,1.5713484026367722764463208046774); | ||
draw(Circle((0,0),15)); | draw(Circle((0,0),15)); | ||
− | draw(A--B);draw(D--E);draw(C--D);draw(C--E);draw(C--F); | + | draw(A--B);draw(D--E);draw(C--D);draw(C--E);draw(C--F);draw(A--D);draw(D--B); |
label("A",A,NE);label("B",B,W);label("C",C,SE);label("D",D,NE);label("E",E,SW);label("O",O,SW);label("F",F,NW); | label("A",A,NE);label("B",B,W);label("C",C,SE);label("D",D,NE);label("E",E,SW);label("O",O,SW);label("F",F,NW); | ||
markscalefactor=0.2; | markscalefactor=0.2; |
Revision as of 21:31, 5 February 2010
Contents
Problem
Let be a diameter of a circle and let 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 ?
. The area of
Solution 1
http://img443.imageshack.us/img443/8034/circlenc1.png
is of diameter and is - = .
is the radius of the circle, so using the Pythagorean theorem height of is $\sqrt{(\frac{1}{2})^2-(\frac{1}{6})^2$ (Error compiling LaTeX. ! Missing } inserted.) = . This is also the height of the .
Area of the is = .
The height of can be found using the area of and as base.
Hence the height of is = .
The diameter is the base for both the triangles and .
Hence, the ratio of the area of to the area of is =
Solution 2
Since and share a base, the ratio of their areas is the ratio of their altitudes. Draw the altitude from to .
.
Since , then . So the ratio of the two altitudes is
See also
2005 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
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 |