Difference between revisions of "1983 AIME Problems/Problem 14"
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== Problem == | == Problem == | ||
− | In the adjoining figure, two circles with radii <math> | + | In the adjoining figure, two circles with radii <math>8</math> and <math>6</math> are drawn with their centers <math>12</math> units apart. At <math>P</math>, one of the points of intersection, a line is drawn in such a way that the chords <math>QP</math> and <math>PR</math> have equal length. Find the square of the length of <math>QP</math>. |
<asy> | <asy> | ||
Line 14: | Line 14: | ||
draw(C1); | draw(C1); | ||
draw(C2); | draw(C2); | ||
− | |||
− | |||
− | |||
draw(Q--R); | draw(Q--R); | ||
label("$Q$",Q,N); | label("$Q$",Q,N); | ||
label("$P$",P,dir(80)); | label("$P$",P,dir(80)); | ||
label("$R$",R,E); | label("$R$",R,E); | ||
− | + | </asy> | |
__TOC__ | __TOC__ |
Revision as of 00:25, 17 May 2015
Problem
In the adjoining figure, two circles with radii and are drawn with their centers units apart. At , one of the points of intersection, a line is drawn in such a way that the chords and have equal length. Find the square of the length of .
Contents
Solution
Solution 1
First, notice that if we reflect over we get . Since we know that is on circle and is on circle , we can reflect circle over to get another circle (centered at a new point with radius ) that intersects circle at . The rest is just finding lengths:
Since is the midpoint of segment , is a median of triangle . Because we know that , , and , we can find the third side of the triangle using Stewart's Theorem or similar approaches. We get . So now we have a kite with , , and , and all we need is the length of the other diagonal . The easiest way it can be found is with the Pythagorean Theorem. Let be the length of . Then
Doing routine algebra on the above equation, we find that , so
Solution 2
Draw additional lines as indicated. Note that since triangles and are isosceles, the altitudes are also bisectors, so let .
Since triangles and are similar. If we let , we have .
Applying the Pythagorean Theorem on triangle , we have . Similarly, for triangle , we have .
Subtracting, .
Solution 3
Let . Angles , , and must add up to . By the Law of Cosines, . Also, angles and equal and . So we have
Taking the of both sides and simplifying using the cosine addition identity gives .
Solution 4
Observe that the length of the area where the two circles intersect can be found explicitly as . Let , then the power of point with regards to the larger circle gives
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |