Difference between revisions of "1983 AIME Problems/Problem 15"
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Next, notice that <math>\angle AOB = \angle AOM - \angle BOM</math>. We can therefore apply the tangent subtraction formula to obtain , <math>\tan AOB =\frac{\tan AOM - \tan BOM}{1 + \tan AOM \cdot \tan AOM} =\frac{\frac 4 3 - \frac 3 4}{1 + \frac 4 3 \cdot \frac 3 4} = \frac{7}{24}</math>. It follows that <math>\sin AOB =\frac{7}{\sqrt{7^2+24^2}} = \frac{7}{25}</math>, resulting in an answer of <math>7 \cdot 25=\boxed{175}</math>. | Next, notice that <math>\angle AOB = \angle AOM - \angle BOM</math>. We can therefore apply the tangent subtraction formula to obtain , <math>\tan AOB =\frac{\tan AOM - \tan BOM}{1 + \tan AOM \cdot \tan AOM} =\frac{\frac 4 3 - \frac 3 4}{1 + \frac 4 3 \cdot \frac 3 4} = \frac{7}{24}</math>. It follows that <math>\sin AOB =\frac{7}{\sqrt{7^2+24^2}} = \frac{7}{25}</math>, resulting in an answer of <math>7 \cdot 25=\boxed{175}</math>. | ||
+ | |||
+ | == Solution 2 == | ||
+ | <asy> | ||
+ | size(10cm); | ||
+ | import olympiad; | ||
+ | pair O = (0,0);dot(O);label("$O$",O,SW); | ||
+ | pair M = (4,0);dot(M);label("$M$",M,SE); | ||
+ | pair N = (4,2);dot(N);label("$N$",N,NE); | ||
+ | draw(circle(O,5)); | ||
+ | pair B = (4,3);dot(B);label("$B$",B,NE); | ||
+ | pair C = (4,-3);dot(C);label("$C$",C,SE); | ||
+ | draw(B--C);draw(O--M); | ||
+ | pair P = (1.5,2);dot(P);label("$P$",P,W); | ||
+ | draw(circle(P,2.5)); | ||
+ | pair A=(3,4);dot(A);label("$A$",A,NE); | ||
+ | draw(O--A); | ||
+ | draw(O--B); | ||
+ | pair Q = (1.5,0); dot(Q); label("$Q$",Q,S); | ||
+ | pair R = (3,0); dot(R); label("$R$",R,S); | ||
+ | draw(P--Q,dotted); draw(A--R,dotted); | ||
+ | pair D=(5,0); dot(D); label("$D$",D,E); | ||
+ | draw(A--D); | ||
+ | </asy> | ||
+ | |||
+ | The above solution works, but is quite messy and somewhat difficult to follow. This solution provides a diagram to scale, and the motivation behind the solution. | ||
+ | |||
+ | First of all, where did the statement "<math>AD</math> is the only chord starting at <math>A</math> and bisected by <math>BC</math> " come from? What is its significance in this problem? What is the criterion for this statement to be true? | ||
+ | |||
+ | We consider the locus of midpoints of the chords from <math>A</math>. It is well known that this is the circle with diameter <math>AO</math>, where <math>O</math> is the center of the circle. The proof is simple: every midpoint of a chord is a dilation of the endpoint with ratio <math>1/2</math> with center <math>A</math>. Thus, the locus is the result of the dilation with ratio <math>1/2</math> of circle <math>O</math> with center <math>A</math>. Let the center of this circle be <math>P</math>. | ||
+ | |||
+ | Aha! Now we see. <math>AD</math> is bisected by <math>BC</math> if they cross at some point <math>N</math> on the circle. Moreover, since <math>AD</math> is the only chord, <math>BC</math> must be tangent to the circle <math>P</math>. | ||
+ | |||
+ | The rest of this problem is straight forward. | ||
+ | |||
+ | Our goal is to find <math>\sin AOB = \sin (AOM - BOM)</math> where <math>M</math> is the midpoint of <math>BC</math>. Then we have <math>BM=3</math> and <math>OM=4</math>. | ||
+ | Let <math>R</math> be the projection of <math>A</math> onto <math>OM</math>, and similarly <math>Q</math> be the projection of <math>P</math> onto <math>OM</math>. Then it remains to find <math>AR</math> so we can use the sine addition formula. | ||
+ | |||
+ | As <math>PN</math> is a radius of circle <math>P</math>, <math>PN=2.5</math>, and similarly, <math>PO=2.5</math>. Since <math>OM=4</math>, <math>OQ=OM-QM=OM-PN=4-2.5=1.5</math>. Thus, <math>PQ=\sqrt{(2.5)^2-1.5^2}=2</math>. | ||
+ | |||
+ | From here, we see that <math>\triangle OAR</math> is a dilation of <math>\triangle OPQ</math> about center <math>O</math> with ratio <math>2</math>, so <math>AR=2PQ=4</math>. | ||
+ | |||
+ | Lastly, we apply the formula: | ||
+ | <cmath> \sin (AOM - BOM) = \sin AOM \cos BOM - \sin BOM \cos AOM = (4/5)(4/5)-(3/5)(3/5)=7/25.</cmath> | ||
+ | |||
+ | Thus, our answer is <math>7*25=\boxed{175}</math>. | ||
== See Also == | == See Also == |
Revision as of 01:18, 17 May 2015
Contents
[hide]Problem
The adjoining figure shows two intersecting chords in a circle, with on minor arc . Suppose that the radius of the circle is , that , and that is bisected by . Suppose further that is the only chord starting at which is bisected by . It follows that the sine of the minor arc is a rational number. If this fraction is expressed as a fraction in lowest terms, what is the product ?
Solution
Let be any fixed point on circle and let be a chord of circle . The locus of midpoints of the chord is a circle , with diameter . Generally, the circle can intersect the chord at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle is tangent to BC at point N.
Let M be the midpoint of the chord . From right triangle , . Thus, .
Notice that the distance equals (Where is the radius of circle P). Evaluating this, . From , we see that
Next, notice that . We can therefore apply the tangent subtraction formula to obtain , . It follows that , resulting in an answer of .
Solution 2
The above solution works, but is quite messy and somewhat difficult to follow. This solution provides a diagram to scale, and the motivation behind the solution.
First of all, where did the statement " is the only chord starting at and bisected by " come from? What is its significance in this problem? What is the criterion for this statement to be true?
We consider the locus of midpoints of the chords from . It is well known that this is the circle with diameter , where is the center of the circle. The proof is simple: every midpoint of a chord is a dilation of the endpoint with ratio with center . Thus, the locus is the result of the dilation with ratio of circle with center . Let the center of this circle be .
Aha! Now we see. is bisected by if they cross at some point on the circle. Moreover, since is the only chord, must be tangent to the circle .
The rest of this problem is straight forward.
Our goal is to find where is the midpoint of . Then we have and . Let be the projection of onto , and similarly be the projection of onto . Then it remains to find so we can use the sine addition formula.
As is a radius of circle , , and similarly, . Since , . Thus, .
From here, we see that is a dilation of about center with ratio , so .
Lastly, we apply the formula:
Thus, our answer is .
See Also
1983 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 14 |
Followed by Last question | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |