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1983 AIME Problems/Problem 15: Difference between revisions

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== Problem ==
== Problem ==
The adjoining figure shows two intersecting [[chord]]s in a [[circle]], with <math>B</math> on minor arc <math>AD</math>. Suppose that the radius of the circle is <math>5</math>, that <math>BC=6</math>, and that <math>AD</math> is [[bisect]]ed by <math>BC</math>. Suppose further that <math>AD</math> is the only chord starting at <math>A</math> which is bisected by <math>BC</math>. It follows that the [[sine]] of the minor arc <math>AB</math> is a rational number. If this fraction is expressed as a fraction <math>\frac{m}{n}</math> in lowest terms, what is the product <math>mn</math>?
The adjoining figure shows two intersecting [[chord]]s in a [[circle]], with <math>B</math> on minor arc <math>AD</math>. Suppose that the radius of the circle is <math>5</math>, that <math>BC=6</math>, and that <math>AD</math> is [[bisect]]ed by <math>BC</math>. Suppose further that <math>AD</math> is the only chord starting at <math>A</math> which is bisected by <math>BC</math>. It follows that the [[sine]] of the minor arc <math>AB</math> is a rational number. If this fraction is expressed as a fraction <math>\frac{m}{n}</math> in lowest terms, what is the product <math>mn</math>?
//Credit to Adamz//
<asy>size(100);
<asy>size(100);
defaultpen(linewidth(.8pt)+fontsize(11pt));
defaultpen(linewidth(.8pt)+fontsize(11pt));
Line 24: Line 25:


== Solution ==
== Solution ==
Let <math>A</math> be any fixed point on [[circle]] <math>O</math> and let <math>AD</math> be a [[chord]] of circle <math>O</math>. The [[locus]] of [[midpoint]]s <math>N</math> of the chord <math>AD</math> is a circle <math>P</math>, with diameter <math>AO</math>. Generally, the circle <math>P</math> can intersect the chord <math>BC</math> at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle <math>P</math> is tangent to BC at point N.
Let M be the midpoint of the chord <math>BC</math>. From [[right triangle]] <math>OMB</math>, <math>OM = \sqrt{OB^2 - BM^2} =4</math>. Thus, <math>\tan \angle BOM = \frac{BM}{OM} = \frac 3 4</math>.
Notice that the distance <math>OM</math> equals <math>PN + PO \cos AOM = r(1 + \cos AOM)</math> (Where <math>r</math> is the radius of circle P). Evaluating this, <math>\cos \angle AOM = \frac{OM}{r} - 1 = \frac{2OM}{R} - 1 = \frac 8 5 - 1 = \frac 3 5</math>. From <math>\cos \angle AOM</math>, we see that <math>\tan \angle AOM =\frac{\sqrt{1 - \cos^2 \angle AOM}}{\cos \angle AOM} = \frac{\sqrt{5^2 - 3^2}}{3} = \frac 4 3</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>
<asy>
size(10cm);
size(10cm);
Line 53: Line 45:
pair D=(5,0); dot(D); label("$D$",D,E);
pair D=(5,0); dot(D); label("$D$",D,E);
draw(A--D);
draw(A--D);
</asy>
</asy>Let <math>A</math> be any fixed point on [[circle]] <math>O</math> and let <math>AD</math> be a [[chord]] of circle <math>O</math>. The [[locus]] of [[midpoint]]s <math>N</math> of the chord <math>AD</math> is a circle <math>P</math>, with diameter <math>AO</math>. Generally, the circle <math>P</math> can intersect the chord <math>BC</math> at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle <math>P</math> is tangent to BC at point N.
 
Let M be the midpoint of the chord <math>BC</math>. From [[right triangle]] <math>OMB</math>, <math>OM = \sqrt{OB^2 - BM^2} =4</math>. Thus, <math>\tan \angle BOM = \frac{BM}{OM} = \frac 3 4</math>.
 
Notice that the distance <math>OM</math> equals <math>PN + PO \cos AOM = r(1 + \cos AOM)</math> (Where <math>r</math> is the radius of circle P). Evaluating this, <math>\cos \angle AOM = \frac{OM}{r} - 1 = \frac{2OM}{R} - 1 = \frac 8 5 - 1 = \frac 3 5</math>. From <math>\cos \angle AOM</math>, we see that <math>\tan \angle AOM =\frac{\sqrt{1 - \cos^2 \angle AOM}}{\cos \angle AOM} = \frac{\sqrt{5^2 - 3^2}}{3} = \frac 4 3</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 1.5 [Motivation]==


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.
The above solution works, but is quite messy and somewhat difficult to follow. This solution provides 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?
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?

Revision as of 22:21, 25 May 2016

Problem

The adjoining figure shows two intersecting chords in a circle, with $B$ on minor arc $AD$. Suppose that the radius of the circle is $5$, that $BC=6$, and that $AD$ is bisected by $BC$. Suppose further that $AD$ is the only chord starting at $A$ which is bisected by $BC$. It follows that the sine of the minor arc $AB$ is a rational number. If this fraction is expressed as a fraction $\frac{m}{n}$ in lowest terms, what is the product $mn$? //Credit to Adamz// [asy]size(100); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=1; pair O1=(0,0); pair A=(-0.91,-0.41); pair B=(-0.99,0.13); pair C=(0.688,0.728); pair D=(-0.25,0.97); path C1=Circle(O1,1); draw(C1); label("$A$",A,W); label("$B$",B,W); label("$C$",C,NE); label("$D$",D,N); draw(A--D); draw(B--C); pair F=intersectionpoint(A--D,B--C); add(pathticks(A--F,1,0.5,0,3.5)); add(pathticks(F--D,1,0.5,0,3.5)); [/asy]

Solution

[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]Let $A$ be any fixed point on circle $O$ and let $AD$ be a chord of circle $O$. The locus of midpoints $N$ of the chord $AD$ is a circle $P$, with diameter $AO$. Generally, the circle $P$ can intersect the chord $BC$ at two points, one point, or they may not have a point of intersection. By the problem condition, however, the circle $P$ is tangent to BC at point N.

Let M be the midpoint of the chord $BC$. From right triangle $OMB$, $OM = \sqrt{OB^2 - BM^2} =4$. Thus, $\tan \angle BOM = \frac{BM}{OM} = \frac 3 4$.

Notice that the distance $OM$ equals $PN + PO \cos AOM = r(1 + \cos AOM)$ (Where $r$ is the radius of circle P). Evaluating this, $\cos \angle AOM = \frac{OM}{r} - 1 = \frac{2OM}{R} - 1 = \frac 8 5 - 1 = \frac 3 5$. From $\cos \angle AOM$, we see that $\tan \angle AOM =\frac{\sqrt{1 - \cos^2 \angle AOM}}{\cos \angle AOM} = \frac{\sqrt{5^2 - 3^2}}{3} = \frac 4 3$

Next, notice that $\angle AOB = \angle AOM - \angle BOM$. We can therefore apply the tangent subtraction formula to obtain , $\tan AOB =\frac{\tan AOM - \tan BOM}{1 + \tan AOM \cdot \tan AOM} =\frac{\frac 4 3 - \frac 3 4}{1 + \frac 4 3 \cdot \frac 3 4} = \frac{7}{24}$. It follows that $\sin AOB =\frac{7}{\sqrt{7^2+24^2}} = \frac{7}{25}$, resulting in an answer of $7 \cdot 25=\boxed{175}$.

Solution 1.5 [Motivation]

The above solution works, but is quite messy and somewhat difficult to follow. This solution provides the motivation behind the solution.

First of all, where did the statement "$AD$ is the only chord starting at $A$ and bisected by $BC$ " 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 $A$. It is well known that this is the circle with diameter $AO$, where $O$ is the center of the circle. The proof is simple: every midpoint of a chord is a dilation of the endpoint with ratio $1/2$ with center $A$. Thus, the locus is the result of the dilation with ratio $1/2$ of circle $O$ with center $A$. Let the center of this circle be $P$.

Aha! Now we see. $AD$ is bisected by $BC$ if they cross at some point $N$ on the circle. Moreover, since $AD$ is the only chord, $BC$ must be tangent to the circle $P$.

The rest of this problem is straight forward.

Our goal is to find $\sin AOB = \sin (AOM - BOM)$ where $M$ is the midpoint of $BC$. Then we have $BM=3$ and $OM=4$. Let $R$ be the projection of $A$ onto $OM$, and similarly $Q$ be the projection of $P$ onto $OM$. Then it remains to find $AR$ so we can use the sine addition formula.

As $PN$ is a radius of circle $P$, $PN=2.5$, and similarly, $PO=2.5$. Since $OM=4$, $OQ=OM-QM=OM-PN=4-2.5=1.5$. Thus, $PQ=\sqrt{(2.5)^2-1.5^2}=2$.

From here, we see that $\triangle OAR$ is a dilation of $\triangle OPQ$ about center $O$ with ratio $2$, so $AR=2PQ=4$.

Lastly, we apply the formula: \[\sin (AOM - BOM) = \sin AOM \cos BOM - \sin BOM \cos AOM = (4/5)(4/5)-(3/5)(3/5)=7/25.\]

Thus, our answer is $7\cdot25=\boxed{175}$.

See Also

1983 AIME (Problems • Answer Key • Resources)
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