1983 AIME Problems/Problem 15

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 central angle of minor arc $AB$ is a rational number. If this number is expressed as a fraction $\frac{m}{n}$ in lowest terms, what is the product $mn$ ?

$[asy]size(100); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=1; pair O1=(0,0); pair A=(-0.91,-0.41); pair B=(-0.99,0.13); pair C=(0.688,0.728); pair D=(-0.25,0.97); path C1=Circle(O1,1); draw(C1); label("$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 1

-Credit to Adamz for diagram- $[asy] size(10cm); import olympiad; pair O = (0,0);dot(O);label("$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 $7 \cdot 25=\boxed{175}$ .

Motivation Behind Solution 1

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 $7\cdot25=\boxed{175}$ .

Solution 2 (with the help of coordinates)

Let the circle be $x^2 + y^2 = 25$ , and its center be labeled $O=(0,0)$ . Since BC=6, we can calculate (by the Pythagorean Theorem) that the distance from $O$ to the line $BC$ is $4$ . So we can let $B=(3,4)$ and $C=(-3,4)$ . Now assume that $A$ is any point on the major arc BC, and $D$ any point on the minor arc BC. We may let $A=(5 \cos \alpha, 5 \sin \alpha)$ , where $\alpha$ is the angle measured from the positive $x$ axis to the ray $OA$ . It will also be convenient to define $\alpha_0$ as the measure of angle XOB.

Firstly, since B must lie in the minor arc AD, we see that $\alpha \in \left(-\frac{\pi}{2}-\alpha_0,\alpha_0\right)$ . However, since the midpoint of AD must lie on BC, and the highest $y$ coordinate of $D$ is $5$ , we see that the $y$ coordinate can't be lower than $3$ , that is, $\alpha \in \left[\sin^{-1}\frac{3}{5},\alpha_0\right)$ .

Secondly, there is a theorem that says that, in a circle, if a chord is bisected by a radius, then the they must be perpendicular. Therefore, suppose that $P$ is the intersection between $AD$ and $BC$ , then $OP$ is perpendicular to $AD$ . So, if $AD$ is the only chord starting at $A$ which is bisected by $BC$ , it means that $P$ is the only point on the chord $BC$ such that $OP$ is perpendicular to $AD$ . Now we are ready to make it an algebraic problem. Suppose $P=(p,4)$ , $p \in (-3,3)$ . The statement $OP$ is perpendicular to $AD$ is equivalent to the following equation:

$-1 = (slope OP)(slope AP)$ $-1 = \frac{4}{p} \cdot \frac{5\sin \alpha - 4}{5\cos \alpha - p}$

It rearranges to the following:

$p^2 - (5\cos \alpha)p + 16 - 20 \sin \alpha = 0$

Given that this equation has only one real root $p\in (-3,3)$ , we study the following function:

$f(x) = x^2 - (5\cos \alpha)x + 16 - 20 \sin \alpha$

First, by the fact that $f(x)$ has real solution, it is good to look at its discriminant: must be non-negative:

$\Delta = (5\cos \alpha)^2 - 4(16-20\sin \alpha)$ $= 25 (1- \sin^2 \alpha) - 64 + 80 \sin \alpha$ $= -25 \sin^2 \alpha + 80\sin \alpha - 39$ $= (13 - 5\sin \alpha)(5\sin \alpha - 3)$

It is obvious that this is non-negative. If it is actually zero, then $\sin \alpha = \frac{3}{5}$ , and $\cos \alpha = \frac{4}{5}$ . In this case, $p = (5\cos \alpha)/2 = 2 \in (-3,3)$ . We found a possible case. So we calculate $\sin(arc AB) = \sin (\alpha_0 - \alpha) = \frac{4}{5}\cdot \frac{4}{5} - \frac{3}{5} \cdot \frac{3}{5} = \frac{7}{25}$ .

Note to Solution 2

Note: As an AIME problem, it is already done since we have found one possible case. However, it takes one more step to complete it if we need to say that this is the unique possibility, without appealing to the AIME Uniqueness Principle.

Suppose that $\Delta > 0$ , which means that there can be two real roots of $f(x)$ , one lying in the interval $(-3,3)$ , but another falling out of it. We also see that the average of the two root is $(5\cos \alpha) / 2$ , which is a quantity greater than 0, so the root outside of $(-3,3)$ must be no less than 3. Spectating the parabolic curve of $f(x)$ , which is a "U shaped" curve hitting the interval $(-3,3)$ once and $[3,\infty)$ another time, it is evident that $f(-3) > 0$ and $f(3)\leq 0$ . We can just work on the second one:

$0 \geq f(3) = 25 - 15\cos \alpha - 20 \sin \alpha$ $3\cos \alpha + 4 \sin \alpha \geq 5$

The only way to satisfy this equation is when $A=B$ (by working on Cauchy-Schwarz inequality, or just plotting the line $3x+4y=5$ to see that point $A$ can't go above this line), which does not make sense from the description of the problem. It means that the point $A$ lies in the half plane above the line $3x+4y=5$ , inclusive, and the half plane below the line $-3x+4y=5$ , exclusive. It is obviously impossible, by drawing the lines and see that the intersection of the two half planes does not share any point with the circle.

1983 AIME (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Last question
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