# 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

### 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 $OMB$, we have $OM = \sqrt{OB^2 - BM^2} =4$. This gives $\tan \angle BOM = \frac{BM}{OM} = \frac 3 4$.

Notice that the distance $OM$ equals $PN + PO \cos \angle AOM = r(1 + \cos \angle AOM)$, where $r$ is the radius of circle $P$.

Hence $$\cos \angle AOM = \frac{OM}{r} - 1 = \frac{2OM}{R} - 1 = \frac 8 5 - 1 = \frac 3 5$$ (where $R$ represents the radius, $5$, of the large circle given in the question). Therefore, since $\angle AOM$ is clearly acute, 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 subtraction formula for $\tan$ to obtain $$\tan \angle AOB =\frac{\tan \angle AOM - \tan \angle BOM}{1 + \tan \angle AOM \cdot \tan \angle BOM} =\frac{\frac 4 3 - \frac 3 4}{1 + \frac 4 3 \cdot \frac 3 4} = \frac{7}{24}$$ It follows that $\sin \angle AOB =\frac{7}{\sqrt{7^2+24^2}} = \frac{7}{25}$, such that the answer is $7 \cdot 25=\boxed{175}$.

### Solution 2

This solution, while similar to Solution 1, is far more motivated and less contrived.

Firstly, we note the statement in the problem that "$AD$ is the only chord starting at $A$ and bisected by $BC$" – what is its significance? 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 scale factor $\frac{1}{2}$ and center $A$. Thus, the locus is the result of the dilation with scale factor $\frac{1}{2}$ and centre $A$ of circle $O$. Let the center of this circle be $P$.

Now, $AD$ 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 straightforward.

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

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

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

Lastly, we apply the formula: $$\sin{\left(\angle AOM - \angle BOM\right)} = \sin \angle AOM \cos \angle BOM - \sin \angle BOM \cos \angle AOM = \left(\frac{4}{5}\right)\left(\frac{4}{5}\right)-\left(\frac{3}{5}\right)\left(\frac{3}{5}\right)=\frac{7}{25}$$ Thus the answer is $7\cdot25=\boxed{175}$.

### Solution 3 (coordinate geometry)




Let the circle have equation $x^2 + y^2 = 25$, with centre $O(0,0)$. Since $BC=6$, we can calculate (by the Pythagorean Theorem) that the distance from $O$ to the line $BC$ is $4$. Therefore, 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 can write $A=(5 \cos \alpha, 5 \sin \alpha)$, where $\alpha$ is the angle measured from the positive $x$ axis to the ray $OA$. It will also be convenient to define $\angle XOB = \alpha_0$.

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 possible $y$-coordinate of $D$ is $5$, we see that the $y$-coordinate cannot 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 they must be perpendicular. Therefore, suppose that $P$ is the intersection point of $AD$ and $BC$, so that by the theorem, $OP$ is perpendicular to $AD$. So, if $AD$ is the only chord starting at $A$ which is bisected by $BC$, this means that $P$ is the only point on the chord $BC$ such that $OP$ is perpendicular to $AD$. Now suppose that $P=(p,4)$, where $p \in (-3,3)$. The fact that $OP$ must be perpendicular to $AD$ is equivalent to the following equation:

$$-1 = \left(\text{slope of } OP\right)\left(\text{slope of } AP\right)$$ which becomes $$-1 = \frac{4}{p} \cdot \frac{5\sin \alpha - 4}{5\cos \alpha - p}$$

This rearranges to

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

Given that this equation must have 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 the equation $f(x)=0$ has real solutions, its discriminant $\Delta$ must be non-negative, so we calculate

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

It is obvious that this is in fact non-negative. If it is actually zero, then $\sin \alpha = \frac{3}{5}$, and $\cos \alpha = \frac{4}{5}$. In this case, $p = \frac{5\cos \alpha}{2} = 2 \in (-3,3)$, so we have found a possible solution. We thus calculate $\sin(\text{central angle of minor arc } AB) = \sin (\alpha_0 - \alpha) = \frac{4}{5}\cdot \frac{4}{5} - \frac{3}{5} \cdot \frac{3}{5} = \frac{7}{25}$ by the subtraction formula for $\sin$. This means that the answer is $7 \cdot 25 = 175$.

### Addendum to Solution 3

Since this is an AIME problem, we can assume that we are done since we have found one possible case. However, in a full-solution contest, we could not assume that the answer is unique, but would need to prove that this is the unique solution. This can be proven as follows.

Suppose that $\Delta > 0$, which would mean that there could be two real roots of $f(x)$, one lying in the interval $(-3,3)$, and another outside of it. We also see, by Vieta's Formulas, that the average of the two roots is $\frac{5\cos \alpha}{2}$, which is non-negative, so the root outside of $(-3,3)$ must be no less than $3$. By considering the graph of $y=f(x)$, which is a "U-shaped" parabola, it is now evident that $f(-3) > 0$ and $f(3)\leq 0$. We can just use the second inequality:

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

The only way for this inequality to be satisfied is when $A=B$ (by applying the Cauchy-Schwarz inequality, or just plotting the line $3x+4y=5$ to see that point $A$ can't be above this line), which does not make sense in the original problem statement. (For it would mean 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. This can be seen to be impossible by drawing the lines and observing that the intersection of the two half-planes does not share any point with the circle.)

### Solution 4

Let the center of the circle be $O$. Fix $B,C,$ and $A$. Then, as $D$ moves around the circle, the locus of the midpoints of $AD$ is clearly a circle. Since the problems gives that $AD$ is the only chord starting at $A$ bisected by $BC$, it follows that the circle with diameter $DO$ and $AO$ is tangent to $BC$.

Now, let the intersection of $BC$ and $AD$ be $E$ and let the midpoint of $AO$ (the center of the circle tangent to $BC$ that we described beforehand) be $F$. Drop the altitude from $O$ to $BC$ and call its intersection with $BC$ $K$. Drop the perpendicular from $F$ to $KO$ and call its intersection with $KO$ $L$. Clearly, $KO = \sqrt{OC^2-KC^2} = \sqrt{5^2-3^2} = 4$ and since $EF$ is radius, it equals $\frac{5}{2}$. The same applies for $FO$, which also equals $\frac{5}{2}$. By the Pythagorean theorem, we deduce that $FL = 2$, so $EK = 2$. This is very important information! Now we know that $BE = 1$, so by Power of a Point, $AE = ED = \sqrt{5}$.

We’re almost there! Since by the Pythagorean theorem, $ED^2 + EO^2 = 5$, we deduce that $EO = 2\sqrt{5}$. $EC=OC=5$, so $\sin (CEO) = \frac{2\sqrt{5}}{5}$. Furthermore, since $\sin (CEO) = \cos(DEC)$, we know that $\cos (DEC) = \frac{2\sqrt{5}}{5}$. By the law of cosines, $$DC^2 = (\sqrt{5})^2 + 5^2 -10\sqrt{5} \cdot \frac{2\sqrt{5}}{5} = 10$$Therefore, $DC = \sqrt{10} \Longleftrightarrow BA = \sqrt{2}$. Now, drop the altitude from $O$ to $BA$ and call its intersection with $BA$ $Z$. Then, by the Pythagorean theorem, $OZ = \frac{7\sqrt{2}}{2}$. Thus, $\sin (BOZ) = \frac{\sqrt{2}}{10}$ and $\cos (BOZ) = \frac{7\sqrt{2}}{10}$. As a result, $\sin (BOA) = \sin (2 BOZ) = 2\sin(BOZ)\cos(BOZ) = \frac{7}{25}$. $7 \cdot 25 = \boxed{175}$.

## Solution 5

Let I be the intersection of AD and BC.

Lemma: $AI = ID$ if and only if $\angle AIO = 90$.

Proof: If AI = ID, we get AO = OD, and thus IO is a perpendicular bisector of AD. If $\angle AIO = 90$, We can get $\triangle AIO \cong \triangle OID$

Let be this the circle with diameter AO.

Thus, we get $\angle AIO = 90$, implying I must lie on $\omega$. I also must lie on BC. If both of these conditions are satisfied, The extension of AI intersecting with circle O will create a point D such that AI = ID. Thus, since AD is the only chord starting at A, there must be only 1 possible I, implying the circle with diameter AO is tangent to BC.

Let Y be a point such that $\angle DOY = 90$, and minor arc DY contains A,B. Let X be the center of $\omega$. Since $\omega$'s radius is $\frac{5}{2}$, the altitude from X to OY is 1.5, since the altitude from I to OY is 4. XO is 2.5, so $sin(XOY) = sin(AOY) = \frac{3}{5}$. $sin(BOD) = \frac{3}{5}$. If we let $sin(\theta) = \frac{3}{5}$, we can find that what we are looking for is $sin(90 - 2\theta)$, which we can evaluate and get $\frac{7}{25} \implies \boxed{175}$

## See Also

 1983 AIME (Problems • Answer Key • Resources) Preceded byProblem 14 Followed byLast question 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
Invalid username
Login to AoPS