Difference between revisions of "1983 AIME Problems/Problem 15"

Revision as of 01:42, 3 April 2022

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

As with some of the other solutions, we analyze this with a locus—but a different one. We'll consider: given a point $P$ and a line $\ell,$ what is the set of points $X$ such that the midpoint of $PX$ lies on line $\ell$ ? The answer to this question is: a line $m$ parallel to $\ell$ , such that $m$ and $P$ are (1) on opposite sides of $\ell$ . and (2) at the same distance from $\ell$ .

$[asy] size(170); pair P = (0,0), L1 = (-3, 0), L2 = (1.5, 1.5), M1 = (-3, 1), M2 = (1.5, 2.5); pair X = .6*M1 + .4*M2, M = (P+X)/2; draw(L1--L2); draw(M1--M2); draw(P--X, dotted); dot(M); dot("$P$", P, S); dot("$X$", X, N); label("$\ell$", L2, E); label("$m$", M2, E); [/asy]$

Applied to this problem, this means that $D$ is the only point that lies on both (1) the given circle, and (2) the line through $D$ parallel to $BC$ . This means that $BC$ is parallel to the tangent to the given circle at $D$ .

If we take $O$ to be the center of the given circle, then this means that $OD$ is perpendicular to $BC$ . Let $M$ be the midpoint of chord $BC,$ and let $N$ be the intersection of $OD$ with the line through $A$ parallel to $BC$ .

$[asy] size(170); pair O = (0,0), D = (0, 5), B = (-3, 4), C = (3, 4), A = (-4, 3), EE = (4, 3); pair M = (B+C)/2, NN = (A+EE)/2; draw(circle(O, 5)); draw(O--D); draw(B--C); draw(A--EE); draw(B--O--A); dot("$O$", O, SE); label("$D$", D, N); label("$B$", B, WNW); label("$A$", A, WNW); label("$C$", C, ENE); label("$M$", M, NE); label("$N$", NN, SE);[/asy]$

Since $BC = 6,$ we know that $BM = 3$ ; since $OB$ (a radius of the circle) is 5, we can conclude that $\triangle BMO$ is a 3-4-5 right triangle. Since $D$ and line $AN$ are equidistant from line $BC,$ we know that $MN = 1$ , and thus $ON = 3$ . This makes $\triangle ANO$ also a 3-4-5 right triangle.

We're looking for $\sin \angle AOB$ , and we can find that using the angle subtraction formula for sine. We have $\begin{align*} \sin \angle AOB &= \sin(\angle AOM - \angle BOM) \\ &= \sin \angle AOM \cos \angle BOM - \cos \angle AOM \sin \angle BOM \\ &= \frac{4}{5} \cdot \frac{4}{5} - \frac{3}{5} \cdot \frac{3}{5} \\ &= \frac{16 - 9}{25} = \frac{7}{25}. \end{align*}$ This is in lowest terms, so our answer is $mn = 7 \cdot 25 = 175$ .

Solution 2

-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 3

This solution, while similar to Solution 2, is arguably 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 4 (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 4

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 5

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, $ED^2 + EO^2 = 25$ , 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 6

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.

Now, create a coordinate system such that the origin is O and the y-axis is perpendicular to BC. The positive x direction is from B to C.

Let Z be (0,5). Let Y be (-5,0). 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 $sin(XOY) = sin(AOY) = \frac{3}{5}$ . $sin(BOZ) = \frac{3}{5}$ . 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 $\frac{7}{25} \implies \boxed{175}$

-Alexlikemath

Solution 7 (No Trig)

Let $O$ be the center of the circle. The locus of midpoints of chords with $A$ as a endpoint is a circle with diameter $\overline{AO}$ . Additionally, this circle must be tangent to $\overline{BC}$ . Let the center of this circle be $P$ . Let $M$ be the midpoint of $BC$ , $N$ be the foot of the perpendicular from $P$ to $\overline{BM}$ , and $K$ be the foot of the perpendicular from $B$ to $\overline{AP}$ . Let $x=BK$ .

From right triangle $BKO$ , we get $KO = \sqrt{25-x^2}$ . Thus, $KP = \sqrt{25-x^2}-\frac52$ .

Since $BO = 5$ , $BM = 3$ , and $\angle BMO$ is right, $MO=4$ . From quadrilateral $MNPO$ , we get $MN = \sqrt{PO^2 - (MO - NP)^2} = \sqrt{(5/2)^2 - (4 - 5/2)^2} = \sqrt{(5/2)^2 - (3/2)^2} = 2$ . Thus, $BN = 1$ .

Since angles $BNP$ and $BKP$ are right, we get $BK^2+KP^2 = BN^2 + NP^2 \implies x^2 + \left(\sqrt{25-x^2}-\frac52\right)^2 = \left(\frac52\right)^2 + 1$ $25 - 5\sqrt{25-x^2} = 1$ $5\sqrt{25-x^2} = 24$ $25(25-x^2) = 24^2$ $25x^2 = 25^2 - 24^2 = 49$ $x = \frac75$ Thus, $\sin \angle AOB = \frac{x}{5} = \frac{7}{25}\implies \boxed{175}$ .

~rayfish

1983 AIME (Problems • Answer Key • Resources)
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