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

m (Solution 6 (No Trig))
(Added new solution, similar to some of the others, but perhaps simpler.)
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=== Solution 1 ===
 
=== Solution 1 ===
 +
 +
As with some of the other solutions, we analyze this with a locus&mdash;but a different one.  We'll consider: given a point <math>P</math> and a line <math>\ell,</math> what is the set of points <math>X</math> such that the midpoint of <math>PX</math> lies on line <math>\ell</math>?  The answer to this question is: a line <math>m</math> parallel to <math>\ell</math>, such that <math>m</math> and <math>P</math> are (1) on opposite sides of <math>\ell</math>. and (2) at the same distance from <math>\ell</math>.
 +
 +
<center><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;
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draw(L1--L2);
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draw(M1--M2);
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draw(P--X, dotted);
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dot(M);
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dot("$P$", P, S);
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dot("$X$", X, N);
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label("$\ell$", L2, E);
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label("$m$", M2, E);
 +
</asy></center>
 +
 +
Applied to this problem, this means that <math>D</math> is the only point that lies on both (1) the given circle, and (2) the line through <math>D</math> parallel to <math>BC</math>.  This means that <math>BC</math> is parallel to the tangent to the given circle at <math>D</math>.
 +
 +
If we take <math>O</math> to be the center of the given circle, then this means that <math>OD</math> is perpendicular to <math>BC</math>.  Let <math>M</math> be the midpoint of chord <math>BC,</math> and let <math>N</math> be the intersection of <math>OD</math> with the line through <math>A</math> parallel to <math>BC</math>.
 +
 +
<center><asy>
 +
size(170);
 +
pair O = (0,0), D = (0, 5), B = (-3, 4), C = (3, 4), A = (-4, 3), EE = (4, 3);
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pair M = (B+C)/2, NN = (A+EE)/2;
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draw(circle(O, 5));
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draw(O--D);
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draw(B--C);
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draw(A--EE);
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draw(B--O--A);
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dot("$O$", O, SE);
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label("$D$", D, N);
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label("$B$", B, WNW);
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label("$A$", A, WNW);
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label("$C$", C, ENE);
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label("$M$", M, NE);
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label("$N$", NN, SE);</asy></center>
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 +
Since <math>BC = 6,</math> we know that <math>BM = 3</math>; since <math>OB</math> (a radius of the circle) is 5, we can conclude that <math>\triangle BMO</math> is a 3-4-5 right triangle.  Since <math>D</math> and line <math>AN</math> are equidistant from line <math>BC,</math> we know that <math>MN = 1</math>, and thus <math>ON = 3</math>.  This makes <math>\triangle ANO</math> also a 3-4-5 right triangle.
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 +
We're looking for <math>\sin \angle AOB</math>, and we can find that using the [[Trigonometric_identities#Angle_addition_identities | angle subtraction formula]] for sine.  We have
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<cmath> \begin{align*}
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\sin \angle AOB &= \sin(\angle AOM - \angle BOM) \
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&= \sin \angle AOM \cos \angle BOM - \cos \angle AOM \sin \angle BOM \
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&= \frac{4}{5} \cdot \frac{4}{5}  - \frac{3}{5} \cdot \frac{3}{5} \
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&= \frac{16 - 9}{25} = \frac{7}{25}. \end{align*} </cmath>
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This is in lowest terms, so our answer is <math>mn = 7 \cdot 25 = 175</math>.
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 +
=== Solution 2 ===
 
-Credit to Adamz for diagram-
 
-Credit to Adamz for diagram-
 
<asy>
 
<asy>
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Next, notice that <math>\angle AOB = \angle AOM - \angle BOM</math>. We can therefore apply the subtraction formula for <math>\tan</math> to obtain <cmath>\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}</cmath> It follows that <math>\sin \angle AOB =\frac{7}{\sqrt{7^2+24^2}} = \frac{7}{25}</math>, such that the answer is <math>7 \cdot 25=\boxed{175}</math>.
 
Next, notice that <math>\angle AOB = \angle AOM - \angle BOM</math>. We can therefore apply the subtraction formula for <math>\tan</math> to obtain <cmath>\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}</cmath> It follows that <math>\sin \angle AOB =\frac{7}{\sqrt{7^2+24^2}} = \frac{7}{25}</math>, such that the answer is <math>7 \cdot 25=\boxed{175}</math>.
  
=== Solution 2 ===
+
=== Solution 3 ===
  
This solution, while similar to Solution 1, is far more motivated and less contrived.
+
This solution, while similar to Solution 2, is arguably more motivated and less contrived.
  
 
Firstly, we note the statement in the problem that "<math>AD</math> is the only chord starting at <math>A</math> and bisected by <math>BC</math>" &ndash; what is its significance? What is the criterion for this statement to be true?
 
Firstly, we note the statement in the problem that "<math>AD</math> is the only chord starting at <math>A</math> and bisected by <math>BC</math>" &ndash; what is its significance? What is the criterion for this statement to be true?
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Thus the answer is <math>7\cdot25=\boxed{175}</math>.
 
Thus the answer is <math>7\cdot25=\boxed{175}</math>.
  
=== Solution 3 (coordinate geometry) ===
+
=== Solution 4 (coordinate geometry) ===
 
  [[File:Aime1983p15s2.png|500px|link=]]
 
  [[File:Aime1983p15s2.png|500px|link=]]
  
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It is obvious that this is in fact non-negative. If it is actually zero, then <math>\sin \alpha = \frac{3}{5}</math>, and <math>\cos \alpha = \frac{4}{5}</math>. In this case, <math>p = \frac{5\cos \alpha}{2} = 2 \in (-3,3)</math>, so we have found a possible solution. We thus calculate <math>\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}</math> by the subtraction formula for <math>\sin</math>. This means that the answer is <math>7 \cdot 25 = 175</math>.
 
It is obvious that this is in fact non-negative. If it is actually zero, then <math>\sin \alpha = \frac{3}{5}</math>, and <math>\cos \alpha = \frac{4}{5}</math>. In this case, <math>p = \frac{5\cos \alpha}{2} = 2 \in (-3,3)</math>, so we have found a possible solution. We thus calculate <math>\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}</math> by the subtraction formula for <math>\sin</math>. This means that the answer is <math>7 \cdot 25 = 175</math>.
  
=== Addendum to Solution 3 ===
+
=== 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.
 
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.
  
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The only way for this inequality to be satisfied is when <math>A=B</math> (by applying the Cauchy-Schwarz inequality, or just plotting the line <math>3x+4y=5</math> to see that point <math>A</math> can't be above this line), which does not make sense in the original problem statement. (For it would mean that the point <math>A</math> lies in the half-plane above the line <math>3x+4y=5</math>, inclusive, and the half-plane below the line <math>-3x+4y=5</math>, 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.)
 
The only way for this inequality to be satisfied is when <math>A=B</math> (by applying the Cauchy-Schwarz inequality, or just plotting the line <math>3x+4y=5</math> to see that point <math>A</math> can't be above this line), which does not make sense in the original problem statement. (For it would mean that the point <math>A</math> lies in the half-plane above the line <math>3x+4y=5</math>, inclusive, and the half-plane below the line <math>-3x+4y=5</math>, 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===
+
===Solution 5===
 
Let the center of the circle be <math>O</math>. Fix <math>B,C,</math> and <math>A</math>. Then, as <math>D</math> moves around the circle, the locus of the midpoints of <math>AD</math> is clearly a circle. Since the problems gives that <math>AD</math> is the only chord starting at <math>A</math> bisected by <math>BC</math>, it follows that the circle with diameter <math>DO</math> and <math>AO</math> is tangent to <math>BC</math>.
 
Let the center of the circle be <math>O</math>. Fix <math>B,C,</math> and <math>A</math>. Then, as <math>D</math> moves around the circle, the locus of the midpoints of <math>AD</math> is clearly a circle. Since the problems gives that <math>AD</math> is the only chord starting at <math>A</math> bisected by <math>BC</math>, it follows that the circle with diameter <math>DO</math> and <math>AO</math> is tangent to <math>BC</math>.
  
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<cmath>DC^2 = (\sqrt{5})^2 + 5^2 -10\sqrt{5} \cdot \frac{2\sqrt{5}}{5} = 10</cmath>Therefore, <math>DC = \sqrt{10} \Longleftrightarrow BA = \sqrt{2}</math>. Now, drop the altitude from <math>O</math> to <math>BA</math> and call its intersection with <math>BA</math> <math>Z</math>. Then, by the Pythagorean theorem, <math>OZ = \frac{7\sqrt{2}}{2}</math>. Thus, <math>\sin (BOZ) = \frac{\sqrt{2}}{10}</math> and <math>\cos (BOZ) = \frac{7\sqrt{2}}{10}</math>. As a result, <math>\sin (BOA) = \sin (2 BOZ) = 2\sin(BOZ)\cos(BOZ) = \frac{7}{25}</math>. <math>7 \cdot 25 = \boxed{175}</math>.
 
<cmath>DC^2 = (\sqrt{5})^2 + 5^2 -10\sqrt{5} \cdot \frac{2\sqrt{5}}{5} = 10</cmath>Therefore, <math>DC = \sqrt{10} \Longleftrightarrow BA = \sqrt{2}</math>. Now, drop the altitude from <math>O</math> to <math>BA</math> and call its intersection with <math>BA</math> <math>Z</math>. Then, by the Pythagorean theorem, <math>OZ = \frac{7\sqrt{2}}{2}</math>. Thus, <math>\sin (BOZ) = \frac{\sqrt{2}}{10}</math> and <math>\cos (BOZ) = \frac{7\sqrt{2}}{10}</math>. As a result, <math>\sin (BOA) = \sin (2 BOZ) = 2\sin(BOZ)\cos(BOZ) = \frac{7}{25}</math>. <math>7 \cdot 25 = \boxed{175}</math>.
  
=== Solution 5 ===
+
=== Solution 6 ===
  
 
[[Image:Dgram.png|thumb|none|800px]]
 
[[Image:Dgram.png|thumb|none|800px]]
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-Alexlikemath
 
-Alexlikemath
  
=== Solution 6 (No Trig) ===
+
=== Solution 7 (No Trig) ===
 
Let <math>O</math> be the center of the circle. The locus of midpoints of chords with <math>A</math> as a endpoint is a circle with diameter <math>\overline{AO}</math>. Additionally, this circle must be tangent to <math>\overline{BC}</math>. Let the center of this circle be <math>P</math>. Let <math>M</math> be the midpoint of <math>BC</math>, <math>N</math> be the foot of the perpendicular from <math>P</math> to <math>\overline{BM}</math>, and <math>K</math> be the foot of the perpendicular from <math>B</math> to <math>\overline{AP}</math>. Let <math>x=BK</math>.
 
Let <math>O</math> be the center of the circle. The locus of midpoints of chords with <math>A</math> as a endpoint is a circle with diameter <math>\overline{AO}</math>. Additionally, this circle must be tangent to <math>\overline{BC}</math>. Let the center of this circle be <math>P</math>. Let <math>M</math> be the midpoint of <math>BC</math>, <math>N</math> be the foot of the perpendicular from <math>P</math> to <math>\overline{BM}</math>, and <math>K</math> be the foot of the perpendicular from <math>B</math> to <math>\overline{AP}</math>. Let <math>x=BK</math>.
  

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)

Aime1983p15s2.png

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

Dgram.png

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

See Also

1983 AIME (ProblemsAnswer KeyResources)
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