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

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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 chords 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 bisected 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 central angle of minor arc <math>AB</math> is a rational number. If this number is expressed as a fraction <math>\frac{m}{n}</math> in lowest terms, what is the product <math>mn</math>?
  
[[Image:1983_AIME-15.png]]
+
<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>
 +
<!-- [[Image:1983_AIME-15.png|200px]] -->
  
 
== 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>.
+
=== 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 <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 midpoints <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 <math>BC</math> at point <math>N</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>
+
Let <math>M</math> be the midpoint of the chord <math>BC</math>. From right triangle <math>OMB</math>, we have <math>OM = \sqrt{OB^2 - BM^2} =4</math>. This gives <math>\tan \angle BOM = \frac{BM}{OM} = \frac 3 4</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>.
+
Notice that the distance <math>OM</math> equals <math>PN + PO \cos \angle AOM = r(1 + \cos \angle AOM)</math>, where <math>r</math> is the radius of circle <math>P</math>.
 +
 
 +
Hence <cmath>\cos \angle AOM = \frac{OM}{r} - 1 = \frac{2OM}{R} - 1 = \frac 8 5 - 1 = \frac 3 5</cmath> (where <math>R</math> represents the radius, <math>5</math>, of the large circle given in the question). Therefore, since <math>\angle AOM</math> is clearly acute, we see that <cmath>\tan \angle AOM =\frac{\sqrt{1 - \cos^2 \angle AOM}}{\cos \angle AOM} = \frac{\sqrt{5^2 - 3^2}}{3} = \frac 4 3</cmath>
 +
 
 +
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 ===
 +
 
 +
This solution, while similar to Solution 1, is far 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?
 +
 
 +
We consider the locus of midpoints of the chords from <math>A</math>. It is well-known that this is the circle with diameter <math>AO</math>, where <math>O</math> is the center of the circle. The proof is simple: every midpoint of a chord is a dilation of the endpoint with scale factor <math>\frac{1}{2}</math> and center <math>A</math>. Thus, the locus is the result of the dilation with scale factor <math>\frac{1}{2}</math> and centre <math>A</math> of circle <math>O</math>. Let the center of this circle be <math>P</math>.
 +
 
 +
Now, <math>AD</math> is bisected by <math>BC</math> if they cross at some point <math>N</math> on the circle. Moreover, since <math>AD</math> is the only chord, <math>BC</math> must be tangent to the circle <math>P</math>.
 +
 
 +
The rest of this problem is straightforward.
 +
 
 +
Our goal is to find <math>\sin \angle AOB = \sin{\left(\angle AOM - \angle BOM\right)}</math>, where <math>M</math> is the midpoint of <math>BC</math>. We have <math>BM=3</math> and <math>OM=4</math>.
 +
Let <math>R</math> be the projection of <math>A</math> onto <math>OM</math>, and similarly let <math>Q</math> be the projection of <math>P</math> onto <math>OM</math>. Then it remains to find <math>AR</math> so that we can use the addition formula for <math>\sin</math>.
 +
 
 +
As <math>PN</math> is a radius of circle <math>P</math>, <math>PN=2.5</math>, and similarly, <math>PO=2.5</math>. Since <math>OM=4</math>, we have <math>OQ=OM-QM=OM-PN=4-2.5=1.5</math>. Thus <math>PQ=\sqrt{2.5^2-1.5^2}=2</math>.
 +
 
 +
Further, we see that <math>\triangle OAR</math> is a dilation of <math>\triangle OPQ</math> about center <math>O</math> with scale factor <math>2</math>, so <math>AR=2PQ=4</math>.
 +
 
 +
Lastly, we apply the formula:
 +
<cmath> \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}</cmath>
 +
Thus the answer is <math>7\cdot25=\boxed{175}</math>.
 +
 
 +
=== Solution 3 (coordinate geometry) ===
 +
[[File:Aime1983p15s2.png|500px|link=]]
 +
 
 +
Let the circle have equation <math>x^2 + y^2 = 25</math>, with centre <math>O(0,0)</math>. Since <math>BC=6</math>, we can calculate (by the Pythagorean Theorem) that the distance from <math>O</math> to the line <math>BC</math> is <math>4</math>. Therefore, we can let <math>B=(3,4)</math> and <math>C=(-3,4)</math>. Now, assume that <math>A</math> is any point on the major arc BC, and <math>D</math> any point on the minor arc BC. We can write <math>A=(5 \cos \alpha, 5 \sin \alpha)</math>, where <math>\alpha</math> is the angle measured from the positive <math>x</math> axis to the ray <math>OA</math>. It will also be convenient to define <math>\angle XOB = \alpha_0</math>.
 +
 
 +
Firstly, since <math>B</math> must lie in the minor arc <math>AD</math>, we see that <math>\alpha \in \left(-\frac{\pi}{2}-\alpha_0,\alpha_0\right)</math>. However, since the midpoint of <math>AD</math> must lie on <math>BC</math>, and the highest possible <math>y</math>-coordinate of <math>D</math> is <math>5</math>, we see that the <math>y</math>-coordinate cannot be lower than <math>3</math>, that is, <math>\alpha \in \left[\sin^{-1}\frac{3}{5},\alpha_0\right)</math>.
 +
 
 +
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 <math>P</math> is the intersection point of <math>AD</math> and <math>BC</math>, so that by the theorem, <math>OP</math> is perpendicular to <math>AD</math>. So, if <math>AD</math> is the only chord starting at <math>A</math> which is bisected by <math>BC</math>, this means that <math>P</math> is the only point on the chord <math>BC</math> such that <math>OP</math> is perpendicular to <math>AD</math>. Now suppose that <math>P=(p,4)</math>, where <math>p \in (-3,3)</math>. The fact that <math>OP</math> must be perpendicular to <math>AD</math> is equivalent to the following equation:
 +
 
 +
<cmath> -1 = \left(\text{slope of } OP\right)\left(\text{slope of } AP\right)</cmath>
 +
which becomes
 +
<cmath> -1 = \frac{4}{p} \cdot \frac{5\sin \alpha - 4}{5\cos \alpha - p}</cmath>
 +
 
 +
This rearranges to
 +
 
 +
<cmath> p^2 - (5\cos \alpha)p + 16 - 20 \sin \alpha = 0</cmath>
 +
 
 +
Given that this equation must have only one real root <math>p\in (-3,3)</math>, we study the following function:
 +
 
 +
<cmath>f(x) =  x^2 - (5\cos \alpha)x + 16 - 20 \sin \alpha</cmath>
 +
 
 +
First, by the fact that the equation <math>f(x)=0</math> has real solutions, its discriminant <math>\Delta</math> must be non-negative, so we calculate
 +
 
 +
<cmath> \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}</cmath>
 +
 
 +
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 ===
 +
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 <math>\Delta > 0</math>, which would mean that there could be two real roots of <math>f(x)</math>, one lying in the interval <math>(-3,3)</math>, and another outside of it. We also see, by [[Vieta's Formulas]], that the average of the two roots is <math>\frac{5\cos \alpha}{2}</math>, which is non-negative, so the root outside of <math>(-3,3)</math> must be no less than <math>3</math>. By considering the graph of <math>y=f(x)</math>, which is a "U-shaped" parabola, it is now evident that <math>f(-3) > 0</math> and <math>f(3)\leq 0</math>. We can just use the second inequality:
 +
 
 +
<cmath>0 \geq  f(3) =  25  - 15\cos \alpha - 20 \sin \alpha</cmath>
 +
so
 +
<cmath>  3\cos \alpha + 4 \sin \alpha  \geq 5  </cmath>
 +
 
 +
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===
 +
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>.
 +
 
 +
Now, let the intersection of <math>BC</math> and <math>AD</math> be <math>E</math> and let the midpoint of <math>AO</math> (the center of the circle tangent to <math>BC</math> that we described beforehand) be <math>F</math>. Drop the altitude from <math>O</math> to <math>BC</math> and call its intersection with <math>BC</math> <math>K</math>. Drop the perpendicular from <math>F</math> to <math>KO</math> and call its intersection with <math>KO</math> <math>L</math>. Clearly, <math>KO = \sqrt{OC^2-KC^2} = \sqrt{5^2-3^2} = 4</math> and since <math>EF</math> is radius, it equals <math>\frac{5}{2}</math>. The same applies for <math>FO</math>, which also equals <math>\frac{5}{2}</math>. By the Pythagorean theorem, we deduce that <math>FL = 2</math>, so <math>EK = 2</math>. This is very important information! Now we know that <math>BE = 1</math>, so by Power of a Point, <math>AE = ED = \sqrt{5}</math>.
 +
 
 +
We’re almost there! Since by the Pythagorean theorem, <math>ED^2 + EO^2 = 25</math>, we deduce that <math>EO = 2\sqrt{5}</math>. <math>EC=OC=5</math>, so <math>\sin (CEO) = \frac{2\sqrt{5}}{5}</math>. Furthermore, since <math>\sin (CEO) = \cos(DEC)</math>, we know that <math>\cos (DEC) = \frac{2\sqrt{5}}{5}</math>. By the law of cosines,
 +
<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 ===
 +
 
 +
[[Image:Dgram.png|thumb|none|800px]]
 +
 
 +
Let I be the intersection of AD and BC.
 +
 +
Lemma: <math>AI = ID</math> if and only if <math>\angle AIO = 90</math>.
 +
 
 +
Proof: If AI = ID, we get AO = OD, and thus IO is a perpendicular bisector of AD. If <math>\angle AIO = 90</math>, We can get <math>\triangle AIO \cong \triangle OID</math>
 +
 
 +
Let  be this the circle with diameter AO.
 +
 
 +
Thus, we get <math>\angle AIO = 90</math>, implying I must lie on <math>\omega</math>. 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 <math>\omega</math>. Since <math>\omega</math>'s radius is <math>\frac{5}{2}</math>, the altitude from X to OY is 1.5, since the altitude from I to OY is 4. XO is 2.5, so <math>sin(XOY) = sin(AOY) = \frac{3}{5}</math>. <math>sin(BOZ) =  \frac{3}{5}</math>. If we let <math>sin(\theta) = \frac{3}{5}</math>, we can find that what we are looking for is <math>sin(90 - 2\theta)</math>, which we can evaluate and get <math>\frac{7}{25} \implies \boxed{175}</math>
 +
 
 +
-Alexlikemath
 +
 
 +
=== Solution 6 (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>.
 +
 
 +
From right triangle <math>BKO</math>, we get <math>KO = \sqrt{25-x^2}</math>. Thus, <math>KP = \sqrt{25-x^2}-\frac52</math>.
 +
 
 +
Since <math>BO = 5</math>, <math>BM = 3</math>, and <math>\angle BMO</math> is right, <math>MO=4</math>. From quadrilateral <math>MNPO</math>, we get <math>MN = \sqrt{PO^2 - (MO - NP)^2} = \sqrt{(5/2)^2 - (4 - 5/2)^2} = \sqrt{(5/2)^2 - (3/2)^2} = 2</math>. Thus, <math>BN = 1</math>.
 +
 
 +
Since angles <math>BNP</math> and <math>BKP</math> are right, we get <cmath>BK^2+KP^2 = BN^2 + NP^2 \implies x^2 + \left(\sqrt{25-x^2}-\frac52\right)^2 = \left(\frac52\right)^2 + 1</cmath>
 +
<cmath>25 - 5\sqrt{25-x^2} = 1</cmath>
 +
<cmath>5\sqrt{25-x^2} = 24</cmath>
 +
<cmath>25(25-x^2) = 24^2</cmath>
 +
<cmath>25x^2 = 25^2 - 24^2 = 49</cmath>
 +
<cmath>x = \frac75</cmath>
 +
Thus, <math>\sin \angle AOB = \frac{x}{5} = \frac{7}{25}\implies \boxed{175}</math>.
 +
 
 +
~rayfish
  
 
== See Also ==
 
== See Also ==

Latest revision as of 01:55, 26 December 2021

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)

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 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, $ED^2 + EO^2 = 25$, 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 5

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 6 (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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