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

Revision as of 20:56, 15 February 2019

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 cannott 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.)

@@ Line 24: / Line 24: @@
 <!-- [[Image:1983_AIME-15.png|200px]] -->
-== Solution 1 ==
+== Solution ==
+=== Solution 1 ===
 -Credit to Adamz for diagram-
 <asy>
@@ Line 46: / Line 48: @@
 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 [[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 <math>BC</math> at point <math>N</math>.
+</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>.
-Let <math>M</math> 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>.
+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>.
-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 <math>P</math>). 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>
+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>.
-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>.
+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>
-=== Motivation Behind Solution 1===
+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>.
-The above solution works, but is quite messy and somewhat difficult to follow. This solution provides the motivation behind the solution.
+=== Solution 2 ===
-First of all, where did the statement "<math>AD</math> is the only chord starting at <math>A</math> and bisected by <math>BC</math> " come from? What is its significance in this problem? What is the criterion for this statement to be true?
+This solution, while similar to Solution 1, is far more motivated and less contrived.
-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 ratio <math>1/2</math> with center <math>A</math>. Thus, the locus is the result of the dilation with ratio <math>1/2</math> of circle <math>O</math> with center <math>A</math>. Let the center of this circle be <math>P</math>.
+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?
-Aha! Now we see. <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>.
+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>.
-The rest of this problem is straight forward.
+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>.
-Our goal is to find <math>\sin AOB = \sin (AOM - BOM)</math> where <math>M</math> is the midpoint of <math>BC</math>. Then we have <math>BM=3</math> and <math>OM=4</math>.
+The rest of this problem is straightforward.
-Let <math>R</math> be the projection of <math>A</math> onto <math>OM</math>, and similarly <math>Q</math> be the projection of <math>P</math> onto <math>OM</math>. Then it remains to find <math>AR</math> so we can use the sine addition formula.
-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>, <math>OQ=OM-QM=OM-PN=4-2.5=1.5</math>. Thus, <math>PQ=\sqrt{(2.5)^2-1.5^2}=2</math>.
+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>.
-From here, we see that <math>\triangle OAR</math> is a dilation of <math>\triangle OPQ</math> about center <math>O</math> with ratio <math>2</math>, so <math>AR=2PQ=4</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 (AOM - BOM) = \sin AOM \cos BOM - \sin BOM \cos AOM = (4/5)(4/5)-(3/5)(3/5)=7/25.</cmath>
+<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>.
-Thus, our answer is <math>7\cdot25=\boxed{175}</math>.
-== Solution 2 (with the help of coordinates) ==
+=== Solution 3 (coordinate geometry) ===
   [[File:Aime1983p15s2.png|500px|link=]]
-Let the circle be <math>x^2 + y^2 = 25</math>, and its center be labeled <math>O=(0,0)</math>. Since BC=6, we can calculate (by the Pythagorean Theorem) that the distance from <math>O</math> to the line <math>BC</math> is <math>4</math>. So 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 may let <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>\alpha_0</math> as the measure of angle XOB.
+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 B must lie in the minor arc AD, we see that <math>\alpha \in \left(-\frac{\pi}{2}-\alpha_0,\alpha_0\right)</math>. However, since the midpoint of AD must lie on BC, and the highest <math>y</math> coordinate of <math>D</math> is <math>5</math>, we see that the <math>y</math> coordinate can't be lower than <math>3</math>, that is,  <math>\alpha \in \left[\sin^{-1}\frac{3}{5},\alpha_0\right)</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 cannott 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 the they must be perpendicular. Therefore, suppose that <math>P</math> is the intersection between <math>AD</math> and <math>BC</math>, then <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>, it 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 we are ready to make it an algebraic problem. Suppose <math>P=(p,4)</math>, <math>p \in (-3,3)</math>. The statement <math>OP</math> is perpendicular to <math>AD</math> is equivalent to the following equation:
+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 = (slope OP)(slope AP)</cmath>
+<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>
-It rearranges to the following:
+This rearranges to
 <cmath> p^2 - (5\cos \alpha)p + 16 - 20 \sin \alpha = 0</cmath>
-Given that this equation has only one real root <math>p\in (-3,3)</math>, we study the following function:
+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 <math>f(x)</math> has real solution, it is good to look at its discriminant: must be non-negative:
+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> \Delta = (5\cos \alpha)^2 - 4(16-20\sin \alpha)</cmath>
+<cmath> \begin{split}\Delta & = (5\cos \alpha)^2 - 4(16-20\sin \alpha) \\
-<cmath>=  25 (1- \sin^2 \alpha)  - 64 + 80 \sin \alpha</cmath>
+& =  25 (1- \sin^2 \alpha)  - 64 + 80 \sin \alpha \\
-<cmath>= -25 \sin^2 \alpha + 80\sin \alpha - 39</cmath>
+& = -25 \sin^2 \alpha + 80\sin \alpha - 39 \\
-<cmath>=  (13 - 5\sin \alpha)(5\sin \alpha - 3)</cmath>
+& =  (13 - 5\sin \alpha)(5\sin \alpha - 3)\end{split}</cmath>
-It is obvious that this is 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 = (5\cos \alpha)/2 = 2 \in (-3,3)</math>. We found a possible case. So we calculate <math>\sin(arc AB) = \sin (\alpha_0 - \alpha) = \frac{4}{5}\cdot \frac{4}{5} - \frac{3}{5} \cdot \frac{3}{5} = \frac{7}{25}</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>.
-=== Note to Solution 2 ===
+=== Addendum to Solution 3 ===
-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''.
+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 means that there can be two real roots of <math>f(x)</math>, one lying in the interval <math>(-3,3)</math>, but another falling out of it. We also see that the average of the two root is <math> (5\cos \alpha) / 2 </math>, which is a quantity greater than 0, so the root outside of <math>(-3,3)</math> must be no less than 3. Spectating the parabolic curve of <math>f(x)</math>, which is a "U shaped" curve hitting the interval <math>(-3,3)</math> once and <math>[3,\infty)</math> another time, it is evident that <math>f(-3) > 0</math> and <math>f(3)\leq 0</math>. We can just work on the second one:
+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 to satisfy this equation is when <math>A=B</math> (by working on Cauchy-Schwarz inequality, or just plotting the line <math>3x+4y=5</math> to see that point <math>A</math> can't go above this line), which does not make sense from the description of the problem.
+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.)
-It means 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. 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.
 == See Also ==

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