Difference between revisions of "2017 AIME I Problems/Problem 8"

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Problem 8

Two real numbers $a$ and $b$ are chosen independently and uniformly at random from the interval $(0, 75)$ . Let $O$ and $P$ be two points on the plane with $OP = 200$ . Let $Q$ and $R$ be on the same side of line $OP$ such that the degree measures of $\angle POQ$ and $\angle POR$ are $a$ and $b$ respectively, and $\angle OQP$ and $\angle ORP$ are both right angles. The probability that $QR \leq 100$ is equal to $\frac{m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .

Solution 1

Noting that $\angle OQP$ and $\angle ORP$ are right angles, we realize that we can draw a semicircle with diameter $\overline{OP}$ and points $Q$ and $R$ on the semicircle. Since the radius of the semicircle is $100$ , if $\overline{QR} \leq 100$ , then $\overarc{QR}$ must be less than or equal to $60^{\circ}$ .

This simplifies the problem greatly. Since the degree measure of an angle on a circle is simply half the degree measure of its subtended arc, the problem is simply asking:

Given $a, b$ such that $0<a, b<75$ , what is the probability that $|a-b| \leq 30$ ? Through simple geometric probability, we get that $P = \frac{16}{25}$ .

The answer is $16+25=\boxed{041}$

~IYN~

Solution 2 (Trig Bash)

Put $\triangle POQ$ and $\triangle POR$ with $O$ on the origin and the triangles on the $1^{st}$ quadrant. The coordinates of $Q$ and $P$ is $(200 \cos^{2}a,200 \cos a\sin a )$ , $(200\cos^{2}b,200\cos(b)\sin b)$ . So $PQ^{2}$ = $(200 \cos^{2} a - 200 \cos^{2} b)^{2} +(200 \cos a \sin a - 200 \cos b \sin b)^{2}$ , which we want to be less then $100^{2}$ . So $(200 \cos^{2} a - 200 \cos^{2} b)^{2} +(200 \cos a \sin a - 200 \cos b \sin b)^{2} <= 100^{2}$ $(\cos^{2} a - \cos^{2} b)^{2} +(\cos a \sin a - \cos b \sin b)^{2} \le \frac{1}{4}$ $\cos^{4} a + \cos^{4} b - 2\cos^{2} a \cos^{2} b +\cos^{2}a \sin^{2} a + \cos^{2} b \sin^{2} b - 2 \cos a \sin a \cos b \sin b \le \frac{1}{4}$ $\cos^{2} a(\cos^{2} a + \sin^{2} a)+\cos^{2} b(\cos^{2} b+\sin^{2} b) - 2\cos^{2} a \cos^{2} b- 2 \cos a \sin a \cos b \sin b \le \frac{1}{4}$ $\cos^{2} a(1-\cos^{2} b)+\cos^{2} b(1-\cos^{2} a) - 2 \cos a \sin a \cos b \sin b \le \frac{1}{4}$ $(\cos a\sin b)^{2} +(\cos b\sin a)^{2} - 2 (\cos a \sin b)(\cos b \sin a)\le \frac{1}{4}$ $(\cos a\sin b-\cos b\sin a)^{2}\le \frac{1}{4}$ $\sin^{2} (b-a) \le \frac{1}{4}$ So we want $-\frac{1}{2} \le \sin (b-a) \le \frac{1}{2}$ , which is equivalent to $-30 \le b-a \le 30$ or $150 \le b-a \le 210$ . The second inequality is impossible so we only consider what the first inequality does to our $75$ by $75$ box in the $ab$ plane. This cuts off two isosceles right triangles from opposite corners with side lengths $45$ from the $75$ by $75$ box. Hence the probability is $1-\frac{45^2}{75^2} = 1- \frac{9}{25}=\frac{16}{25}$ and the answer is $16+25 = \boxed{41}$

Solution by Leesisi

Solution 3 (Quicker Trig)

$[asy] pair O, P, Q, R; draw(circle(O, 10)); O = (10, 0); P = (-10, 0); Q = (10*cos(pi/3), 10*sin(pi/3)); R = (10*cos(5*pi/6), 10*sin(5*pi/6)); dot(Q); dot(O); dot(P); dot(R); draw(P--O--Q--P--R--O); draw(Q--R, red); label("$O$", O, 2*E); label("$P$", P, 2*W); label("$Q$", Q, NE); label("$R$", R, NW); label("$200$", (0,0), 2*S); label("$x$", (Q+R)/2, N); draw(rightanglemark(O, Q, P, 38)); draw(rightanglemark(O, R, P, 38)); [/asy]$ Let $QR=x.$ Since we are given many angles in the problem, we can compute the lengths of some of the lines in terms of trigonometric functions: $OQ = 200 \cos a, PQ = 200 \sin a, PR = 200 \sin b, OR = 200 \cos b.$ Now observe that quadrilateral $OQRP$ is a cyclic quadrilateral. Thus, we are able to apply Ptolemy's Theorem to it: $200 x + (200 \cos a) (200 \sin b) = (200 \sin a) (200 \cos b),$ $x + 200 (\cos a \sin b) = 200 (\sin a \cos b),$ $x = 200(\sin a \cos b - \sin b \cos a),$ $x = 200 \sin(a-b).$ We want $|x| \le 100$ (the absolute value comes from the fact that $a$ is not necessarily greater than $b,$ so we cannot assume that $Q$ is to the right of $R$ as in the diagram), so we substitute: $|200 \sin(a-b)| \le 100,$ $|\sin(a-b)| \le \frac{1}{2},$ $|a-b| \le 30 ^\circ,$ $-30 \le a-b \le 30.$ By simple geometric probability (see Solution 2 for complete explanation), $\frac{m}{n} = 1 - \frac{2025}{5625} = 1 - \frac{9}{25} = \frac{16}{25},$ so $m+n = \boxed{041}.$

~burunduchok

Solution 4

Scale the circle down from radius 100 (diameter 200) to radius 6 (diameter 12). Then we want the probability that $PQ \le 6$ . Now note that all possible $P$ and $Q$ lie on a $5\pi$ interval on the circumference of the circle. But for $PQ<6$ , $P$ and $Q$ must be less than $2\pi$ apart on the circumference of the circle. Simple geometric probability gives us $\frac{16}{25}$ , so the answer is $41$ . (Professor-Mom)

@@ Line 20: / Line 20: @@
 <cmath>(\cos^{2} a - \cos^{2} b)^{2} +(\cos a \sin a - \cos b \sin b)^{2} \le \frac{1}{4} </cmath>
 <cmath>\cos^{4} a + \cos^{4} b - 2\cos^{2} a \cos^{2} b +\cos^{2}a \sin^{2} a + \cos^{2} b \sin^{2} b - 2 \cos a \sin a \cos b \sin b \le \frac{1}{4} </cmath>
-<cmath>\cos^{2} a(\cos^{2} a + \sin^{2} a)+\cos^{2} b(cos^{2} b+sin^{2} b) - 2\cos^{2} a \cos^{2} b- 2 \cos a \sin a \cos b \sin b \le \frac{1}{4} </cmath>
+<cmath>\cos^{2} a(\cos^{2} a + \sin^{2} a)+\cos^{2} b(\cos^{2} b+\sin^{2} b) - 2\cos^{2} a \cos^{2} b- 2 \cos a \sin a \cos b \sin b \le \frac{1}{4} </cmath>
-<cmath>\cos^{2} a(1-\cos^{2} b)+\cos^{2} b(1-cos^{2} a) - 2 \cos a \sin a \cos b \sin b \le \frac{1}{4} </cmath>
+<cmath>\cos^{2} a(1-\cos^{2} b)+\cos^{2} b(1-\cos^{2} a) - 2 \cos a \sin a \cos b \sin b \le \frac{1}{4} </cmath>
 <cmath>(\cos a\sin b)^{2} +(\cos b\sin a)^{2} - 2 (\cos a \sin b)(\cos b \sin a)\le \frac{1}{4} </cmath>
 <cmath>(\cos a\sin b-\cos b\sin a)^{2}\le \frac{1}{4} </cmath>

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