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

(→‎Solution 2 (Trig Bash))
(→‎Solution 2 (Trig Bash))
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==Solution 2 (Trig Bash)==
 
==Solution 2 (Trig Bash)==
 
Put <math>\triangle POQ</math> and <math>\triangle POR</math> with <math>O</math> on the origin and the triangles on the 1st quadrant
 
Put <math>\triangle POQ</math> and <math>\triangle POR</math> with <math>O</math> on the origin and the triangles on the 1st quadrant
−
the coordinates of <math>Q</math> and <math>P</math> is <math>(200 cos^{2} a,200 cos a*sin a)</math>, <math>(200 cos^{2} b ,200 cos b*sin b )</math>. So <math>PQ^{2}</math> = <math>(200 cos^{2} a - 200cos^{2} b)^{2} +(200 cos a*sin a - 200 cos b * sin b)^{2}</math>, which we want to be less then <math>100^{2}</math>.
+
the coordinates of <math>Q</math> and <math>P</math> is <math>(200 \cos^{2}a,200 \cos a\sin a )</math>, <math>(200\cos^{2}b,200\cos(b)*\sin b)</math>. So <math>PQ^{2}</math> = <math>(200 \cos^{2} a - 200 \cos^{2} b)^{2} +(200 \cos a \sin a - 200 \cos b \sin b)^{2}</math>, which we want to be less then <math>100^{2}</math>.
−
So <math>(200 cos^{2} a - 200cos^{2} b)^{2} +(200 cos a*sin a - 200 cos b * sin b)^{2} <= 100^{2} </math>
+
So <math>(200 \cos^{2} a - 200 \cos^{2} b)^{2} +(200 \cos a \sin a - 200 \cos b \sin b)^{2} <= 100^{2} </math>
−
<cmath>(cos^{2} a - cos^{2} b)^{2} +(cos a*sin a - cos b * sin b)^{2} \le \frac{1}{4} </cmath>
+
<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 - 2cos^{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^{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(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} - 2 (\cos a \sin b)(\cos b \sin a)+(\cos b\sin a)^{2} \le \frac{1}{4} </cmath>
 +
<cmath>(\cos a\sin b-\cos b\sin a)^{2}\le \frac{1}{4} </cmath>
 +
So we want <math> -\frac{1}{2} \le \sin b-a \le \frac{1}{2} </math> or <math> -30 \le b-a \le 30</math>. So as the above solution states, the probability is <math>\frac{16}{25}</math> and the answer is <math>16+25 = \box{41}</math>
  
 
==See Also==
 
==See Also==
 
{{AIME box|year=2017|n=I|num-b=7|num-a=9}}
 
{{AIME box|year=2017|n=I|num-b=7|num-a=9}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 14:41, 10 March 2017

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 1st 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 $(200 \cos^{2} a - 200 \cos^{2} b)^{2} +(200 \cos a \sin a - 200 \cos b \sin b)^{2} <= 100^{2}$ \[(\cos^{2} a - \cos^{2} b)^{2} +(\cos a \sin a - \cos b \sin b)^{2} \le \frac{1}{4}\] \[\cos^{4} a + \cos^{4} b - 2\cos^{2} a \cos^{2} b +\cos^{2}a \sin^{2} a + \cos^{2} b \sin^{2} b - 2 \cos a \sin a \cos b \sin b \le \frac{1}{4}\] \[\cos^{2} a(\cos^{2} a + \sin^{2} a)+\cos^{2} b(cos^{2} b+sin^{2} b) - 2\cos^{2} a \cos^{2} b- 2 \cos a \sin a \cos b \sin b \le \frac{1}{4}\] \[\cos^{2} a(1-\cos^{2} b)+\cos^{2} b(1-cos^{2} a) - 2 \cos a \sin a \cos b \sin b \le \frac{1}{4}\] \[(\cos a\sin b)^{2} - 2 (\cos a \sin b)(\cos b \sin a)+(\cos b\sin a)^{2} \le \frac{1}{4}\] \[(\cos a\sin b-\cos b\sin a)^{2}\le \frac{1}{4}\] So we want $-\frac{1}{2} \le \sin b-a \le \frac{1}{2}$ or $-30 \le b-a \le 30$. So as the above solution states, the probability is $\frac{16}{25}$ and the answer is $16+25 = \box{41}$ (Error compiling LaTeX. Unknown error_msg)

See Also

2017 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 7 		Followed by
Problem 9
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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