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

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(→‎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> is <math>(200cos(a))</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 - 200cos^{2} a)^{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} a)^{2} +(200 cos a*sin a - 200 cos b * sin b)^{2} ≤ 100 </math>
 +
<math>(200 cos^{2} a - 200cos^{2} a)^{2} +(200 cos a*sin a - 200 cos b * sin b)^{2} ≤  </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 13:16, 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 $PQ^{2}$ = $(200 cos^{2} a - 200cos^{2} a)^{2} +(200 cos a*sin a - 200 cos b * sin b)^{2}$, which we want to be less then $100^{2}$. So 

$(200 cos^{2} a - 200cos^{2} a)^{2} +(200 cos a*sin a - 200 cos b * sin b)^{2} ≤ 100$ (Error compiling LaTeX. Unknown error_msg)
$(200 cos^{2} a - 200cos^{2} a)^{2} +(200 cos a*sin a - 200 cos b * sin b)^{2} ≤$ (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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