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Difference between revisions of "2017 AIME II Problems/Problem 3"
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Could someone help me out with this? Thanks! - The Turtle
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Could someone help me out with the Asymptote? Thanks! - The Turtle
  
 
The set of all points closer to point <math>B</math> than to point <math>A</math> lie to the right of the perpendicular bisector of <math>AB</math> (line <math>PZ</math> in the diagram), and the set of all points closer to point <math>B</math> than to point <math>C</math> lie below the perpendicular bisector of <math>BC</math> (line <math>PX</math> in the diagram). Therefore, the set of points inside the triangle that are closer to <math>B</math> than to either vertex <math>A</math> or vertex <math>C</math> is bounded by quadrilateral <math>BXPZ</math>. Because <math>X</math> is the midpoint of <math>BC</math> and <math>Z</math> is the midpoint of <math>AB</math>, <math>X=(10,5)</math> and <math>Z=(6,0)</math>. The coordinates of point <math>P</math> is the solution to the system of equations defined by lines <math>PX</math> and <math>PZ</math>. Using the point-slope form of a linear equation and the fact that the slope of the line perpendicular to a line with slope <math>m</math> is <math>-\frac{1}{m}</math>, the equation for line <math>PX</math> is <math>y=\frac{2}{5}x+1</math> and the equation for line <math>PZ</math> is <math>x=6</math>. The solution of this system is <math>P=\left(6,\frac{17}{5}\right)</math>. Using the shoelace formula on quadrilateral <math>BXPZ</math> and triangle <math>ABC</math>, the area of quadrilateral <math>BXPZ</math> is <math>\frac{109}{5}</math> and the area of triangle <math>ABC</math> is <math>60</math>. Finally, the probability that a randomly chosen point inside the triangle is closer to vertex <math>B</math> than to vertex <math>A</math> or vertex <math>C</math> is the ratio of the area of quadrilateral <math>BXPZ</math> to the area of <math>ABC</math>, which is <math>\frac{\frac{109}{5}}{60}=\frac{109}{300}</math>. The answer is <math>109+300=\boxed{409}</math>.
 
The set of all points closer to point <math>B</math> than to point <math>A</math> lie to the right of the perpendicular bisector of <math>AB</math> (line <math>PZ</math> in the diagram), and the set of all points closer to point <math>B</math> than to point <math>C</math> lie below the perpendicular bisector of <math>BC</math> (line <math>PX</math> in the diagram). Therefore, the set of points inside the triangle that are closer to <math>B</math> than to either vertex <math>A</math> or vertex <math>C</math> is bounded by quadrilateral <math>BXPZ</math>. Because <math>X</math> is the midpoint of <math>BC</math> and <math>Z</math> is the midpoint of <math>AB</math>, <math>X=(10,5)</math> and <math>Z=(6,0)</math>. The coordinates of point <math>P</math> is the solution to the system of equations defined by lines <math>PX</math> and <math>PZ</math>. Using the point-slope form of a linear equation and the fact that the slope of the line perpendicular to a line with slope <math>m</math> is <math>-\frac{1}{m}</math>, the equation for line <math>PX</math> is <math>y=\frac{2}{5}x+1</math> and the equation for line <math>PZ</math> is <math>x=6</math>. The solution of this system is <math>P=\left(6,\frac{17}{5}\right)</math>. Using the shoelace formula on quadrilateral <math>BXPZ</math> and triangle <math>ABC</math>, the area of quadrilateral <math>BXPZ</math> is <math>\frac{109}{5}</math> and the area of triangle <math>ABC</math> is <math>60</math>. Finally, the probability that a randomly chosen point inside the triangle is closer to vertex <math>B</math> than to vertex <math>A</math> or vertex <math>C</math> is the ratio of the area of quadrilateral <math>BXPZ</math> to the area of <math>ABC</math>, which is <math>\frac{\frac{109}{5}}{60}=\frac{109}{300}</math>. The answer is <math>109+300=\boxed{409}</math>.

Revision as of 15:27, 23 March 2017

Problem

A triangle has vertices $A(0,0)$, $B(12,0)$, and $C(8,10)$. The probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to either vertex $A$ or vertex $C$ can be written as $\frac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$.

Solution

[asy] pair A,B,C,D,X,Z,P; A=(0,0); B=(12,0); C=(8,10); X=(10,5); Z=(6,0); P=(6,3.4); fill(B--X--P--Z--cycle,lightgray); draw(A--B--C--cycle); dot(A); label("$A$",A,SW); dot(B); label("$B$",B,SE); dot(C); label("$C$",C,N); draw(X--P,dashed); draw(Z--P,dashed); dot(X); label("$X$",X,NE); dot(Z); label("$Z$",Z,S); dot(P); label("$P$",P,NW); [/asy]

Could someone help me out with the Asymptote? Thanks! - The Turtle

The set of all points closer to point $B$ than to point $A$ lie to the right of the perpendicular bisector of $AB$ (line $PZ$ in the diagram), and the set of all points closer to point $B$ than to point $C$ lie below the perpendicular bisector of $BC$ (line $PX$ in the diagram). Therefore, the set of points inside the triangle that are closer to $B$ than to either vertex $A$ or vertex $C$ is bounded by quadrilateral $BXPZ$. Because $X$ is the midpoint of $BC$ and $Z$ is the midpoint of $AB$, $X=(10,5)$ and $Z=(6,0)$. The coordinates of point $P$ is the solution to the system of equations defined by lines $PX$ and $PZ$. Using the point-slope form of a linear equation and the fact that the slope of the line perpendicular to a line with slope $m$ is $-\frac{1}{m}$, the equation for line $PX$ is $y=\frac{2}{5}x+1$ and the equation for line $PZ$ is $x=6$. The solution of this system is $P=\left(6,\frac{17}{5}\right)$. Using the shoelace formula on quadrilateral $BXPZ$ and triangle $ABC$, the area of quadrilateral $BXPZ$ is $\frac{109}{5}$ and the area of triangle $ABC$ is $60$. Finally, the probability that a randomly chosen point inside the triangle is closer to vertex $B$ than to vertex $A$ or vertex $C$ is the ratio of the area of quadrilateral $BXPZ$ to the area of $ABC$, which is $\frac{\frac{109}{5}}{60}=\frac{109}{300}$. The answer is $109+300=\boxed{409}$.

See Also

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

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