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 1

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

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 $109+300=\boxed{409}$ .

Solution 2

Since we know the coordinates of all three vertices of the triangle, we can find the side lengths: $AB=12$ , $AC=2\sqrt{41}$ , and $BC=2\sqrt{29}$ . We notice that the point where the three distances are the same is the circumcenter - so we use one of the triangle area formulas to find the circumradius, since we know what the area is. $\frac{12 \cdot 2\sqrt{41} \cdot 2\sqrt{29}}{4 \cdot R}=\frac{12 \cdot 10}{2}.$ We rearrange to get $R=\frac{\sqrt{41} \cdot \sqrt{29}}{5}.$ $[asy] draw((0,0)--(12,0)--(8,10)--(0,0)); draw((6,0)--(6,3.4)--(10,5)); draw((6,3.4)--(4,5)); label("$A$", (0,0), SW); label("$B$", (12,0), SE); label("$C$", (8, 10), N); label("$P$", (6, 3.4), NNE); label("$R$", (10, 5), NE); label("$S$", (6, 0), S); label("$T$", (4, 5), NW); [/asy]$ We know that $AP=\frac{\sqrt{41} \cdot \sqrt{29}}{5}$ , and $AS=6$ , so using the Pythagorean Theorem gives $SP=\frac{17}{5}$ . This means $[ASP]=[BSP]=\frac{17}{5} \cdot 6 \cdot \frac{1}{2} = \frac{51}{5}$ . Similarly, we know that $BP=\frac{\sqrt{41} \cdot \sqrt{29}}{5}$ , and $BR=\sqrt{29}$ , so we get that $PR=\frac{4\sqrt{29}}{5}$ , and so $[BRP]=[CRP]=\frac{4\sqrt{29}}{5} \cdot \sqrt{29} \cdot \frac{1}{2} = \frac{58}{5}$ . Lastly, we know that $CP=\frac{\sqrt{41} \cdot \sqrt{29}}{5}$ , and $CT=\sqrt{41}$ , so we get that $PT=\frac{2\sqrt{41}}{5}$ , and $[ATP]=[CTP]=\frac{2\sqrt{41}}{5} \cdot \sqrt{41} \cdot \frac{1}{2} = \frac{41}{5}$ . Therefore, our answer is $\frac{51+58}{2(51+58+41)}=\frac{109}{300} \rightarrow \boxed{409}$ .

Solution 3

To start the problem, identify the two midpoints that connect $AB$ and $BC$ . This is because the midpoints of such lines is the mark at which the point will sway closer to vertex $A$ / $C$ or vertex $B$ . The midpoint of $AB$ is $(6,0)$ , and the midpoint of $BC$ is $(10,5)$ . Then, determine the line at which the distance between vertex $B$ and vertex $C$ are the same. Assuming that $x$ is the real value of $x$ and $y$ is the real value of $y$ , we can create a simple equation:

$\sqrt{|x-8|^2 + (10-y)^2}$ = $\sqrt{(12-x)^2 + y^2}$

where the left side of the equation is for the distance to vertex $C$ and the right side of the equation is the distance to vertex $B$ .

Squaring both sides and then distributing, we get

$x^2 - 16x + 64 + y^2 - 20y + 100 = x^2 - 24x + 144 + y^2$ .

Notice that $(x-8)^2 = (-x+8)^2$ , and thus there is no need to create another equation.

Simplifying, we get $8x + 20 = 20y$ .

Divide both sides by 20, then simplify, and the line that represents equivalent distance between vertex $B$ and vertex $C$ is $y = \frac{2}{5} + 1$ .

This line starts at the midpoint of $BC$ , which is $(10,5)$ , and ends at the line $x=6$ , as $x=6$ represents equivalent distance between vertex $A$ and vertex $B$ . Plug in $x=6$ to the equation $y = \frac{2}{5} + 1$ , and we get $y = \frac{17}{5}$ . Now that we have our four points that are $(6,0), (6,\frac{17}{5}), (10,5)$ , and $(12,0)$ , we can calculate the area of the quadrilateral in which a point is closer to vertex $B$ as opposed to either vertex $A$ or vertex $C$ . Simply draw a square that has the points $(6,0), (6,5), (12,5)$ and $(12,0)$ , and then subtract the two triangles that appear in between.

Thus, the area of the quadrilateral is $6\cdot5 - (\frac{\frac{8}{5}\cdot4}{2} + \frac{5\cdot2}{2}) \rightarrow 30 - \frac{41}{5} = \frac{109}{5}$ . Since the problem asks us for the probability that a point chosen inside the triangle is inside the quadrilateral, and because the area of $\triangle ABC$ is $\frac{12\cdot10}{2} = 60$ , the probability is $\frac{\frac{109}{5}}{60}=\frac{109}{300}$ , which means the final answer is $109+300=\boxed{409}$ .

2017 AIME II (Problems • Answer Key • Resources)
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2017 AIME II Problems/Problem 3

Contents

Problem

Solution 1

Solution 2

Solution 3

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