Difference between revisions of "2017 AIME II Problems/Problem 3"
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A triangle has vertices <math>A(0,0)</math>, <math>B(12,0)</math>, and <math>C(8,10)</math>. The probability that a randomly chosen point inside the triangle is closer to vertex <math>B</math> than to either vertex <math>A</math> or vertex <math>C</math> can be written as <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>. | A triangle has vertices <math>A(0,0)</math>, <math>B(12,0)</math>, and <math>C(8,10)</math>. The probability that a randomly chosen point inside the triangle is closer to vertex <math>B</math> than to either vertex <math>A</math> or vertex <math>C</math> can be written as <math>\frac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>. | ||
− | ==Solution== | + | ==Solution 1== |
<asy> | <asy> | ||
pair A,B,C,D,X,Z,P; | pair A,B,C,D,X,Z,P; | ||
Line 25: | Line 25: | ||
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>. | ||
+ | |||
+ | ==Solution 2== | ||
+ | |||
+ | Since we know the coordinates of all three vertices of the triangle, we can find the side lengths: <math>AB=12</math>, <math>AC=2\sqrt{41}</math>, and <math>BC=2\sqrt{29}</math>. 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. | ||
+ | <cmath>\frac{12 \cdot 2\sqrt{41} \cdot 2\sqrt{29}}{4 \cdot R}=\frac{12 \cdot 10}{2}.</cmath> | ||
+ | We rearrange to get | ||
+ | <cmath>R=\frac{\sqrt{41} \cdot \sqrt{29}}{5}.</cmath> | ||
+ | <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 <math>AP=\frac{\sqrt{41} \cdot \sqrt{29}}{5}</math>, and <math>AS=6</math>, so using the Pythagorean Theorem gives <math>SP=\frac{17}{5}</math>. This means <math>[ASP]=[BSP]=\frac{17}{5} \cdot 6 \cdot \frac{1}{2} = \frac{51}{5}</math>. | ||
+ | Similarly, we know that <math>BP=\frac{\sqrt{41} \cdot \sqrt{29}}{5}</math>, and <math>BR=\sqrt{29}</math>, so we get that <math>PR=\frac{4\sqrt{29}}{5}</math>, and so <math>[BRP]=[CRP]=\frac{4\sqrt{29}}{5} \cdot \sqrt{29} \cdot \frac{1}{2} = \frac{58}{5}</math>. | ||
+ | Lastly, we know that <math>CP=\frac{\sqrt{41} \cdot \sqrt{29}}{5}</math>, and <math>CT=\sqrt{41}</math>, so we get that <math>PT=\frac{2\sqrt{41}}{5}</math>, and <math>[ATP]=[CTP]=\frac{2\sqrt{41}}{5} \cdot \sqrt{41} \cdot \frac{1}{2} = \frac{41}{5}</math>. | ||
+ | Therefore, our answer is <math>\frac{51+58}{2(51+58+41)}=\frac{109}{300} \rightarrow \boxed{409}</math>. | ||
=See Also= | =See Also= | ||
{{AIME box|year=2017|n=II|num-b=2|num-a=4}} | {{AIME box|year=2017|n=II|num-b=2|num-a=4}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 07:28, 24 March 2017
Contents
[hide]Problem
A triangle has vertices , , and . The probability that a randomly chosen point inside the triangle is closer to vertex than to either vertex or vertex can be written as , where and are relatively prime positive integers. Find .
Solution 1
The set of all points closer to point than to point lie to the right of the perpendicular bisector of (line in the diagram), and the set of all points closer to point than to point lie below the perpendicular bisector of (line in the diagram). Therefore, the set of points inside the triangle that are closer to than to either vertex or vertex is bounded by quadrilateral . Because is the midpoint of and is the midpoint of , and . The coordinates of point is the solution to the system of equations defined by lines and . Using the point-slope form of a linear equation and the fact that the slope of the line perpendicular to a line with slope is , the equation for line is and the equation for line is . The solution of this system is . Using the shoelace formula on quadrilateral and triangle , the area of quadrilateral is and the area of triangle is . Finally, the probability that a randomly chosen point inside the triangle is closer to vertex than to vertex or vertex is the ratio of the area of quadrilateral to the area of , which is . The answer is .
Solution 2
Since we know the coordinates of all three vertices of the triangle, we can find the side lengths: , , and . 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. We rearrange to get We know that , and , so using the Pythagorean Theorem gives . This means . Similarly, we know that , and , so we get that , and so . Lastly, we know that , and , so we get that , and . Therefore, our answer is .
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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