Difference between revisions of "2017 AIME II Problems/Problem 3"
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− | Could someone help me out with | + | 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 16:27, 23 March 2017
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
[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,SW);
dot(B);
label("
",B,SE);
dot(C);
label("
",C,N);
draw(X--P,dashed);
draw(Z--P,dashed);
dot(X);
label("
",X,NE);
dot(Z);
label("
",Z,S);
dot(P);
label("
",P,NW);
[/asy]
Could someone help me out with the Asymptote? Thanks! - The Turtle
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
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.