Difference between revisions of "2017 AIME II Problems/Problem 3"
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Revision as of 02:39, 7 November 2020
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
.
Solution 3
To start the problem, identify the two midpoints that connect and
. This is because the midpoints of such lines is the mark at which the point will sway closer to vertex
/
or vertex
. The midpoint of
is
, and the midpoint of
is
. Then, determine the line at which the distance between vertex
and vertex
are the same. Assuming that
is the real value of
and
is the real value of
, we can create a simple equation:
=
where the left side of the equation is for the distance to vertex and the right side of the equation is the distance to vertex
.
Squaring both sides and then distributing, we get
.
Notice that , and thus there is no need to create another equation.
Simplifying, we get .
Divide both sides by 20, then simplify, and the line that represents equivalent distance between vertex and vertex
is
.
This line starts at the midpoint of , which is
, and ends at the line
, as
represents equivalent distance between vertex
and vertex
. Plug in
to the equation
, and we get
. Now that we have our four points that are
, and
, we can calculate the area of the quadrilateral in which a point is closer to vertex
as opposed to either vertex
or vertex
. Simply draw a rectangle that has the points
and
, and then subtract the two triangles that appear in between.
Thus, the area of the quadrilateral is .
Since the problem asks us for the probability that a point chosen inside the triangle is inside the quadrilateral, and because the area of
is
, the probability is
, which means the final answer is
.
Solution by IronicNinja~
Solution 4
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.