2020 AIME I Problems/Problem 13
Contents
Problem
Point lies on side
of
so that
bisects
The perpendicular bisector of
intersects the bisectors of
and
in points
and
respectively. Given that
and
the area of
can be written as
where
and
are relatively prime positive integers, and
is a positive integer not divisible by the square of any prime. Find
Solution 1
Points are defined as shown. It is pretty easy to show that by spiral similarity at
by some short angle chasing. Now, note that
is the altitude of
, as the altitude of
. We need to compare these altitudes in order to compare their areas. Note that Stewart's theorem implies that
, the altitude of
. Similarly, the altitude of
is the altitude of
, or
. However, it's not too hard to see that
, and therefore
. From here, we get that the area of
is
, by similarity. ~awang11
Solution 2(coord bash + basic geometry)
Let lie on the x-axis and
be the origin.
is
. Use Heron's formula to compute the area of triangle
. We have
. and
. We now find the altitude, which is
, which is the y-coordinate of
. We now find the x-coordinate of
, which satisfies
, which gives
since the triangle is acute. Now using the Angle Bisector Theorem, we have
and
to get
. The coordinates of D are
.
Since we want the area of triangle
, we will find equations for perpendicular bisector of AD, and the other two angle bisectors. The perpendicular bisector is not too challenging: the midpoint of AD is
and the slope of AD is
. The slope of the perpendicular bisector is
. The equation is(in point slope form)
.
The slope of AB, or in trig words, the tangent of
is
.
Finding
and
. Plugging this in to half angle tangent, it gives
as the slope of the angle bisector, since it passes through
, the equation is
.
Similarly, the equation for the angle bisector of
will be
.
For
use the B-angle bisector and the perpendicular bisector of AD equations to intersect at
.
For
use the C-angle bisector and the perpendicular bisector of AD equations to intersect at
.
The area of AEF is equal to
since AD is the altitude of that triangle with EF as the base, with
being the height.
and
, so
which gives
. NEVER overlook coordinate bash in combination with beginner synthetic techniques.~vvluo
Solution 3 (Coordinate Bash + Trig)
Let and
be the line
.
We compute that
, so
.
Thus,
lies on the line
. The length of
at a point
is
, so
.
We now have the coordinates ,
and
.
We also have
by the angle-bisector theorem and
by taking the midpoint.
We have that because
,
by half angle formula.
We also compute , so
.
Now, has slope
, so it's perpendicular bisector has slope
and goes through
.
We find that this line has equation .
As , we have that line
has form
.
Solving for the intersection point of these two lines, we get
and thus
We also have that because ,
has form
.
Intersecting the line and the perpendicular bisector of
yields
.
Solving this, we get and so
.
We now compute .
We also have
.
As , we have
.
The desired answer is ~Imayormaynotknowcalculus
Solution 4 (Barycentric Coordinates)
As usual, we will use homogenized barycentric coordinates.
We have that will have form
. Similarly,
has form
and
has form
.
Since
and
, we also have
.
It remains to determine the equation of the line formed by the perpendicular bisector of
.
This can be found using EFFT. Let a point on
have coordinates
.
We then have that the displacement vector
and that the displacement vector
has form
.
Now, by EFFT, we have
.
This equates to
.
Now, intersecting this with , we have
,
, and
.
This yields
,
, and
, or
.
Similarly, intersecting this with , we have
,
, and
.
Solving this, we obtain
,
, and
, or
.
We finish by invoking the Barycentric Distance Formula twice; our first displacement vector being .
We then have
, thus
.
Our second displacement vector is .
As a result,
, so
.
As , the desired area is
. ~Imayormaynotknowcalculus
Remark: The area of can also be computed using the Barycentric Area Formula, although it may increase the risk of computational errors; there are also many different ways to proceed once the coordinates are determined.
Solution 5 (geometry+trig)
To get the area of , we try to find
and
.
Since is the angle bisector, we can get that
and
. By applying Stewart's Theorem, we can get that
. Therefore
.
Since is the perpendicular bisector of
, we know that
. Since
is the angle bisector of
,
we know that
. By applying the Law of Sines to
and
, we know that
. Since
is not equal to
and therefore these two triangles are not congruent, we know that
and
are supplementary. Then we know that
and
are also supplementary. Given that
, we can get that
is half of
. Similarly, we have
is half of
.
By applying the Law of Cosines, we get , and then
. Similarly, we can get
and
. Based on some trig identities, we can compute that
, and
.
Finally, the area of equals
. Therefore, the final answer is
. ~xamydad
Remark: I didn't figure out how to add segments ,
,
and
. Can someone please help add these segments?
(Added :) ~Math_Genius_164)
Solution 6
First and foremost as
is the perpendicular bisector of
. Now note that quadrilateral
is cyclic, because
and
. Similarly quadrilateral
is cyclic,
Let
,
,
be the
,
, and
excenters of
respectively. Then it follows that
. By angle bisector theorem we have
. Now let the feet of the perpendiculars from
and
to
be
and
resptively. Then by tangents we have
From the previous ratios,
Similarly we can find that
and
and thus
-tkhalid
Solution 7 -Leonard_my_dude
Trig values we use here:
First let the incenter be . Let
be the midpoint of minor arc
on
and let
be the foot of
to
.
We can find using Stewart's Theorem: from Angle Bisector Theorem
and
. Then it is easy to find that
.
Now we trig bash for . Notice that
from the Incenter Excenter Lemma. We obtain that
. To get
we angle chase to get
. Then
gives
. This means
.
Now let . It is easy to angle chase
and
. Since
, we compute that
which implies
which gives an answer of
. ~Leonard_my_dude~
See Also
2020 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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