2023 AIME I Problems/Problem 8
Problem
Rhombus has
There is a point
on the incircle of the rhombus such that the distances from
to the lines
and
are
and
respectively. Find the perimeter of
Diagram
~MRENTHUSIASM
Solution 1
This solution refers to the Diagram section.
Solution 2
Label the points of the rhombus to be ,
,
, and
and the center of the incircle to be
so that
,
, and
are the distances from point
to side
, side
, and
respectively. Through this, we know that the distance from the two pairs of opposite lines of rhombus
is
and circle
has radius
.
Call the feet of the altitudes from P to side , side
, and side
to be
,
, and
respectively. Additionally, call the feet of the altitudes from
to side
, side
, and side
to be
,
, and
respectively.
Draw a line segment from to
so that it is perpendicular to
. Notice that this segment length is equal to
and is
by Pythagorean Theorem.
Similarly, perform the same operations with side to get
.
By equal tangents, . Now, label the length of segment
and
.
Using Pythagorean Theorem again, we get
Which also gives us and
.
Since the diagonals of the rhombus intersect at and are angle bisectors and are also perpendicular to each other, we can get that
~Danielzh
Solution 3
Denote by the center of
.
We drop an altitude from
to
that meets
at point
.
We drop altitudes from
to
and
that meet
and
at
and
, respectively.
We denote
.
We denote the side length of
as
.
Because the distances from to
and
are
and
, respectively, and
, the distance between each pair of two parallel sides of
is
.
Thus,
and
.
We have
Thus, .
In , we have
.
Thus,
Taking the imaginary part of this equation and plugging and
into this equation, we get
We have
Because is on the incircle of
,
. Plugging this into
, we get the following equation
By solving this equation, we get and
.
Therefore,
.
Therefore, the perimeter of is
.
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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.