Difference between revisions of "2023 AIME I Problems/Problem 8"
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Note that <math>\angle RXZ = \angle TZX = 90^\circ</math> by the properties of tangents, so <math>RTZX</math> is a rectangle. It follows that the diameter of <math>\odot O</math> is <math>XZ = RT = 25.</math> | Note that <math>\angle RXZ = \angle TZX = 90^\circ</math> by the properties of tangents, so <math>RTZX</math> is a rectangle. It follows that the diameter of <math>\odot O</math> is <math>XZ = RT = 25.</math> | ||
+ | |||
+ | Let <math>x=PQ</math> and <math>y=RX=TZ.</math> | ||
+ | We apply the Power of a Point Theorem to <math>R</math> and <math>T:</math> | ||
+ | <cmath>\begin{align*} | ||
+ | y^2 &= 9(9+x), \\ | ||
+ | y^2 &= 16(16-x). | ||
+ | \end{align*}</cmath> | ||
+ | We solve this system of equations to get <math>x=7</math> and <math>y=12.</math> Alternatively, we can find these results by the symmetry on rectangle <math>RTZX</math> and semicircle <math>\widehat{XPZ}.</math> | ||
==Solution 2== | ==Solution 2== |
Revision as of 18:31, 9 February 2023
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.
Let be the incenter of
for which
is tangent to
and
at
and
respectively. Moreover, suppose that
and
are the feet of the perpendiculars from
to
and
respectively, such that
intersects
at
and
We obtain the following diagram:
Note that
by the properties of tangents, so
is a rectangle. It follows that the diameter of
is
Let and
We apply the Power of a Point Theorem to
and
We solve this system of equations to get
and
Alternatively, we can find these results by the symmetry on rectangle
and semicircle
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.