Difference between revisions of "2023 AIME I Problems/Problem 8"
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+ | ==Solution 2== | ||
+ | Label the points of the rhombus to be <math>X</math>, <math>Y</math>, <math>Z</math>, and <math>W</math> and the center of the incircle to be <math>O</math> so that <math>9</math>, <math>5</math>, and <math>16</math> are the distances from point <math>P</math> to side <math>ZW</math>, side <math>WX</math>, and <math>XY</math> respectively. Through this, we know that the distance from the two pairs of opposite lines of rhombus <math>XYZW</math> is <math>25</math> and circle <math>O</math> has radius <math>\frac{25}{2}</math>. | ||
+ | |||
+ | Call the feet of the altitudes from P to side <math>ZW</math>, side <math>WX</math>, and side <math>XY</math> to be <math>A</math>, <math>B</math>, and <math>C</math> respectively. Additionally, call the feet of the altitudes from <math>O</math> to side <math>ZW</math>, side <math>WX</math>, and side <math>XY</math> to be <math>D</math>, <math>E</math>, and <math>F</math> respectively. | ||
+ | |||
+ | Draw a line segment from <math>P</math> to <math>\overline{OD}</math> so that it is perpendicular to <math>\overline{OD}</math>. Notice that this segment length is equal to <math>AD</math> and is <math>\sqrt{(\frac{25}{2})^2-(\frac{7}{2})^2}=12</math> by Pythagorean Theorem. | ||
+ | |||
+ | Similarly, perform the same operations with side <math>WX</math> to get <math>BE=10</math>. | ||
+ | |||
+ | By equal tangents, <math>WD=WE</math>. Now, label the length of segment <math>WA=n</math> and <math>WB=n+2</math>. | ||
+ | |||
+ | Using Pythagorean Theorem again, we get | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | WA^2+PA^2&=WB^2+PB^2 | ||
+ | \\ | ||
+ | n^2+9^2&=(n+2)^2+5^2 | ||
+ | \\ | ||
+ | n&=13. | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | Which also gives us <math>\tan{\angle{OWX}}=\frac{1}{2}</math> and <math>OW=\frac{25\sqrt{5}}{2}</math>. | ||
+ | |||
+ | Since the diagonals of the rhombus intersect at <math>O</math> and are angle bisectors and are also perpendicular to each other, we can get that | ||
+ | |||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | \frac{OX}{OW}&=\tan{\angle{OWX}} \\ | ||
+ | OX&=\frac{25\sqrt{5}}{4} \\ | ||
+ | WX^2&=OW^2+OX^2 \\ | ||
+ | WX&=\frac{125}{4} \\ | ||
+ | 4WX&=\boxed{125}. | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | |||
+ | ~Danielzh | ||
==Solution 3== | ==Solution 3== | ||
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==See also== | ==See also== |
Revision as of 16:56, 9 February 2023
Contents
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 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.