2023 AIME I Problems/Problem 8
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
Solution 1
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 16 and 9, 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 (1), 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)
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 $\sqrt{(\frac{25}{2})^2-(\frac^2{7}{2})^2}=12$ (Error compiling LaTeX. Unknown error_msg) 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 \tan{\angle{OWX}}=\frac{1}{2} and
Since the diagonals of the rhombus intersect at and are angle bisectors and are also perpendicular to each other, we can get that
$$ (Error compiling LaTeX. Unknown error_msg) \begin{align*} \frac{OX}{OW}=\tan{\angle{OWX}} \\ OX=\frac{25\sqrt{5}}{4} \\ WX^2=OW^2+OX^2 \\ WX=125/4 \\ 4WX= \end{align*} $$ (Error compiling LaTeX. Unknown error_msg)
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 |
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