Difference between revisions of "2023 AIME I Problems/Problem 8"
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</cmath> | </cmath> | ||
− | Which also gives us \tan{\angle{OWX}}=\frac{1}{2} and <math>OW=\frac{25\sqrt{5}}{2}</math> | + | 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 | 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*} | \begin{align*} | ||
\frac{OX}{OW}=\tan{\angle{OWX}} | \frac{OX}{OW}=\tan{\angle{OWX}} | ||
Line 104: | Line 104: | ||
WX=125/4 | WX=125/4 | ||
\\ | \\ | ||
− | 4WX= | + | 4WX=\boxed{125} |
\end{align*} | \end{align*} | ||
− | + | </cmath> | |
==See also== | ==See also== |
Revision as of 09:09, 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
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 and
Since the diagonals of the rhombus intersect at and are angle bisectors and are also perpendicular to each other, we can get that
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.