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
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
We extend beyond to intersect and at and respectively, where So, we have and On the other hand, we have by the Pythagorean Theorem on right Together, we conclude that Therefore, points and must be collinear.
Let be the foot of the perpendicular from to Note that as shown below: As and by the AA Similarity, we conclude that The ratio of similitude is We get from which
Finally, the perimeter of is
~MRENTHUSIASM (inspired by awesomeming327. and WestSuburb)
Solution 2
This solution refers to the Diagram section.
Define points and as Solution 1 does. Moreover, let be the foot of the perpendicular from to be the foot of the perpendicular from to and be the foot of the perpendicular from to
We obtain the following diagram: Note that the diameter of is so It follows that:
- In right we have by symmetry, from which by the Pythagorean Theorem.
- In right we have by symmetry, from which by the Pythagorean Theorem.
Since and we conclude that We apply the Sine of a Sum Formula: Note that from which We solve this equation to get
Finally, the perimeter of is
~MRENTHUSIASM (credit given to TheAMCHub)
Solution 3
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 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 perpendicular from to 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
Solution 4
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)
Solution 5
The center of the incircle is Denote the points in which the incircle meets and as and respectively. Next, also denote the base of the perpendicular from to and as and respectively.
We can easily see that the radius of the circle is Using this and Pythagorus on right and we find that and
Since by properties of circle tangents, we can deduce by the above information that Doing Pythagorus on right and we find that (because ) From solving the just derived equations, we find that and
Next, we use Pythagorus on right (we can see it's right because of properties of rhombuses.) We get We know By Pythagorus on and we also know and Substituting these in, we have Solving for we get Now we find that each side of the rhombus The perimeter of the rhombus would be that times Our final answer is
~s214425
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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