Difference between revisions of "2024 AIME I Problems/Problem 5"
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We find that <cmath>\angle GAB = 90-\angle DAG = 90 - (180 - \angle GHD) = \angle DHE.</cmath> | We find that <cmath>\angle GAB = 90-\angle DAG = 90 - (180 - \angle GHD) = \angle DHE.</cmath> | ||
− | Let <math>x = DE</math> and <math>T = FG \cap AB</math>. By similar triangles <math>\triangle DHE \sim \triangle | + | Let <math>x = DE</math> and <math>T = FG \cap AB</math>. By similar triangles <math>\triangle DHE \sim \triangle GAT</math> we have <math>\frac{DE}{EH} = \frac{GT}{AT}</math>. Substituting lengths we have <math>\frac{x}{17} = \frac{16 + 17}{184 + x}.</math> Solving, we find <math>x = 3</math> and thus <math>CE = 107 - 3 = \boxed{104}.</math> |
~AtharvNaphade ~coolruler ~eevee9406 | ~AtharvNaphade ~coolruler ~eevee9406 | ||
Latest revision as of 09:28, 6 October 2024
Contents
[hide]Problem
Rectangles and are drawn such that are collinear. Also, all lie on a circle. If ,,, and , what is the length of ?
Video Solution & More by MegaMath
https://www.youtube.com/watch?v=A-awfSnHceE
Solution 1
We use simple geometry to solve this problem.
We are given that , , , and are concyclic; call the circle that they all pass through circle with center . We know that, given any chord on a circle, the perpendicular bisector to the chord passes through the center; thus, given two chords, taking the intersection of their perpendicular bisectors gives the center. We therefore consider chords and and take the midpoints of and to be and , respectively.
We could draw the circumcircle, but actually it does not matter for our solution; all that matters is that , where is the circumradius.
By the Pythagorean Theorem, . Also, . We know that , and ; ; ; and finally, . Let . We now know that and . Recall that ; thus, . We solve for :
The question asks for , which is .
~Technodoggo
Solution 2
Suppose . Extend and until they meet at . From the Power of a Point Theorem, we have . Substituting in these values, we get . We can use guess and check to find that , so .
~alexanderruan
~diagram by Technodoggo
Solution 3
We find that
Let and . By similar triangles we have . Substituting lengths we have Solving, we find and thus ~AtharvNaphade ~coolruler ~eevee9406
Solution 4
One liner:
~Bluesoul
Explanation
Let intersect at (using the same diagram as Solution 2).
The formula calculates the distance from to (or ), , then shifts it to and the finds the distance from to , . minus that gives , and when added to , half of , gives
Solution 5
Let This means that Since quadrilateral is cyclic,
Let Then, with side ratio Also, since Using the similar triangles, we have and
Since we want we only need to solve for in this system of equations. Solving yields so
~PureSwag
Solution 6
Using a ruler (also acting as a straight edge), draw the figure to scale with one unit = 1mm. With a compass, draw circles until you get one such that are on the edge of the drawn circle. From here, measuring with your ruler should give
Note: 1 mm is probably the best unit to use here just for convenience (drawing all required parts of the figure fits into a normal-sized scrap paper 8.5 x 11); also all lines can be drawn with a standard 12-inch ruler
~kipper
Video Solution with Circle Properties
https://youtu.be/1LWwJeFpU9Y
~Veer Mahajan
Video Solution 1 by OmegaLearn.org
Video Solution 2
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Fast Video Solution by Do Math or Go Home
https://www.youtube.com/watch?v=Hz3PGY_a9Hc
See also
2024 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 4 |
Followed by Problem 6 | |
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.