Difference between revisions of "2024 AIME I Problems/Problem 5"
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</asy> | </asy> | ||
− | ==Solution 1 | + | ==Solution 1== |
Suppose <math>DE=x</math>. Extend <math>AD</math> and <math>GH</math> until they meet at <math>P</math>. From the [[Power of a Point Theorem]], we have <math>(PH)(PG)=(PD)(PA)</math>. Substituting in these values, we get <math>(x)(x+184)=(17)(33)</math>. Using simple guess and check, we find that <math>x=3</math> so <math>EC=\boxed{104}</math>. | Suppose <math>DE=x</math>. Extend <math>AD</math> and <math>GH</math> until they meet at <math>P</math>. From the [[Power of a Point Theorem]], we have <math>(PH)(PG)=(PD)(PA)</math>. Substituting in these values, we get <math>(x)(x+184)=(17)(33)</math>. Using simple guess and check, we find that <math>x=3</math> so <math>EC=\boxed{104}</math>. | ||
+ | <asy> | ||
+ | import graph; | ||
+ | unitsize(0.1cm); | ||
+ | |||
+ | pair A = (0,0);pair B = (107,0);pair C = (107,16);pair D = (0,16);pair E = (3,16);pair F = (187,16);pair G = (187,33);pair H = (3,33);pair P = (0,33); | ||
+ | label("$A$", A, SW);label("$B$", B, SE);label("$C$", C, N);label("$D$", D, W);label("$E$", E, S);label("$F$", F, SE);label("$G$", G, NE);label("$H$", H, N);label("$P$", P, NW); | ||
+ | draw(E--D--A--B--C--E--H--G--F--C); | ||
+ | draw(D--P--H, dashed); | ||
+ | </asy> | ||
~alexanderruan | ~alexanderruan |
Revision as of 19:59, 2 February 2024
Problem
Rectangles and are drawn such that are collinear. Also, all lie on a circle. If ,,, and , what is the length of ?
Solution 1
Suppose . Extend and until they meet at . From the Power of a Point Theorem, we have . Substituting in these values, we get . Using simple guess and check, we find that so .
~alexanderruan
Solution 2
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 3
First, draw a line from to . is then a cyclic quadrilateral.
The triangle formed by and and the intersection between lines and is similar to triangle .
Solving similarity ratios gives , so . ~coolruler ~eevee9406
Solution 4
One liner:
~Bluesoul
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.