Difference between revisions of "2024 AIME I Problems/Problem 5"
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− | 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>. | + | 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)=561</math>. We know that <math>EC</math> must be an integer, so we can substitute small integer values for <math>x</math>. <math>x=3</math> works, so <math>EC=\boxed{104}</math>. |
<asy> | <asy> | ||
import graph; | import graph; |
Revision as of 23:10, 4 February 2024
Contents
Problem
Rectangles and
are drawn such that
are collinear. Also,
all lie on a circle. If
,
,
, and
, what is the length of
?
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
:
\begin{align*} (x+92)^2+8^2&=25^2+92^2 \\ (x+92)^2&=625+(100-8)^2-8^2 \\ &=625+10000-1600+64-64 \\ &=9025 \\ x+92&=95 \\ x&=3. \\ \end{align*}
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 know that
must be an integer, so we can substitute small integer values for
.
works, so
.
~alexanderruan
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
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.