Difference between revisions of "2024 AIME I Problems/Problem 5"
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One liner: <math>107-\sqrt{92^2+25^2-8^2}+92=\boxed{104}</math> | One liner: <math>107-\sqrt{92^2+25^2-8^2}+92=\boxed{104}</math> | ||
− | ~Bluesoul | + | ~Bluesoul |
+ | ===Explanation=== | ||
+ | Let <math>OP</math> intersect <math>DF</math> at <math>T</math> (using the same diagram as Solution 2). | ||
+ | |||
+ | The formula calculates the distance from <math>O</math> to <math>H</math> (or <math>G</math>), <math>\sqrt{92^2+25^2}</math>, then shifts it to <math>OD</math> and the finds the distance from <math>O</math> to <math>Q</math>, \sqrt{92^2+25^2-8^2}. <math>107</math> minus that gives <math>CT</math>, and when added to <math>92</math>, half of <math>FE=TE</math>, gives <math>CT+TE=CE</math> | ||
==See also== | ==See also== |
Revision as of 21:07, 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
:
\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 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
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
, \sqrt{92^2+25^2-8^2}.
minus that gives
, and when added to
, half of
, gives
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.