2024 AIME I Problems/Problem 5: Difference between revisions
Technodoggo (talk | contribs) mNo edit summary |
Another solution, im bad at editing |
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~Technodoggo | ~Technodoggo | ||
Someone please edit this solution. I'm bad at latex | |||
Draw a line from A to G | |||
Draw cyclic quad | |||
Triangle formed by A, G, and intersection between lines AB and GF is similar to triangle DHE | |||
Solving similarity ratios gives DE = 3, so CE = 107 - 3 = 104 | |||
~coolruler | |||
==See also== | ==See also== | ||
Revision as of 14:41, 2 February 2024
We use simple geometry to solve this problem.
![[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); label("$A$", A, SW);label("$B$", B, SE);label("$C$", C, N);label("$D$", D, NW);label("$E$", E, S);label("$F$", F, SE);label("$G$", G, NE);label("$H$", H, NW); draw(E--D--A--B--C--E--H--G--F--C); /*Diagram by Technodoggo*/ [/asy]](http://latex.artofproblemsolving.com/d/d/b/ddb848f758af38daab590bdc84340229457e95a1.png)
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.
![[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); label("$A$", A, SW);label("$B$", B, SE);label("$C$", C, N);label("$D$", D, NW);label("$E$", E, S);label("$F$", F, SE);label("$G$", G, NE);label("$H$", H, NW); draw(E--D--A--B--C--E--H--G--F--C); pair P = (95, 33);pair Q = (0, 8); dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H);dot(P);dot(Q); label("$P$", P, N);label("$Q$", Q, W); draw(Q--(107,8));draw(P--(95,0)); pair O = (95,8); dot(O);label("$O$", O, NW); /*Diagram by Technodoggo*/ [/asy]](http://latex.artofproblemsolving.com/6/c/7/6c7e2633768e591b7e2c76a92223f1bdbfc08859.png)
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
Someone please edit this solution. I'm bad at latex
Draw a line from A to G
Draw cyclic quad
Triangle formed by A, G, and intersection between lines AB and GF is similar to triangle DHE
Solving similarity ratios gives DE = 3, so CE = 107 - 3 = 104 ~coolruler
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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: Unable to save thumbnail to destination
