Difference between revisions of "2024 AIME I Problems/Problem 5"

Revision as of 19:40, 2 February 2024

Problem

Rectangles $ABCD$ and $EFGH$ are drawn such that $D,E,C,F$ are collinear. Also, $A,D,H,G$ all lie on a circle. If $BC=16$ , $AB=107$ , $FG=17$ , and $EF=184$ , what is the length of $CE$ ?

Solution 1 (need help with diagram)

Suppose $DE=x$ . Extend $AD$ and $GH$ until they meet at $P$ . From the Power of a Point Theorem, we have $(PH)(PG)=(PD)(PA)$ . Substituting in these values, we get $(x)(x+184)=(17)(33)$ . Using simple guess and check, we find that $x=3$ so $EC=\boxed{104}$ . -alexanderruan

Solution 2

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$", H, NW); draw(E--D--A--B--C--E--H--G--F--C); /*Diagram by Technodoggo*/ [/asy]$

We are given that $A$ , $D$ , $H$ , and $G$ are concyclic; call the circle that they all pass through circle $\omega$ with center $O$ . 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 $HG$ and $AD$ and take the midpoints of $HG$ and $AD$ to be $P$ and $Q$ , 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$", O, NW); /*Diagram by Technodoggo*/ [/asy]$

We could draw the circumcircle, but actually it does not matter for our solution; all that matters is that $OA=OH=r$ , where $r$ is the circumradius.

By the Pythagorean Theorem, $OQ^2+QA^2=OA^2$ . Also, $OP^2+PH^2=OH^2$ . We know that $OQ=DE+HP$ , and $HP=\dfrac{184}2=92$ ; $QA=\dfrac{16}2=8$ ; $OP=DQ+HE=8+17=25$ ; and finally, $PH=92$ . Let $DE=x$ . We now know that $OA^2=(x+92)^2+8^2$ and $OH^2=25^2+92^2$ . Recall that $OA=OH$ ; thus, $OA^2=OH^2$ . We solve for $x$ :

\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 $CE$ , which is $CD-x=107-3=\boxed{104}$ .

~Technodoggo

Solution 3

First, draw a line from $A$ to $G$ . $ADHG$ is then a cyclic quadrilateral.

The triangle formed by $A$ and $G$ and the intersection between lines $AB$ and $GF$ is similar to triangle $DHE$ .

Solving similarity ratios gives $DE=3$ , so $CE=107-3=\boxed{104}$ . ~coolruler ~eevee9406

Solution 4

One liner: $107-\sqrt{92^2+25^2-8^2}+92=\boxed{104}$

~Bluesoul

@@ Line 65: / Line 65: @@
 Solving similarity ratios gives <math>DE=3</math>, so <math>CE=107-3=\boxed{104}</math>.
 ~coolruler ~eevee9406
+==Solution 4==
+One liner: <math>107-\sqrt{92^2+25^2-8^2}+92=\boxed{104}</math>
+~Bluesoul
 ==See also==

2024 AIME I (Problems • Answer Key • Resources)
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