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

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We use simple geometry to solve this problem.
  
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<asy>
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import graph;
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unitsize(0.1cm);
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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);
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draw(E--D--A--B--C--E--H--G--F--C);
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/*Diagram by Technodoggo*/
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</asy>
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 +
We are given that <math>A</math>, <math>D</math>, <math>H</math>, and <math>G</math> are concyclic; call the circle that they all pass through circle <math>\omega</math> with center <math>O</math>. 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 <math>HG</math> and <math>AD</math> and take the midpoints of <math>HG</math> and <math>AD</math> to be <math>P</math> and <math>Q</math>, 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);
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draw(E--D--A--B--C--E--H--G--F--C);
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pair P = (95, 33);pair Q = (0, 8);
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dot(A);dot(B);dot(C);dot(D);dot(E);dot(F);dot(G);dot(H);dot(P);dot(Q);
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label("$P$", P, N);label("$Q$", Q, W);
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draw(Q--(107,8));draw(P--(95,0));
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pair O = (95,8);
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dot(O);label("$O$", O, NW);
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/*Diagram by Technodoggo*/
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</asy>
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We could draw the circumcircle, but actually it does not matter for our solution; all that matters is that <math>OA=OH=r</math>, where <math>r</math> is the circumradius.
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 +
By the Pythagorean Theorem, <math>OQ^2+QA^2=OA^2</math>. Also, <math>OP^2+PH^2=OH^2</math>. We know that <math>OQ=DE+HP</math>, and <math>HP=\dfrac{184}2=92</math>; <math>QA=\dfrac{16}2=8</math>; <math>OP=DQ+HE=8+17=25</math>; and finally, <math>PH=92</math>. Let <math>DE=x</math>. We now know that <math>OA^2=(x+92)^2+8^2</math> and <math>OH^2=25^2+92^2</math>. Recall that <math>OA=OH</math>; thus, <math>OA^2=OH^2</math>. We solve for <math>x</math>:
 +
 +
\begin{align*}
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(x+92)^2+8^2&=25^2+92^2 \\
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(x+92)^2&=625+(100-8)^2-8^2 \\
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&=625+10000-1600+64-64 \\
 +
&=9025 \\
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x+92&=95 \\
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x&=3. \\
 +
\end{align*}
 +
 +
The question asks for <math>CE</math>, which is <math>CD-x=107-3=\boxed{104}</math>.

Revision as of 13:40, 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]

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 $CD-x=107-3=\boxed{104}$.

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