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

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Let <math>\angle{DHE} = \theta.</math> This means that <math>DE = 17\tan{\theta}.</math> Since quadrilateral <math>ADHG</math> is cyclic, <math>\angle{DAG} = 180 - \angle{DHG} = 90 - \theta.</math>
 
Let <math>\angle{DHE} = \theta.</math> This means that <math>DE = 17\tan{\theta}.</math> Since quadrilateral <math>ADHG</math> is cyclic, <math>\angle{DAG} = 180 - \angle{DHG} = 90 - \theta.</math>
  
−
Let <math>X = AG \cap DF.</math> Then, <math>\Delta DXA \~ \Delta FXG,</math> with side ratio <math>16:17.</math> Also, since <math>\angle{DAG} = 90 - \theta, \angle{DXA} = \angle{FXG} = \theta.</math> Using the similar triangles, we have <math>\tan{\theta} = \frac{16}{DX} = \frac{17}{FX}</math> and <math>DX + FX = DE + EF = 17\tan{\theta} + 184.</math>
+
Let <math>X = AG \cap DF.</math> Then, <math>\Delta DXA \sim \Delta FXG,</math> with side ratio <math>16:17.</math> Also, since <math>\angle{DAG} = 90 - \theta, \angle{DXA} = \angle{FXG} = \theta.</math> Using the similar triangles, we have <math>\tan{\theta} = \frac{16}{DX} = \frac{17}{FX}</math> and <math>DX + FX = DE + EF = 17\tan{\theta} + 184.</math>
  
 
Since we want <math>CE = CD - DE = 107 - 17\tan{\theta},</math> we only need to solve for <math>\tan{\theta}</math> in this system of equations. Solving yields <math>\tan{\theta} = \frac{3}{17},</math> so <math>CE = \boxed{104.}</math>
 
Since we want <math>CE = CD - DE = 107 - 17\tan{\theta},</math> we only need to solve for <math>\tan{\theta}</math> in this system of equations. Solving yields <math>\tan{\theta} = \frac{3}{17},</math> so <math>CE = \boxed{104.}</math>

Revision as of 19:04, 3 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 $CE$?

[asy] import graph; unitsize(0.1cm); pair A = (0,0);pair B = (70,0);pair C = (70,16);pair D = (0,16);pair E = (3,16);pair F = (90,16);pair G = (90,33);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); [/asy]

Solution 1

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 $EC=\boxed{104}$. [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);pair P = (0,33); label("$A$", A, SW);label("$B$", B, SE);label("$C$", C, N);label("$D$", D, W);label("$E$", E, S);label("$F$", F, SE);label("$G$", G, NE);label("$H$", H, N);label("$P$", P, NW); draw(E--D--A--B--C--E--H--G--F--C); draw(D--P--H, dashed); [/asy]

~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

We find that \[\angle GAB = 90-\angle DAG = 90 - (180 - \angle GHD) = \angle DHE.\]

Let $x = DE$ and $T = FG \cap AB$. By similar triangles $\triangle DHE \sim \triangle GAB$ we have $\frac{DE}{EH} = \frac{GT}{AT}$. Substituting lengths we have $\frac{x}{17} = \frac{16 + 17}{184 + x}.$ Solving, we find $x = 3$ and thus $CE = 107 - 3 = \boxed{104}.$ ~AtharvNaphade ~coolruler ~eevee9406

Solution 4

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

~Bluesoul

Explanation

Let $OP$ intersect $DF$ at $T$ (using the same diagram as Solution 2).

The formula calculates the distance from $O$ to $H$ (or $G$), $\sqrt{92^2+25^2}$, then shifts it to $OD$ and the finds the distance from $O$ to $Q$, $\sqrt{92^2+25^2-8^2}$. $107$ minus that gives $CT$, and when added to $92$, half of $FE=TE$, gives $CT+TE=CE$

Solution 5

Let $\angle{DHE} = \theta.$ This means that $DE = 17\tan{\theta}.$ Since quadrilateral $ADHG$ is cyclic, $\angle{DAG} = 180 - \angle{DHG} = 90 - \theta.$

Let $X = AG \cap DF.$ Then, $\Delta DXA \sim \Delta FXG,$ with side ratio $16:17.$ Also, since $\angle{DAG} = 90 - \theta, \angle{DXA} = \angle{FXG} = \theta.$ Using the similar triangles, we have $\tan{\theta} = \frac{16}{DX} = \frac{17}{FX}$ and $DX + FX = DE + EF = 17\tan{\theta} + 184.$

Since we want $CE = CD - DE = 107 - 17\tan{\theta},$ we only need to solve for $\tan{\theta}$ in this system of equations. Solving yields $\tan{\theta} = \frac{3}{17},$ so $CE = \boxed{104.}$

~PureSwag

Video Solution with Circle Properties

https://youtu.be/1LWwJeFpU9Y
~Veer Mahajan

Video Solution 1 by OmegaLearn.org

https://youtu.be/Ss-u5auH4fE

See also

2024 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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All AIME Problems and Solutions

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