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

Latest revision as of 09:28, 6 October 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 $CE$ ?

$[asy] import graph; unitsize(0.1cm); pair A = (0,0);pair B = (70,0);pair C = (70,16);pair D = (0,16);pair E = (3,16);pair F = (90,16);pair G = (90,33);pair H = (3,33); dot(A^^B^^C^^D^^E^^F^^G^^H); label("$A$", A, S);label("$B$", B, S);label("$C$", C, N);label("$D$", D, N);label("$E$", E, S);label("$F$", F, S);label("$G$", G, N);label("$H$", H, N); draw(E--D--A--B--C--E--H--G--F--C); [/asy]$

Video Solution & More by MegaMath

https://www.youtube.com/watch?v=A-awfSnHceE

Solution 1

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 2

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 $(x)(x+184)=(17)(33)=561$ . We can use guess and check to 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$", P, NW); draw(E--D--A--B--C--E--H--G--F--C); draw(D--P--H, dashed); /*graph originally by Technodoggo, revised by alexanderruan*/ [/asy]$

~alexanderruan

~diagram by 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 $\triangle DHE \sim \triangle GAT$ 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

Solution 6

Using a ruler (also acting as a straight edge), draw the figure to scale with one unit = 1mm. With a compass, draw circles until you get one such that $A,D,H,G$ are on the edge of the drawn circle. From here, measuring with your ruler should give $CE = \boxed{104.}$

Note: 1 mm is probably the best unit to use here just for convenience (drawing all required parts of the figure fits into a normal-sized scrap paper 8.5 x 11); also all lines can be drawn with a standard 12-inch ruler

~kipper

Video Solution with Circle Properties

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

Video Solution 1 by OmegaLearn.org

https://youtu.be/Ss-u5auH4fE

Video Solution 2

https://youtu.be/R6dkIKuZHsM

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Fast Video Solution by Do Math or Go Home

https://www.youtube.com/watch?v=Hz3PGY_a9Hc

@@ Line 11: / Line 11: @@
 draw(E--D--A--B--C--E--H--G--F--C);
 </asy>
+==Video Solution & More by MegaMath==
+https://www.youtube.com/watch?v=A-awfSnHceE
 ==Solution 1==
@@ Line 66: / Line 68: @@
 ==Solution 2==
-Suppose <math>DE=x</math>. Extend <math>AD</math> and <math>GH</math> until they meet at <math>P</math>. From the [[Power of a Point Theorem]], we have <math>(PH)(PG)=(PD)(PA)</math>. Substituting in these values, we get <math>(x)(x+184)=(17)(33)=561</math>. We know that <math>EC</math> must be an integer, so we can use guess and check on <math>x</math>. <math>x=3</math> works, so <math>EC=\boxed{104}</math>.
+Suppose <math>DE=x</math>. Extend <math>AD</math> and <math>GH</math> until they meet at <math>P</math>. From the [[Power of a Point Theorem]], we have <math>(PH)(PG)=(PD)(PA)</math>. Substituting in these values, we get <math>(x)(x+184)=(17)(33)=561</math>. We can use guess and check to find that <math>x=3</math>, so <math>EC=\boxed{104}</math>.
 <asy>
 import graph;
@@ Line 81: / Line 83: @@
 ~alexanderruan
+~diagram by Technodoggo
 ==Solution 3==
 We find that <cmath>\angle GAB = 90-\angle DAG = 90 - (180 - \angle GHD) = \angle DHE.</cmath>
-Let <math>x = DE</math> and <math>T = FG \cap AB</math>. By similar triangles <math>\triangle DHE \sim \triangle GAB</math> we have <math>\frac{DE}{EH} = \frac{GT}{AT}</math>. Substituting lengths we have <math>\frac{x}{17} = \frac{16 + 17}{184 + x}.</math> Solving, we find <math>x = 3</math> and thus <math>CE = 107 - 3 = \boxed{104}.</math>
+Let <math>x = DE</math> and <math>T = FG \cap AB</math>. By similar triangles <math>\triangle DHE \sim \triangle GAT</math> we have <math>\frac{DE}{EH} = \frac{GT}{AT}</math>. Substituting lengths we have <math>\frac{x}{17} = \frac{16 + 17}{184 + x}.</math> Solving, we find <math>x = 3</math> and thus <math>CE = 107 - 3 = \boxed{104}.</math>
 ~AtharvNaphade ~coolruler ~eevee9406
@@ Line 107: / Line 110: @@
 ~PureSwag
+==Solution 6==
+Using a ruler (also acting as a straight edge), draw the figure to scale with one unit = 1mm. With a compass, draw circles until you get one such that <math>A,D,H,G</math> are on the edge of the drawn circle. From here, measuring with your ruler should give <math>CE = \boxed{104.}</math>
+Note: 1 mm is probably the best unit to use here just for convenience (drawing all required parts of the figure fits into a normal-sized scrap paper 8.5 x 11); also all lines can be drawn with a standard 12-inch ruler
+~kipper
 ==Video Solution with Circle Properties==

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