Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(Created blank page)
 
m (→‎Solution 3)
 
(40 intermediate revisions by 16 users not shown)
Line 1: Line 1:
 +
==Problem==
 +
Rectangles <math>ABCD</math> and <math>EFGH</math> are drawn such that <math>D,E,C,F</math> are collinear. Also, <math>A,D,H,G</math> all lie on a circle. If <math>BC=16</math>,<math>AB=107</math>,<math>FG=17</math>, and <math>EF=184</math>, what is the length of <math>CE</math>?
  
 +
<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 <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);
 +
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 <math>OA=OH=r</math>, where <math>r</math> is the circumradius.
 +
 +
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*}
 +
(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 <math>CE</math>, which is <math>CD-x=107-3=\boxed{104}</math>.
 +
 +
~Technodoggo
 +
 +
 +
==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 can use guess and check to find that <math>x=3</math>, so <math>EC=\boxed{104}</math>.
 +
<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 <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 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
 +
 +
==Solution 4==
 +
 +
One liner: <math>107-\sqrt{92^2+25^2-8^2}+92=\boxed{104}</math>
 +
 +
~Bluesoul
 +
===Explanation===
 +
Let <math>OP</math> intersect <math>DF</math> at <math>T</math> (using the same diagram as Solution 2).
 +
 +
The formula calculates the distance from <math>O</math> to <math>H</math> (or <math>G</math>), <math>\sqrt{92^2+25^2}</math>, then shifts it to <math>OD</math> and the finds the distance from <math>O</math> to <math>Q</math>, <math>\sqrt{92^2+25^2-8^2}</math>. <math>107</math> minus that gives <math>CT</math>, and when added to <math>92</math>, half of <math>FE=TE</math>, gives <math>CT+TE=CE</math>
 +
 +
==Solution 5==
 +
 +
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 \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>
 +
 +
~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==
 +
https://youtu.be/1LWwJeFpU9Y
 +
<br>~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
 +
 +
==See also==
 +
{{AIME box|year=2024|n=I|num-b=4|num-a=6}}
 +
 +
{{MAA Notice}}

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2024_AIME_I_Problems/Problem_5&oldid=229093"