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| − | ==Problem==
| + | 1+1=2 |
| − | 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>?
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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 = (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);
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| − | dot(A^^B^^C^^D^^E^^F^^G^^H);
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| − | 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);
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| − | draw(E--D--A--B--C--E--H--G--F--C);
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| − | </asy>
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| − | ==Video Solution & More by MegaMath==
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| − | https://www.youtube.com/watch?v=A-awfSnHceE
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| − | | |
| − | ==Solution 1==
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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);
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| − | 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.
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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);
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| − | 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>:
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| − | | |
| − | \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 \\
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| − | &=9025 \\
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| − | x+92&=95 \\
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| − | x&=3. \\
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| − | \end{align*}
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| − | | |
| − | The question asks for <math>CE</math>, which is <math>CD-x=107-3=\boxed{104}</math>.
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| − | | |
| − | ~Technodoggo
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| − | | |
| − | ==Solution 2==
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| − | 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>.
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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);pair P = (0,33);
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| − | 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);
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| − | draw(E--D--A--B--C--E--H--G--F--C);
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| − | draw(D--P--H, dashed);
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| − | | |
| − | /*graph originally by Technodoggo, revised by alexanderruan*/
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| − | </asy>
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| − | | |
| − | ~alexanderruan
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| − | | |
| − | ~diagram by Technodoggo
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| − | | |
| − | ==Solution 3==
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| − | We find that <cmath>\angle GAB = 90-\angle DAG = 90 - (180 - \angle GHD) = \angle DHE.</cmath>
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| − | | |
| − | 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>
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| − | ~AtharvNaphade ~coolruler ~eevee9406
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| − | | |
| − | ==Solution 4==
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| − | | |
| − | One liner: <math>107-\sqrt{92^2+25^2-8^2}+92=\boxed{104}</math>
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| − | | |
| − | ~Bluesoul
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| − | ===Explanation===
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| − | Let <math>OP</math> intersect <math>DF</math> at <math>T</math> (using the same diagram as Solution 2).
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| − | 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>
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| − | | |
| − | ==Solution 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>
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| − | 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>
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| − | | |
| − | 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>
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| − | | |
| − | ~PureSwag
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| − | | |
| − | ==Solution 6==
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| − | 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>
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| − | 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
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| − | | |
| − | ~kipper
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| − | | |
| − | ==Video Solution with Circle Properties==
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| − | https://youtu.be/1LWwJeFpU9Y
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| − | <br>~Veer Mahajan
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| − | | |
| − | ==Video Solution 1 by OmegaLearn.org==
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| − | https://youtu.be/Ss-u5auH4fE
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| − | | |
| − | ==Video Solution 2==
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| − | https://youtu.be/R6dkIKuZHsM
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| − | | |
| − | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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| − | | |
| − | ==Fast Video Solution by Do Math or Go Home==
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| − | https://www.youtube.com/watch?v=Hz3PGY_a9Hc
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| − | | |
| − | ==See also==
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| − | {{AIME box|year=2024|n=I|num-b=4|num-a=6}}
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| − | | |
| − | {{MAA Notice}}
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