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Difference between revisions of "2020 AIME I Problems/Problem 1"

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== Problem ==
 
== Problem ==
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Let <math>ABCD</math> be a [[parallelogram]].  Extend <math>\overline{DA}</math> through <math>A</math> to a point <math>P,</math> and let <math>\overline{PC}</math> meet <math>\overline{AB}</math> at <math>Q</math> and <math>\overline{DB}</math> at <math>R.</math>  Given that <math>PQ = 735</math> and <math>QR = 112,</math> find <math>RC.</math>
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In <math>\triangle ABC</math> with <math>AB=AC,</math> point <math>D</math> lies strictly between <math>A</math> and <math>C</math> on side <math>\overline{AC},</math> and point <math>E</math> lies strictly between <math>A</math> and <math>B</math> on side <math>\overline{AB}</math> such that <math>AE=ED=DB=BC.</math> The degree measure of <math>\angle ABC</math> is <math>\tfrac{m}{n},</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n.</math>
 +
 
 +
== Solution 1==
 +
<asy>
 +
size(10cm);
 +
pair A, B, C, D, F;
 +
A = (0, tan(3 * pi / 7));
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B = (1, 0);
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C = (-1, 0);
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F = rotate(90/7, A) * (A - (0, 2));
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D = rotate(900/7, F) * A;
 +
 
 +
draw(A -- B -- C -- cycle);
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draw(F -- D);
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draw(D -- B);
 +
 
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label("$A$", A, N);
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label("$B$", B, E);
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label("$C$", C, W);
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label("$D$", D, W);
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label("$E$", F, E);
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</asy>
 +
 
 +
If we set <math>\angle{BAC}</math> to <math>x</math>, we can find all other angles through these two properties:
 +
1. Angles in a triangle sum to <math>180^{\circ}</math>.
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2. The base angles of an isosceles triangle are congruent.
 +
 
 +
Now we angle chase. <math>\angle{ADE}=\angle{EAD}=x</math>, <math>\angle{AED} = 180-2x</math>, <math>\angle{BED}=\angle{EBD}=2x</math>, <math>\angle{EDB} = 180-4x</math>, <math>\angle{BDC} = \angle{BCD} = 3x</math>, <math>\angle{CBD} = 180-6x</math>. Since <math>AB = AC</math> as given by the problem, <math>\angle{ABC} = \angle{ACB}</math>, so <math>180-4x=3x</math>. Therefore, <math>x = 180/7^{\circ}</math>, and our desired angle is <cmath>180-4\left(\frac{180}{7}\right) = \frac{540}{7}</cmath> for an answer of <math>\boxed{547}</math>.
 +
 
 +
See here for a video solution: https://youtu.be/4e8Hk04Ax_E
 +
 
 +
==Solution 2==
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Let <math>\angle{BAC}</math> be <math>x</math> in degrees. <math>\angle{ADE}=x</math>.
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By Exterior Angle Theorem on triangle <math>AED</math>, <math>\angle{BED}=2x</math>.
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By Exterior Angle Theorem on triangle <math>ADB</math>, <math>\angle{BDC}=3x</math>.
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This tells us <math>\angle{BCA}=\angle{ABC}=3x</math> and <math>3x+3x+x=180</math>.
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Thus <math>x=\frac{180}{7}</math> and we want <math>\angle{ABC}=3x=\frac{540}{7}</math> to get an answer of <math>\boxed{547}</math>.
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==Solution 3 (Official MAA)==
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Let <math>x = \angle ABC = \angle ACB</math>. Because <math>\triangle BCD</math> is isosceles, <math>\angle CBD = 180^\circ - 2x</math>. Then
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<cmath>\angle DBE = x - \angle CBD = x - (180^\circ - 2x) = 3x - 180^\circ\!.</cmath>Because <math>\triangle EDA</math> and <math>\triangle DBE</math> are also isosceles,
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<cmath>\angle BAC =\frac12(\angle EAD + \angle ADE) = \frac12(\angle BED)= \frac12(\angle DBE)</cmath>
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<cmath>= \frac12 (3x - 180^\circ) = \frac32x-90^\circ\!.</cmath>
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Because <math>\triangle ABC</math> is isosceles, <math>\angle BAC</math> is also <math>180^\circ-2x</math>, so <math>\frac32x - 90^\circ = 180^\circ - 2x</math>, and it follows that
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<math>\angle ABC = x = \left(\frac{540}7\right)^\circ</math>. The requested sum is <math>540+7 = 547</math>.
 +
 
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<asy>
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unitsize(4 cm);
  
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== Solution ==
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pair A, B, C, D, E;
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=== Solution 1 ===
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real a = 180/7;
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[[Image:AIME_1998-6.png|350px]]
 
  
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There are several [[similar triangles]]. <math>\triangle PAQ\sim \triangle PDC</math>, so we can write the [[proportion]]:
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A = (0,0);
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B = dir(180 - a/2);
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C = dir(180 + a/2);
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D = extension(B, B + dir(270 + a), A, C);
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E = extension(D, D + dir(90 - 2*a), A, B);
  
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<div style="text-align:center;">
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draw(A--B--C--cycle);
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<math>\frac{AQ}{CD} = \frac{PQ}{PC} = \frac{735}{112 + 735 + RC} = \frac{735}{847 + RC}</math>
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draw(B--D--E);
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</div>
 
  
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Also, <math>\triangle BRQ\sim DRC</math>, so:
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label("$A$", A, dir(0));
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label("$B$", B, NW);
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label("$C$", C, SW);
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label("$D$", D, S);
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label("$E$", E, N);
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</asy>
  
−
<div style="text-align:center;">
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https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_25
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<math>\frac{QR}{RC} = \frac{QB}{CD} = \frac{112}{RC} = \frac{CD - AQ}{CD} = 1 - \frac{AQ}{CD}</math><br />
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(Almost Mirrored)
  
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<math>\frac{AQ}{CD} = 1 - \frac{112}{RC} = \frac{RC - 112}{RC}</math>
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See here for a video solution:
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</div>
 
  
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Substituting,
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https://youtu.be/4XkA0DwuqYk
  
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<div style="text-align:center;">
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==Video solution==
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<math>\frac{AQ}{CD} = \frac{735}{847 + RC} = \frac{RC - 112}{RC}</math><br />
 
  
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<math>735RC = (RC + 847)(RC - 112)</math><br />
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https://youtu.be/IH7yM3L5xjA
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<math>0 = RC^2 - 112\cdot847</math>
 
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</div>
 
  
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Thus, <math> RC = \sqrt{112*847} = 308</math>.
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https://youtu.be/mgRNqSDCvgM ~yofro
  
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=== Solution 2 ===
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==See Also==
  
−
We have <math>\triangle BRQ\sim \triangle DRC</math> so <math>\frac{112}{RC} = \frac{BR}{DR}</math>. We also have <math>\triangle BRC \sim \triangle DRP</math> so <math>\frac{ RC}{847} = \frac {BR}{DR}</math>. Equating the two results gives <math>\frac{112}{RC} =  \frac{ RC}{847}</math> and so <math>RC^2=112*847</math> which solves to <math>RC=\boxed{481}</math>
+
{{AIME box|year=2020|n=I|before=First Problem|num-a=2}}
 +
[[Category:Introductory Geometry Problems]]
 +
{{MAA Notice}}

Revision as of 15:51, 14 December 2020

Problem

In $\triangle ABC$ with $AB=AC,$ point $D$ lies strictly between $A$ and $C$ on side $\overline{AC},$ and point $E$ lies strictly between $A$ and $B$ on side $\overline{AB}$ such that $AE=ED=DB=BC.$ The degree measure of $\angle ABC$ is $\tfrac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n.$

Solution 1

[asy] size(10cm); pair A, B, C, D, F; A = (0, tan(3 * pi / 7)); B = (1, 0); C = (-1, 0); F = rotate(90/7, A) * (A - (0, 2)); D = rotate(900/7, F) * A; draw(A -- B -- C -- cycle); draw(F -- D); draw(D -- B); label("$A$", A, N); label("$B$", B, E); label("$C$", C, W); label("$D$", D, W); label("$E$", F, E); [/asy]

If we set $\angle{BAC}$ to $x$, we can find all other angles through these two properties: 1. Angles in a triangle sum to $180^{\circ}$. 2. The base angles of an isosceles triangle are congruent.

Now we angle chase. $\angle{ADE}=\angle{EAD}=x$, $\angle{AED} = 180-2x$, $\angle{BED}=\angle{EBD}=2x$, $\angle{EDB} = 180-4x$, $\angle{BDC} = \angle{BCD} = 3x$, $\angle{CBD} = 180-6x$. Since $AB = AC$ as given by the problem, $\angle{ABC} = \angle{ACB}$, so $180-4x=3x$. Therefore, $x = 180/7^{\circ}$, and our desired angle is \[180-4\left(\frac{180}{7}\right) = \frac{540}{7}\] for an answer of $\boxed{547}$.

See here for a video solution: https://youtu.be/4e8Hk04Ax_E

Solution 2

Let $\angle{BAC}$ be $x$ in degrees. $\angle{ADE}=x$. By Exterior Angle Theorem on triangle $AED$, $\angle{BED}=2x$. By Exterior Angle Theorem on triangle $ADB$, $\angle{BDC}=3x$. This tells us $\angle{BCA}=\angle{ABC}=3x$ and $3x+3x+x=180$. Thus $x=\frac{180}{7}$ and we want $\angle{ABC}=3x=\frac{540}{7}$ to get an answer of $\boxed{547}$.

Solution 3 (Official MAA)

Let $x = \angle ABC = \angle ACB$. Because $\triangle BCD$ is isosceles, $\angle CBD = 180^\circ - 2x$. Then \[\angle DBE = x - \angle CBD = x - (180^\circ - 2x) = 3x - 180^\circ\!.\]Because $\triangle EDA$ and $\triangle DBE$ are also isosceles, \[\angle BAC =\frac12(\angle EAD + \angle ADE) = \frac12(\angle BED)= \frac12(\angle DBE)\] \[= \frac12 (3x - 180^\circ) = \frac32x-90^\circ\!.\] Because $\triangle ABC$ is isosceles, $\angle BAC$ is also $180^\circ-2x$, so $\frac32x - 90^\circ = 180^\circ - 2x$, and it follows that $\angle ABC = x = \left(\frac{540}7\right)^\circ$. The requested sum is $540+7 = 547$.

[asy] unitsize(4 cm); pair A, B, C, D, E; real a = 180/7; A = (0,0); B = dir(180 - a/2); C = dir(180 + a/2); D = extension(B, B + dir(270 + a), A, C); E = extension(D, D + dir(90 - 2*a), A, B); draw(A--B--C--cycle); draw(B--D--E); label("$A$", A, dir(0)); label("$B$", B, NW); label("$C$", C, SW); label("$D$", D, S); label("$E$", E, N); [/asy]

https://artofproblemsolving.com/wiki/index.php/1961_AHSME_Problems/Problem_25 (Almost Mirrored)

See here for a video solution:

https://youtu.be/4XkA0DwuqYk

Video solution

https://youtu.be/IH7yM3L5xjA

https://youtu.be/mgRNqSDCvgM ~yofro

See Also

2020 AIME I (Problems • Answer Key • Resources)
Preceded by
First Problem 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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