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

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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(\frac{180}{7}) = \frac{540}{7}</cmath> for an answer of <math>\boxed{547}</math>.
 
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(\frac{180}{7}) = \frac{540}{7}</cmath> for an answer of <math>\boxed{547}</math>.
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-molocyxu
  
 
==See Also==
 
==See Also==

Revision as of 16:18, 12 March 2020

Note: Please do not post problems here until after the AIME.

Problem

In $\triangle ABC$ with $AB=BC,$ 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

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 isoceles 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 \[180-4(\frac{180}{7}) = \frac{540}{7}\] for an answer of $\boxed{547}$.

-molocyxu

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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