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

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== Solution ==
 
== Solution ==
Solution 1 (Angle Chasing)
 
 
If we set <math>\angle{BAC}</math> to <math>x</math>, we can find all other angles through these two properties:
 
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>.
 
1. Angles in a triangle sum to <math>180^{\circ}</math>.

Revision as of 16:09, 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 $\boxed{547}$.

See Also

2020 AIME I (ProblemsAnswer KeyResources)
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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