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

m (→‎Solution 1)
m (→‎Solution 2)
Line 36: Line 36:
 
By Exterior Angle Theorem on triangle <math>AED</math>, <math>\angle{BED}=2x</math>.
 
By Exterior Angle Theorem on triangle <math>AED</math>, <math>\angle{BED}=2x</math>.
 
By Exterior Angle Theorem on triangle <math>ADB</math>, <math>\angle{BDC}=3x</math>.
 
By Exterior Angle Theorem on triangle <math>ADB</math>, <math>\angle{BDC}=3x</math>.
−
This tells us <math>\angle{BCA}=\angle{ABC}=3x</math>
+
This tells us <math>\angle{BCA}=\angle{ABC}=3x</math> and <math>3x+3x+x=180</math>.
 
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>.
 
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>.
  

Revision as of 21:40, 14 March 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 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 \[180-4\left(\frac{180}{7}\right) = \frac{540}{7}\] for an answer of $\boxed{547}$.

-molocyxu

Solution 2

Let $\angle{BAC}$ be $x$. $\angle{ADE}=x$. 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 $\boxed{547}$.

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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