2020 AIME I Problems/Problem 1

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 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

Solution 4 (writing equations)

graph soon

We write equations based on the triangle sum of angles theorem. There are angles that do not need variables as the less variables the better.

$\begin{align*} \angle A &= y \\ \angle B &= x+z \\ \angle C &= \frac{180-x}{2} \\ \end{align*}$

Then, using triangle sum of angles theorem, we find that

$\begin{align*} \angle A + \angle B + \angle C = x+y+z+\frac{180-x}{2}=180 \\ \end{align*}$

Now we just need to find the variables.

$\begin{align*} (180-2y)+z = 180& \\ (180-2z)+y+\frac{180-x}{2} = 180& \\ \end{align*}$

Notice how all the equations equal 180. We can use this to write

$\begin{align*} (180-2y)+z = (180-2z)+y+\frac{180-x}{2}=x+y+z+\frac{180-x}{2} \\ \end{align*}$

Simplifying, we get

$\begin{align*} (180-2y)+z=(180-2z)+y+\frac{180-x}{2} \\ 360-4y+2z=360-4z+2y+180-x \\ \end{align*}$

$\begin{align*} 6z=6y+180-x \\ x=6y-6z+180 \\ \end{align*}$

$\begin{align*} (180-2y)+z=6y-6z+180+y+z+\frac{180-(6y-6+180)}{2} \\ 360-4y+2z=12y-12z+360+2y+2z+180-6y+6z-180 \\ \end{align*}$

$\begin{align*} 6z=12y& \\ z=2y& \\ \end{align*}$

Theres more. We are at a dead end right now because we forgot that the problem states that the triangle is isosceles. With this, we can write the equation

$\begin{align*} \frac{180-x}{2}=x+z \\ \end{align*}$

Substituting $z$ with $2y$ , we get

$\begin{align*} \frac{180-x}{2}=x+2y \\ 180-x=2x+4y \\ \end{align*}$ $\begin{align*} 180-(6y-6z+180)=2(6y-6z+180)+4y& \\ 180-6y+12y-180=12y-24y+360+4y& \\ \end{align*}$ $\begin{align*} 6y=-8y+360& \\ \end{align*}$

With this, we get

$\begin{align*} y=\frac{180}{7} \\ x=\frac{180}{7} \\ z=\frac{360}{7} \\ \end{align*}$

And a final answer of $\frac{180}{7}+\frac{360}{7} = \frac{540}{7} = \boxed{547}$ .

~orenbad

Video Solution by OmegaLearn

https://youtu.be/O_o_-yjGrOU?t=333

~ pi_is_3.14

Video solution

https://youtu.be/IH7yM3L5xjA

https://youtu.be/mgRNqSDCvgM ~yofro

Solution without words

vladimir.shelomovskii@gmail.com, vvsss

2020 AIME I (Problems • Answer Key • Resources)
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