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 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

2020 AIME I (Problems • Answer Key • Resources)
Preceded by First Problem	Followed by Problem 2
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