AIME 2020 I · 第 1 题

Solution 1

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 $540+7 = 547$.

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.

$$\begin{aligned} \angle A &= y \\ \angle B &= x+z \\ \angle C &= \frac{180-x}{2} \\ \end{aligned}$$ Then, using triangle sum of angles theorem, we find that

$$\begin{aligned} \angle A + \angle B + \angle C = x+y+z+\frac{180-x}{2}=180 \\ \end{aligned}$$ Now we just need to find the variables.

$$\begin{aligned} (180-2y)+z = 180& \\ (180-2z)+y+\frac{180-x}{2} = 180& \\ \end{aligned}$$ Notice how all the equations equal 180. We can use this to write

$$\begin{aligned} (180-2y)+z = (180-2z)+y+\frac{180-x}{2}=x+y+z+\frac{180-x}{2} \\ \end{aligned}$$ Simplifying, we get

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

$$\begin{aligned} \frac{180-x}{2}=x+z \\ \end{aligned}$$ Substituting $z$ with $2y$, we get

$$\begin{aligned} \frac{180-x}{2}=x+2y \\ 180-x=2x+4y \\ \end{aligned}$$ $$\begin{aligned} 180-(6y-6z+180)=2(6y-6z+180)+4y& \\ 180-6y+12y-180=12y-24y+360+4y& \\ \end{aligned}$$ $$\begin{aligned} 6y=-8y+360& \\ \end{aligned}$$ With this, we get

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