Track / Stage 5 / 312 of 400 #912 of 1964

Problem 912

AIME late

Geometry Difficulty 5.8 Find the answer

5. In a convex 2018-gon $A_{1} A_{2} \ldots A_{2018}$ (not necessarily regular), the sides $A_{1} A_{2}$ and $A_{3} A_{4}$ are extended to intersect at point $B_{2}$ ; the same is done with the pairs of sides $A_{2} A_{3}$ and $A_{4} A_{5}$ (resulting in point $B_{3}$ ), $\ldots, A_{2017} A_{2018}$ and $A_{1} A_{2}$ (resulting in point $B_{2018}$ ), $A_{2018} A_{1}$ and $A_{2} A_{3}$ (resulting in point $B_{1}$ ).

In the end, a "star" $A_{1} B_{1} A_{2} B_{2} A_{3} B_{3} A_{4} \ldots B_{2018} A_{1}$ is obtained. Find the sum of the angles $B_{1}, B_{2}, \ldots, B_{2018}$ of this "star".

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

5. It remains to substitute the numerical values:

$S_{1}=\frac{240 \sqrt{13}-39}{800}, \quad S_{2}=\frac{39}{400}, \quad S_{3}=\frac{120 \sqrt{13}-39}{200}, \quad S_{4}=\frac{1239-120 \sqrt{13}}{200}$

ANSWER. The ratio of the specified areas is $\frac{1239-120 \sqrt{13}}{120 \sqrt{13}-39}$ .

PROBLEM 14 (FOR 10TH GRADE). The integer part $\lfloor x\rfloor$ of a real number $x$ is defined as the greatest integer $M$ such that $M \leq x$ . Solve the equation $\sqrt{\lfloor x / 3-2\rfloor}=\lfloor\sqrt{x / 3-2}\rfloor$ .

SOLUTION. Introduce the variable $y=x / 3-2$ . In this case, $y \geq 0$ . The equation will take the form

$\sqrt{\lfloor y\rfloor}=\lfloor\sqrt{y}\rfloor \text {. }$

If $y=n^{2}$ , where $n$ is an integer, then the left and right sides of equation (1) are equal to $n$ . Let $n^{2}<y<(n+1)^{2}$ . Then $n<\sqrt{y}<n+1$ and the right side is $\lfloor y\rfloor=n$ . Equation (1) becomes $\sqrt{\lfloor y\rfloor}=n$ . Squaring both sides: $\lfloor y\rfloor=n^{2}$ . This is equivalent to the inequalities $n^{2} \leq y<n^{2}+1$ .

Thus, $y \in \bigcup_{n=0}^{\infty}\left[n^{2} ; n^{2}+1\right)$ .

Returning to the original variable $x=3(y+2)$ , we get

ANSWER: $x \in \bigcup_{n=0}^{\infty}\left[3\left(n^{2}+2\right) ; 3\left(n^{2}+3\right)\right)$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.