Maths Olympiad Prep

Track / Stage 5 / 375 of 400 #975 of 1964

Problem 975

AIME late
Geometry Difficulty 5.8 Find the answer

In a convex 2018-gon A1A2A2018A_{1} A_{2} \ldots A_{2018} (not necessarily regular), the sides A1A2A_{1} A_{2} and A3A4A_{3} A_{4} are extended to intersect at point B2B_{2}; the same is done with the pairs of sides A2A3A_{2} A_{3} and A4A5A_{4} A_{5} (resulting in point B3B_{3}), ,A2017A2018\ldots, A_{2017} A_{2018} and A1A2A_{1} A_{2} (resulting in point B2018B_{2018}), A2018A1A_{2018} A_{1} and A2A3A_{2} A_{3} (resulting in point B1B_{1}).

In the end, a "star" A1B1A2B2A3B3A4B2018A1A_{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 B1,B2,,B2018B_{1}, B_{2}, \ldots, B_{2018} of this "star".

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Official solution

5. It remains to substitute the numerical values:

S1=2401339800,S2=39400,S3=1201339200,S4=123912013200 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 1239120131201339\frac{1239-120 \sqrt{13}}{120 \sqrt{13}-39}.

PROBLEM 14 (FOR 10TH GRADE). The integer part x\lfloor x\rfloor of a real number xx is defined as the greatest integer MM such that MxM \leq x. Solve the equation x/32=x/32\sqrt{\lfloor x / 3-2\rfloor}=\lfloor\sqrt{x / 3-2}\rfloor.

SOLUTION. Introduce the variable y=x/32y=x / 3-2. In this case, y0y \geq 0. The equation will take the form

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

If y=n2y=n^{2}, where nn is an integer, then the left and right sides of equation (1) are equal to nn. Let n2<y<(n+1)2n^{2}<y<(n+1)^{2}. Then n<y<n+1n<\sqrt{y}<n+1 and the right side is y=n\lfloor y\rfloor=n. Equation (1) becomes y=n\sqrt{\lfloor y\rfloor}=n. Squaring both sides: y=n2\lfloor y\rfloor=n^{2}. This is equivalent to the inequalities n2y<n2+1n^{2} \leq y<n^{2}+1.

Thus, yn=0[n2;n2+1)y \in \bigcup_{n=0}^{\infty}\left[n^{2} ; n^{2}+1\right).

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

ANSWER: xn=0[3(n2+2);3(n2+3))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 reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.