Track / Stage 5 / 123 of 400 #723 of 1964

Problem 723

AIME late

Geometry Difficulty 5.3 Find the answer

Given that the number of integer points (points with integer coordinates) on the closed region (including the boundary) enclosed by the circle $x^{2}+y^{2}=8$ is one-fifth of the number of integer points on the closed region (including the boundary) enclosed by the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{4}=1$ , then the range of the positive real number $a$ is $\qquad$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Official solution

荟䝴 $\quad 22 \leq a21$ , 与 $\frac{2 m}{\sqrt{3}} \leq a<[a]+1=21$ contradicts; if $[a]=21$ , then $125=5+2 \times 21+4 m \Rightarrow 4 m=78$ , which is impossible;
Therefore, $[a]=22, 125=5+2 \times 22+4 m \Rightarrow m=19$ , it is not hard to verify
$\frac{2 \times 19}{\sqrt{3}}<22 \leq a<[a]+1=23<\frac{2 \times 20}{\sqrt{3}}$

Thus $22 \leq a<23$ .

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.