Track / Stage 5 / 80 of 400 #680 of 1964

Problem 680

AIME late

Geometry Difficulty 5.3 Find the answer

8. 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. Spacing and $ signs are ignored.

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 and solution reproduced as published; topic, difficulty and ordering added by this site.