Olympiad Maths Prep

Track / Stage 4 / 172 of 340 #432 of 2000

Problem 432

AMC 12 late, AIME early
Combinatorics Difficulty 4.9 Find the answer

3. Given that the ellipse C1C_{1} and the hyperbola C2C_{2} share foci F1(3,0),F2(3,0)F_{1}(3,0), F_{2}(-3,0), and have coincident minor axes. Then the number of lattice points inside the region enclosed by the intersection points of C1C_{1} and C2C_{2} is \qquad

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Official solution

3.25.

From the problem, we can let
C1:x2m+9+y2m=1,C2:x29my2m=1, C_{1}: \frac{x^{2}}{m+9}+\frac{y^{2}}{m}=1, C_{2}: \frac{x^{2}}{9-m}-\frac{y^{2}}{m}=1,

where 0<m<90<m<9.
Let the intersection point be P(x0,y0)P\left(x_{0}, y_{0}\right). Then, since point PP lies on C1C_{1} and C2C_{2}, we have x02m+9+x029m=2\frac{x_{0}^{2}}{m+9}+\frac{x_{0}^{2}}{9-m}=2, which simplifies to x02=9m29x_{0}^{2}=9-\frac{m^{2}}{9}.
Thus, y02=m29y_{0}^{2}=\frac{m^{2}}{9}, where x02,y020,9x_{0}^{2}, y_{0}^{2} \neq 0,9.
Therefore, x02+y02=9x_{0}^{2}+y_{0}^{2}=9.
From the sketch (omitted), it is known that there are 25 lattice points within the enclosed region.

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