Maths Olympiad Prep

Library / /40 of 128

Geometry Difficulty 5.1 AIME, harder Prove it Philippines

Problem:

Given O(0,0)O(0,0) and Q(1,2)Q(1,2), consider the set of points PP for which OP:PQ=5:4O P: P Q=5: 4. It is known that this set is a circle. Find its center.

Solution

Solution:

Suppose we have P(x,y)P(x, y). This implies in particular that 4OP=5PQ4 O P=5 P Q, or 16OP2=25PQ216 O P^{2}=25 P Q^{2}. Hence, we have
16(x2+y2)=25(x−1)2+25(y−2)2 16\left(x^{2}+y^{2}\right)=25(x-1)^{2}+25(y-2)^{2}
This simplifies to
9x2−50x+25+9y2−100y+100=0 9 x^{2}-50 x+25+9 y^{2}-100 y+100=0
and, completing squares,
(3x−253)2+(3y−503)2=(253)2+(503)2−25−100=20009 \left(3 x-\frac{25}{3}\right)^{2}+\left(3 y-\frac{50}{3}\right)^{2}=\left(\frac{25}{3}\right)^{2}+\left(\frac{50}{3}\right)^{2}-25-100=\frac{2000}{9}
Our center is then given by (259,509)\left(\frac{25}{9}, \frac{50}{9}\right).

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.