We can view these conditions as a geometry diagram as seen below. So, we know that fe=43 (since e=a−b=43c−43d=43f and we know that e2+f2=15 (since this is a2+c2−b2+d2). Also, note that ac+bd−ad−bc=(a−b)(c−d)=ef. So, solving for e and f, we find that e2+f2=225, so 16e2+16f2=3600, so (4e)2+(4f)2=3600, so (3f)2+(4f)2=3600, so f2(32+42)=3600, so 25f2=3600, so f2=144 and f=12. Thus, e=4312=9. Therefore, ef=9∗12=1 0 8}$.
A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.
Solution
The value of ef is 108.
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: Omni-MATH,
licensed Apache-2.0.
Statement reproduced verbatim; metadata (topic, difficulty) added by this project.