AMC 10 · 2024 · #25

Test the rule on actual quadruples. Take (a,b,c,d)=(1,1,1,-1), which has a+d=0 and bc=1 ≠ -a²=-1, so it should count: f(x)=(x+1)/(x-1) gives f(2)=3 and f(3)=2, and f(0)=-1 and f(-1)=0, so points come in swapped pairs exactly as symmetry demands. Now take (1,-1,1,-1), which has a+d=0 but bc=-1=-a², one of the 38 removals: f(x)=(x-1)/(x-1)=1, a horizontal line with a hole, whose mirror image is a vertical line — correctly rejected, and this is the case that separates the answer choices. A branch-one quadruple with c=0, say (2,3,0,-2), gives f(x)=(2x+3)/-2=-x-3/2, a line of slope -1; it meets y=x at a right angle, so reflecting in y=x swaps its two halves and returns the same line — symmetric, as the count claims. A quadruple failing a+d=0, say (2,1,1,1), gives f(f(x))=(5x+3)/(3x+2) ≠ x — correctly rejected. Next a size check: there are 11⁴-11=14630 allowed quadruples, and 1292/14630≈ 8.8%, which sits just under the 1/11≈ 9.1% you would expect from the single equation a+d=0 — the small shortfall is the 38 removals, and the sign is right. Finally the answer choices themselves are a check, because each one is a specific slip: 1330 is 1320+10, forgetting to remove the 38 constants; 1320 is branch one alone with neither the removals nor the line y=x; and 1282 is branch one done correctly but with the 10 copies of y=x overlooked. The correct total sits at 1292, choice (B).