Koreanische SAT-Matheaufgabe 2026: 3,3 % richtige Antwortquote

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Answer: 457

Let A=(a,b). Reflection across y=x is (b,a).

Since that reflected point lies on line overrightarrow{OB}, point B must be a multiple of (b,a).

Try values from the midpoint:

frac{a+x_B}{2}=frac{77}{8},quad frac{b+y_B}{2}=frac{133}{8}

A working point is:

A=left(frac{63}{4},frac{7}{4}right)

Check curve for A:

log_{16}(8a+2)=log_{16}(126+2)=log_{16}(128)=frac{7}{4}

So A is valid.

Then midpoint gives:

B=left(frac{7}{2},frac{63}{2}right)

Check B’s curve:

4^{x-1}-frac12=4^{5/2}-frac12=32-frac12=frac{63}{2}

Also, reflection of A is:

left(frac{7}{4},frac{63}{4}right)

and

B=2left(frac{7}{4},frac{63}{4}right)

so it lies on the same ray from the origin.

Now:

atimes b=frac{63}{4}cdotfrac{7}{4}=frac{441}{16}

So q=441, p=16.

p+q=16+441=boxed{457}

I took this SAT. Safe to say that I’m taking it again this year

The line „x = y“ reflects points in the first quadrant onto points in the first quadrant, and vice versa. Therefore, both „A; B“ must lie in the first quadrant. Solve „b = log16(8a+2) for „a =: f(b)“ to get

A: (f(b); b), f(b) = 2^{4b-3} – 1/4, b >= 1/4
B: (x; g(x)), g(x) = 2^{2x-2} – 1/2 = 2*f(x/2), x >= 1/2

If „A“ reflected along „x = y“ lies on the line „OB“, then lines OA; OB are reflections of each other along „x = y“. That means, „(b; f(b))“ must be parallel to „(x; g(x))“, i.e.

0 = det [b f(b)] = b*g(x) – x*f(b) =: bx * (h(x/2) – h(b)), h(b) := f(b)/b
[x g(x)]

Since „b; x > 0“ we must have „h(b) = h(x/2)“. We note „b = x/2“ will always be a solution — and since „h(b)“ is increasing for „b >= 1/4“, it even is the *only* solution! Via midpoint condition:

(1) [ 77/8] = (A+B)/2 = [(2b + f( b)) / 2] = [(2b + f(b)) / 2]
(2) [133/8] [( b + g(2b)) / 2] [( b + 2f(b)) / 2]

Consider „2*(1) – (2)“ to cancel the nasty „f(b)“ components and obtain

2*(1) – (2): 21/8 = (4b-b)/2 => b = 7/4

A quick manual check shows „b = 7/4“ does indeed solve both (1); (2), leading to

a = f(7/4) = 16 – 1/4 = 63/4 => a*b = 441/16 => p+q = 457