Question 16·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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In the system of equations below, $k$ is a positive integer.

\begin{aligned} x^2+y^2&=50 \\[1.4em] xy+12x-12y&=k \end{aligned}

The system has exactly one ordered-pair solution $(x,y)$ , where both $x$ and $y$ are real. What is the value of $k$ ?

Your answer

An exactly one solution condition is a cue to look for symmetry: if every solution has a partner, the only possible single solution is a point that pairs with itself. A Desmos trial can reveal the pairing, but algebra must prove it. Then use the circle to limit the possible values of $x-y$ ; a tempting maximum may lie outside that limit.

Hints

Hint 1
Under the reflection $(x,y)\mapsto(-y,-x)$ , check what happens to $xy$ and $12x-12y$ . If both equations stay true, an only solution must be paired with itself.
Hint 2
A point paired with itself has $y=-x$ . Substitute that into the circle equation to find the candidates. Which candidate makes $k$ negative, contrary to the question?
Hint 3
Finding a candidate isn't enough: another point might give the same $k$ . Let $d=x-y$ and use the circle to bound it. Is the resulting expression for $k$ increasing throughout the allowed range?

Step-by-step

Use symmetry, then prove uniqueness
Step 1
Look for a pairing
Try $k=50$ , using the $50$ already in the circle equation. Type all three lines in Desmos and click the two intersections. They are about $(6.02,3.71)$ and $(-3.71,-6.02)$ . The second looks like $(-y,-x)$ from the first; now check whether that reflection always works.
Step 2
Prove what one solution requires
Replacing $(x,y)$ with $(-y,-x)$ leaves $x^2+y^2$ unchanged. It also leaves the second expression unchanged: $(-y)(-x)+12(-y)-12(-x)=xy+12x-12y$ . So every solution has a reflected partner. If there's exactly one ordered pair, it must be its own partner: $y=-x$ .
Step 3
Find the points that pair with themselves
Put $y=-x$ into the circle: $x^2+(-x)^2=50$ . Combine the squares: $2x^2=50$ . Divide by $2$ : $x^2=25$ . Take both square roots: $x=\pm5$ . The only possible single-solution points are $(5,-5)$ and $(-5,5)$ .
Step 4
Use the sign of the constant
At $(-5,5)$ , the three terms $xy$ , $12x$ , and $-12y$ are all negative. Their sum cannot be a positive integer, so only $(5,-5)$ remains as a candidate. We still need to rule out a different point giving that candidate's $k$ .
Step 5
Bound the difference on the circle
Let $d=x-y$ . Adding the squares gives $(x+y)^2+d^2=2(x^2+y^2)=100$ . Since a square can't be negative, $d^2\le100$ , so $-10\le d\le10$ . This circle restriction matters when you check whether another point could have the same $k$ .
Step 6
Write the second equation using that difference
Expand $d^2$ and use the circle equation: $d^2=x^2+y^2-2xy=50-2xy$ . Solve for $xy$ : $xy=25-\frac{d^2}{2}$ . Replace $xy$ and $x-y$ in the second equation: $k=25-\frac{d^2}{2}+12d$ . Define this expression as $f(t)$ in Desmos, where $t$ stands for the difference; we'll evaluate it after checking its allowed range.
Step 7
Show the candidate's value cannot occur twice
Half of $24$ is $12$ , and $12^2/2=72$ , so complete the square in the expression for $k$ by adding and subtracting $72$ : $f(d)=25+72-\frac{d^2-24d+144}{2}$ . Factor the square and combine constants: $f(d)=97-\frac{(12-d)^2}{2}$ . Throughout $-10\le d\le10$ , $12-d$ stays positive and shrinks as $d$ grows. So $f(d)$ strictly increases: only $d=10$ , which gives $(5,-5)$ , can produce $f(10)$ . The parabola's turning point at $d=12$ lies outside the circle's range.
Step 8
Evaluate the one possible value
Type $f(10)$ . Desmos displays $95$ . This is a positive integer, and the preceding step shows it occurs at only one point on the circle, $(5,-5)$ . So the required value of $k$ is $95$ . Grid in 95.

Question 16·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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