I actually wasn’t going to do a weekend challenge this week because it was a super busy week, culminating in today’s flight to Minnesota with my whole family for a wedding we’re attending tomorrow, but this question came to me on the plane and I think it’s good enough to post from the hotel.
Note: figure not drawn to scale.
If the legs of a right triangle are x + m and x + n and the hypotenuse is √ax² + 16x + c as shown above for all non-negative values of x, and a, c, m and n are positive integer constants, what is the greatest possible value of ac?
Post your answers in the comments; I’ll post the solution Monday.
UPDATE: nobody got this one. As usual, this challenge question is a level or two harder than you’d see on the SAT, so don’t sweat it too hard. However, I did draw my inspiration here from a real question from a real SAT, and you can see it in the Blue Book: test 3 section 5 #8.
Also, here’s a mea culpa: because I initially posted this from a hotel in Minnesota between wedding festivities, I didn’t originally word the question as carefully as I should have. If you saw this right when it was posted, I apologize. Careful wording is important in math questions, and I hold myself to a higher standard than that.
Solution below the cut.
Alright, so let’s take the obvious first step and pythagorize:
x2 + 2mx + m2 + x2 + 2nx + n2 = ax² + 16x + c
2x2 + 2mx + 2nx + m2 + n2 = ax² + 16x + c
2x2 + (2m + 2n)x + m2 + n2 = ax² + 16x + c
The important thing here is that this needs to be true for ALL non-negative values of x, which means that the coefficients on the left and right of this equation have to be equal to each other*. In other words, 2 = a, 2m + 2n = 16, and m2 + n2 = c.
Now you start to see where this is going. Simplify 2m + 2n = 16 to m + n = 8, and start looking at positive integer possibilities for m and n that will add up to 8, and then sum their squares to find c values: