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.

Prize this week: It won’t raaaeeaaaaiiiiin on your wedding day.
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?

The triangle is green because weddings are festive.

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:

(x + m)2 + (x + n)2 = (√ax² + 16x + c)2
x2 + 2mx + m2 + x2 + 2nx + n2ax² + 16x + c
Now do a little rearranging:

2x2 + 2mx + 2nx + m2 + n2ax² + 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 = a2m + 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 values:

7 + 1 = 8; 72 + 12 = 50
6 + 2 = 8; 62 + 22 = 40
5 + 3 = 8; 52 + 32 = 34
4 + 4 = 8; 42 + 42 = 32
Greatest value for c is 50. Since we saw before that the value of a is 2, the greatest possible value of ac is 100. Phew.
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* I’m not going to do it rigorously here, but I know this is a sticking point for a lot of people, so you can prove this to yourself by plugging in different values for x and seeing what happens. For example: if you say x = 1, then your left side will be 2 + 2m + 2n + m2 + n2 and your right side will be a + 16 + c. Then you could say m = 2 and n = 3, and your left side would be 2 + 4 + 6 + 4 + 9 = 25. You could then set a = 5 and c = 4 to make the right side = 25 as well. Those same numbers will NOT work when you set x = 2. If the equation has to work for all values of x, then the coefficients of the terms containing x have to be equal on either side of the equation.