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:
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Comments (3)
Well, that explains it. I set x=1 and saw that the other integers could be as high as I wanted them. I figured I was just missing something…
Yeah. This is one of the most difficult things to wrap your head around that the SAT will occasionally toss into a test. Usually (as in the Blue Book question I referenced above) it’ll be multiple choice, and thus susceptible to backsolve, but for the Weekend Challenge I wanted to take it up a couple notches. 🙂
It isn’t too hard to explain the seemingly mysterious fact that if two quadratics are equal for all values of x, their coefficients must match (though, as you say, it is obviously beyond the scope of most SAT reasoning, in drawing on a deep idea from precalculus). Suppose we have ax^2+bx+c=dx^2+ex+f. Collecting terms on the left, we have (a-d)x^2+(b-e)x+(c-f)=0, and this must be true for all values of x — i.e. the function (a-d)x^2+(b-e)x+(c-f) must have infinitely many roots. If the coefficients weren’t all 0, the left hand side would be a nontrivial polynomial of degree 2, which by the fundamental theorem of algebra can’t have more than 2 roots. This is a contradiction, so the coefficients must all be 0, i.e. a-d=0, b-e=0, and c-f=0. And this shows a=d, b=e, and c=f. I think this is accessible to most high school students who have had a course that covers roots of polynomials.
Of course, there’s also a less elegant algebraic, rather than precalculus, perspective here. Plugging in different values of x gives a system of equations. When x=0, we have
c-f=0
which means c=f.
When x=1, we have
a+b-d-e=0
which means a+b=d+e.
When x=2, we have
4a+2b-4d-2e=0
But 4a+2b-4d-2e=2a+2(a+b)-2d-2(d+e) and since a+b=d+e from above, this third equation just says 2a-2d=0, or a=d. Back substituting into the second equation gives b=e.