A Putnam Geometry Problem

[jay_shark]

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Grrrr, jay.

This was one of the years I took the Putnam, and I felt this was an extremely easy question - until finding out, after I took the test, that I had interpreted the question quite differently than the writers intended it (apparently a majority, but not all, of my fellow test-takers understood it as intended.)

I remain as annoyed today as I was the day after I found out.

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I also misinterpreted the problem the first time I read it . I'm not sure if it's done intentionally but many of their problems are poorly worded .

Here is the solution I came up with that doesn't require any trigonometry or calculus . It can be explained to a grade 11 student as long as he's familiar with circle geometry .

Label the points A(0,0) B(x,0) C(x,y) D(x,-x) . We wish to maximize the area of triangle ACD since it is equivalent to x*y/2 + x*x/2 which is just alpha/2 . We know that x^2 + y^2 =1 and is constant for all x,y . Angle ADC is 45 degrees (or pi/4). Therefore their exists a circle of fixed radius that circumscribes the triangle ADC with constant chord 1 and a subtended angle of 45 degrees . So it's clear that the maximum area of a triangle in a circle is when we take the triangle to be isosceles and so angle ACD=angle CAD =67.5 which implies that angle CAD = 22.5 degrees or pi/8 .

[CSLewisJr]

Here's what I came up with:



We want to maximize shaded area R, which is the difference in area between the large rectangle and the two squares. R = |DC|*|CB| = sqrt(2)*|EC|*|CB|
The perimeter of triangle BCE is greatest when |EC| = |CB|, and |EB| is constant, so this maximizes |EC| + |CB|. By AM/GM, |EC|*|CB| and R also greatest when the triangle is isosceles. The area of the larger square is 1/2 + sqrt(20)/20.