olymmath

Putnam 1965

William Lowell Putnam Mathematical Competition 1965.

Session A

A1

  • Session A

Let ABCABC be a triangle with angle A<∠C<90∘<∠BA < \angle C < 90^\circ < \angle B. Consider the bisectors of the external angles at AA and BB, each measured from the vertex to the opposite side (extended). Suppose both of these line-segments are equal to ABAB. Compute the angle AA.

Answer, solutions and source

A2

  • Session A

Given any positive integer nn find the value of \n∑r=0⌊(n−1)/2⌋{n−2rn(nr)}2,\sum_{r=0}^{\lfloor (n-1)/2 \rfloor} \left\{\frac{n - 2r}{n}\binom{n}{r}\right\}^2,\n where ⌊x⌋\lfloor x \rfloor means the greatest integer not exceeding xx, and (nr)\binom{n}{r} is the binomial coefficient "nn choose rr," with the convention (00)=1\binom{0}{0} = 1. Return your final answer in binomial coefficient form with all other multiplicants reduced to the lowest form.

Answer, solutions and source

A5

  • Session A

In how many ways can the integers from 1 to nn be ordered subject to the condition that, except for the first integer on the left, every integer differs by 1 from some integer to the left of it?

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Session B

B1

  • Session B

Evaluate lim⁡n→∞∫01∫01⋯∫01cos⁡2(π2n(x1+x2+⋯+xn))dx1dx2⋯dxn.\lim_{n \to \infty} \int_0^1 \int_0^1 \cdots \int_0^1 \cos^2 \left(\frac{\pi}{2n}(x_1 + x_2 + \cdots + x_n)\right) dx_1 dx_2 \cdots dx_n.

Answer, solutions and source