1. Given any reals A, B, C, show that An2 + Bn + C < n! for all sufficiently large integers n.
2. A triangle lies entirely inside a polygon. Show that its perimeter does not exceed the perimeter of the polygon.
3. Show that a triangle inscribed in a parallelogram has area at most half that of the parallelogram.
Solutions
Problem 1
Given any reals A, B, C, show that An2 + Bn + C < n! for all sufficiently large integers n.
Solution
n! > 2n(n-1)(n-2) for n > 5. But (n-1)/n ≥ 5/6, (n-2)/n ≥ 4/6 for n > 5, so 2(n-1)(n-2) > n2 and hence n! > n3 for n > 5.
Problem 2
A triangle lies entirely inside a polygon. Show that its perimeter does not exceed the perimeter of the polygon.
Solution
Take the triangle to be ABC. Let the rays CA, AB, BC meet the polygon at P, Q, R respectively (P may coincide with A etc). Let [PQ] denote the path along the sides of the polygon from P to Q. Similarly [QR] and [RP]. Then AQ is a straight line so it is ≤ AP + [PQ]. Similarly, BR ≤ BQ + [QR], CP ≤ CR + [RP]. Adding, AQ + BR + CP ≤ AP + BQ + CR + perim poly. But AQ = AB + BQ, BR = BC + CR, CP = CA + AP, so perim triangle ≤ perim polygon.
Problem 3
Show that a triangle inscribed in a parallelogram has area at most half that of the parallelogram.
Solution
There are three cases to consider. In the first two cases the triangle has base ≤ b and height ≤ h, so its area ≤ ½bh. In the third case we can divide the triangle into two triangles ABD and BCD using a line through B parallel to a side of the parallelogram. We then apply the second case to each triangle separately.
