4th Eötvös Competition Problems 1897

1.  ABC is a right-angled triangle. Show that sin A sin B sin(A - B) + sin B sin C sin(B - C) + sin C sin A sin(C - A) + sin(A - B) sin(B - C) sin(C - A) = 0.

2.  ABC is an arbitrary triangle. Show that sin(A/2) sin(B/2) sin(C/2) < 1/4.

3.  The line L contains the distinct points A, B, C, D in that order. Construct a rectangle whose sides (or their extensions) intersect L at A, B, C, D and such that the side which intersects L at C has length k. How many such rectangles are there? 

Solutions

 

Problem 1

ABC is a right-angled triangle. Show that sin A sin B sin(A - B) + sin B sin C sin(B - C) + sin C sin A sin(C - A) + sin(A - B) sin(B - C) sin(C - A) = 0.

Solution

We have 1st term = sin B cos B, 2nd term = -sin B cos B cos 2B, 3rd term = -sin B cos B, 4th term = sin B cos B cos 2B. 

Problem 2

ABC is an arbitrary triangle. Show that sin(A/2) sin(B/2) sin(C/2) < 1/4.

Solution

wlog A ≤ B ≤ C. Then A/2 ≤ 30^o, so sin A/2 ≤ 1/2. Also B/2 is acute and B/2 = (90^o-C/2)-A/2 < 90^o-C/2. Hence sin B/2 sin C/2 < sin(90^o-C/2) sin C/2 = cos C/2 sin C/2 = 1/2 sin C ≤ 1/2. 

Problem 3

The line L contains the distinct points A, B, C, D in that order. Construct a rectangle whose sides (or their extensions) intersect L at A, B, C, D and such that the side which intersects L at C has length k. How many such rectangles are there?

Solution

Draw a line through B parallel to the rectangle sides length k. Suppose it meets the side through A at X. Then ∠AXB = 90o and BX = k. So if AB ≤ k then there is no rectangle. If AB > k, then X must lie on the circle diameter AB and on the circle center B radius k. There are two such points and hence two possible rectangles. Having constructed X, the rest follows easily.