You are given two line segments, one is four times as long as the the other. The longer line segment is equal to the perimeter of an isosceles triangle. The shorter is equal to an altitude of the triangle.
Find a straightedge and compass construction of the triangle if the given altitude is perpendicular to the base.
Show that a straightedge and compass construction of the triangle is impossible if the given altitude is perpendicular to one of the congruent legs.
First, consider what the dimensions are to be for the isosceles triangle:
Let's call the length of the shorter line 1 unit.
The semiperimeter of the isosceles triangle is 2, and half the isosceles triangle is therefore a right triangle with altitude 1, with the hypotenuse and the base adding up to 2. We'll find the base of the right triangle, which is half the base of the isosceles triangle:
1 + x^2 = (2  x)^2
1 + x^2 = 4  4x + x^2
1 = 4  4x
4x = 3
x = 3/4
So first construct straight line and then another one perpendicular to it.
Lay off 3 of the shorter segment's lengths along the first straight line (or one of the longer segment's length, and backtrack by the shorter length), then bisect the distance from the intersection of the two full lines to the point 3 units removed. Then bisect the segment from the intersection to the first midpoint. Now you have a point 3/4 units from the right angle formed by the two lines.
Lay off 1 shorter length on the perpendicular line. Connect that endpoint to the previously found point at 3/4 distance on the other line.
Repeat the above on the other side of the perpendicular lines' intersection. Then the construction is complete.

Posted by Charlie
on 20100701 12:47:30 