1. ## Construction

Hi everyone,

This one is giving me a hard time---

You are given a fixed triangle T with base B. Show that is is always possible to construct, with straightedge and compass, a straight line that is parallel to B and that divides T into two parts of equal area. Similarly, divide T into five parts of equal area.

Anyone have/know a solution to this? I think one of the scale factors is 1/sqrt 2...but how would you properly construct what needs to be constructed??

2. Originally Posted by harold
Hi everyone,

This one is giving me a hard time---

You are given a fixed triangle T with base B. Show that is is always possible to construct, with straightedge and compass, a straight line that is parallel to B and that divides T into two parts of equal area.
You have triangle ABC, the base is AC.

Put a point on AB, label it X

Put a point on BC, label it Y

Now we have the information that line XY is parrallel to line AC

try proving that the trapezoid AXYC can have the same area as triangle BYX

3. Originally Posted by harold
Hi everyone,

This one is giving me a hard time---

You are given a fixed triangle T with base B. Show that is is always possible to construct, with straightedge and compass, a straight line that is parallel to B and that divides T into two parts of equal area. Similarly, divide T into five parts of equal area.

Anyone have/know a solution to this? I think one of the scale factors is 1/sqrt 2...but how would you properly construct what needs to be constructed??
Let's use Quick's notation. Let ABC be any triangle and consider the side AC as the "base." We wish to draw a line segment through the triangle such that the line is parallel to AC. Call the intersection of this line segment with the side AB X and the intersection of the line segment with the side CB Y. Further, assume that the line segment XY divides the triangle into two equal areas.

Now, since the line segment XY is parallel to AC we have that triangle ABC is similar to triangle XBY, thus their proportions are in ratio to each other. Specifically, since triangle XBY is half the area of triangle ABC we know that (AB)^2 = 2 * (XB)^2. (Where "(AB)" is the length of the segment AB.) Thus (XB) = (AB)/sqrt(2). All we need to do is construct a line with the length (AB)/sqrt(2) and mark that distance off on the line AB. Then we can simply draw a line through X parallel to AC and the intersection of this line with CB will be our point Y.

So how to construct (AB)/sqrt(2)? It makes the diagram rather messy, but it can be done this way. We want to construct a square with a diagonal of length (AB). So construct a line perpendicular to AB at point A. Now bisect this right angle, giving a 45 degree angle and extend this line. Call this line 1. Do the same for point B: make a perpendicular, bisect the right angle and extend the line. Call this line 2. (We want to construct line 2 such that it intersects line 1!) Call the point of intersection of lines 1 and 2 point D. ADB is an isosceles right triangle with sides AD and BD equal. Thus (AD) = (AB)/sqrt(2).

Now that we have the length we need, we can copy out that length along the line BA (we want to measure it from point B to point X, not from A to X).

As I mentioned before, now that we have point X we can construct a line parallel to AC through X. This line will intersect CB at the required point Y.

-Dan

I can't do the diagrams. If someone that can would be so kind as to provide diagrams for this to add clarity I would be appreciative. Even give rep points (IF I'm allowed.) I assume I can "thank" anyone.

4. Given the triangle any line parallel to the base cuts of a triangle similar to
the original. If the line divides the original triangle into two pieces of equal
area, the the little triangle that is cut off is half the area of the full triangle.

But area of similar figures scale as the square of the equivalent linear
dimensions of the figures. So the little triangle has sides of 1/sqrt(2) times
that of the original triangle, but this is equal to sqrt(2)/2 the side of the original triangle. So to show that the line exists we need only
show that we can construct points on the sides other than the base which
are a distance x*sqrt(2)/2 from the apex of the triangle down the edge
where x is the length of the edge.

But we know how to divide a length into two pieces of equal length so for
each of the edges construct a segment of length x/2 where x is the length of
the edge.

We also know how to construct a segment of length sqrt(2) times a given
length (construct a square of side equal to the given length, and the
diagonal is of the required length).

Hence for any line segment of length x we can construct with straight
edge and compass a segment of length x sqrt(2)/2.

Thus we can construct the required line parallel to the base which divides
the area into two equal pieces.

RonL