How about a Geometry Problem that's bugged me for years?

Question

You have a Triangle ABC. You have three lines drawn from vertex A to segment BC... AX is an altitude. AY is an angle bisector. AZ is a median. Segments AX, AY, and AZ quadrasect Angle BAC.

Your task: Prove that Triangle ABC is a right triangle.

I think this is kinda tough, so good luck.

Answer 1

Lets call angle BAC = 4x.
Firstly, draw in the altitude. We know that this splits BAC into x and 3x; so angle ABC = 90-x and angle ACB = 90-3x.

Now draw the triangle again with the median drawn in. Now this splits BAC into 3x and x, and since we know angles ACB and ABC in terms of x, we can write all 6 angles in the diagram in terms of x.

Now lets say BC has length 2y, and AM (the median) has length m.

By the sin rule, in triangle ABM we have m/sin(90-x) = y/sin(3x), so m/y = sin(90-x)/sin(3x).
In triangle ACM we have m/sin(90-3x) = y/sin(x), so m/y = sin(90-3x)/sin(x).

So we have sin(90-x)/sin(3x) = sin(90-3x)/sin(x).
Remember that sin(90-a) = cos(a). So we have cos(x)/sin(3x) = cos(3x)/sin(x), or sin(x)cos(x) = sin(3x)cos(3x).

Now note that sin(2a) = 2sin(a)cos(a), so the above reduces to 0.5sin(2x) = 0.5sin(6x), or sin(2x) = sin(6x).

Thus we must have 2x = pi - 6x, or x = pi/8.

Thus angle BAC = 4x = pi/2, or 90 degrees.
As a side, we also know angle ABC = 67.5 degrees, and ACB = 22.5 degrees.
(edit - I am highly intrigued by the 'simple' proof mentioned by a previous answerer. Highly intrigued, and also highly doubtful :) It is very difficult to get any isosceles triangle using the median, as the two equal sides are on the same line. I'm reasonably convinced they must have made a mistake, but hopefully they can prove me wrong.. perhaps by telling me where the isosceles triangle is?)

Responding to ironduke below.. uh, the question is not to prove that the three segments quadrasect angle BAC if ABC is a right triangle; its the other way around (if they *do* quadrasect the angle, then ABC must be a right triangle). You're trying to prove something completely different (and obviously false).

Answer 2

One problem is that you have nnot excluded an isoceles right triangle. If the angle is right, then the altitude , median and angle bisector are all the same line. Thus the cannot possibly qudrasect angle A. This is a counterexample that shows what you are trying to prove is not true in the case of the isoceles right triangle.

It can be shown that there is no triangle that meets your criteria.
Basically the proof goes like this:
angle C = angle CAZ +angle ZAY+ angle YAX -90.
angle AZY = angle YAX + ZAY - 90
Subtracting, we get angle C- angle AZY = angle CAZ,
or angle AZY = angle C - angle CAZ
But angle AZY = Angle C + angle CAZ because an extrior angle is equal to the sum of the no-adjacent interior angles.
Therefore , by reductio ad absurdum , we have shown your theorem or hypothesis to be false.

A neater problem is to show that the angle bisector always falls between the altitude and the median for any triangles except isosceles and equilateral.

Answer 3

This is surprisingly easy. When I read it at first, I thought it would be very hard, but a simple diagram and using properties of isoceles and right triangles can help you show that 5 times angle BAX plus angle B (closest to the altitude) equals 180.** Then, you can use the fact that BAX is a right triangle to show that angle B plus BAX equals 90. Subtract, and you get 4 times angle BAX equals 90. This means that angle A must be a right angle. It makes a lot more sense with a diagram but I am devoid of a scanner. Good question, though. I love Geometry.

**This is done by using the rule that an exterior angle of a triangle is equal to the sum of opposite interior angles. In this case, angle AZC equals 3 times angle BAX plus angle A. Add 2 times angle BAX due to the given and the defintion of an icoceles triangle, and then use the definion of a triangle to show that these must add to 180.

Answer 4

Still working on the problem. Will update the answer when I figure it out. Sometime, a good geometry problem deserve a good night sleep. As for the diagram needing a scanner, use Microsoft paint to draw it. It might take a while, but it's better than finding a scanner at this hour.


XR

Answer 5

Help me out with the geometrical definitions of altitude and median for the triangle.

Is angle BAC general? I.e. it can be any of the angles, not necessarily the right angle.