EMAT 6600

Problem Solving


 

 


An Exploration of Polyas Problem

For a certain given triangle ABC, from any vertex construct an angle bisector, the median, and the altitude. Given that the vertex angle is dived into four equal angles, find the size of the angle at the vertex, as depicted below.

 

fig 1

 

Let us name the new points.

Since all the angles from the vertex are congruent, AE is the bisector of BAD, AD is the bisector of EAF, and AF is of DAC. So, DABE ≈ DEAD by ASA. Thus, we can state that DBAD is an isosceles triangle.

With this note in mind, for both triangles, angles EBA and BAE sum up to 90 and angles ADE and EAD sum up to 90. However, angles BAD, ADB and DBA sum up to 180. Also, 4(BAE) < 180, which implies that BAE < 60.

Now, since we know the sum of the angles of a triangle equal 180, we can use that to set up some equations/inequalities. This is very elementary.

Suppose the following: x + y = 90

x + 4y 180

where x is the measure of the angles ABE and ADE and y is the measure of the angles BAE and EAD. From there, we obtain that y 30 and x 60.

With a bit of reasoning, we see that 30 and 60 would not work, for 60 + 4(30) =180 and not less.

So the angle A can measure from 4 to 116

or A = {x| 4 < x < 116}.

 


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