Janet Kaplan

The Star of David

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I have named this construction the "Star of David" because that is, in fact, its shape when the angles, lines and the resulting triangles are perfectly symmetrical.

But our challenge is to find the relationship between the angles of Triangle ABC and those of Triangle LMN in all cases. We are given these parameters to begin our construction:

Construct the internal angle bisectors of acute Triangle ABC and extend them to meet the Circumcircle at points L,M and N, respectively. Triangle ABC is colored blue; Triangle LMN is colored orange. For a GSP construction that can be manipulated, CLICK HERE.

Go into the GSP construction of the Star of David and you will notice that the angles of the triangles have been measured. Move the vertices of the blue triangle (ABC) all around to see the angle measurements change. Only these vertices can be moved, as the vertices of the orange triangle are dependent upon the construction of the blue. Make both triangles equilateral by getting as close as you can to 60 degree measurements for each angle. Its symmetry is beautiful, but it doesn't tell us anything about the relationship between the angles of the two triangles.

So what is the relationship? I began with equilateral triangles and changed the size of the angles from there. I noticed that as I increased the size of one of the blue angles, the size of the corresponding orange angle decreased by a factor of one-half. For example, when I increased <A by 20 degrees to 80 degrees, and kept <B and <C constant, <L (its opposite, so to speak) decreased by 10 degrees to 50 degrees. Decreasing <A to 40 degrees increased <L by 1/2 or 10 degrees.

Interestingly, when I moved any of the blue vertices such that the size of the orange angles changed, the opposite thing happened! That is, increasing orange <M by 20 degrees, made its ‘complement’ blue angle <B decrease by a factor of 2 (40 degrees) to 20 degrees.

The following chart diagrams this relationship.

Change in Blue Angle             Resulting Change in Orange Angle

Increase                                   Decrease by factor of 1/2

Decrease                                 Increase by factor of 1/2

Change in Orange Angle         Resulting Change in Blue Angle

Increase                                  Decrease by factor of 2

Decrease                                 Increase by factor of 2

There are additional observations that follow directly from the above. When the size of one angle is changed, the other two angles in the same triangle adjust as well, such that 180 degrees are preserved. For that matter, as <B and <C decrease to comply with the increase of <A, their complements in the orange triangle <M and <N increase, taking up the degrees lost by <L. The relationship is perfectly balanced in the sense that the 2:1 ratio between the blue angles and their complimentary orange angles continues to hold regardless of how the shapes of the triangles change.

We also know that as the size of the inscribed angles change, the arc that they subtend changes accordingly. That is, an angle increasing from 60 to 80 degrees will now subtend an arc of 160 degrees, while its complimentary angle will subtend an arc of 100 degrees.

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From our observations we can see there is a 2 to 1 relationship between the angles in the two triangles. How can we prove this to be the case?

PROOF

Construct segments LC and MC. Remember that the dashed blue lines are the angle bisectors of the blue triangle.

Arc LC is subtended by both 1/2 <A and a portion of <N. Therefore, 1/2 <A and that portion of <N are congruent.

Arc MC is subtended by both 1/2 <B and the other portion of <N. Therefore, 1/2 <A and that portion of <N are congruent.

From this it follows that <N = <A + <B.

We also know that 90 degrees = 1/2 (<A + <B + <C) or 180 - <C = <A + <B or 90 - 1/2 <C = <A + <B = <N.

Now that we have the basic equalities set, if we increase <N by 10 degrees, such that <N' = 70, then 90 - 1/2 <C must also equal 70. This can only be true when <C is decreased by 20 degrees, such that <C' = 50 degrees.

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