Math 487 Lab 10 (Altitudes, Dilations and Euler line)

It will be helpful to have a circumcircle tool available.

Dilating a circle by scaling ratio -1/2

• Construct a triangle ABC and also construct its 3 altitude lines.
• Also construct the midpoints of the sides of ABC.
• Next draw a point P and dilate the whole figure with center P and ratio -1/2.
• Move P around to study the nature of the new figure A'B'C'. In particular, notice that P is on segment AA' and that AP/AA' = 2/3.

• Now move P in a position so that the dilation image A' coincides with the midpoint of BC.
• Where are the points B' and C' now? What point in the triangle is the point P?
• You can lock P in this position by constructing the centroid G of triangle ABC and merging P to G.

So now A', B', and C' are the midpoints of the sides of the triangle ABC.

Proof that the altitudes of ABC are concurrent

• The images of the altitudes of ABC are now 3 special lines of A'B'C'. What are they?

Since we know that the perpendicular bisectors of A'B'C' are concurrent at a point O, this means that the altitudes of ABC must be concurrent at a point that we will call H. Why? And moreover, O = H'. THIS IS A PROOF THAT THE ALTTITUDES OF ABC ARE CONCURRENT.

• Also, we can see that G is on segment HO, and HG/HO = 2/3. Can you see why?

The Euler Line

• Construct the circumcircle of ABC. Then dilate it by ratio -1/2 and center G (former P) to get the circumcircle of A'B'C'.
• Also dilate the circumcenter O of ABC to get the center B of the circumcircle of A'B'C'.

So now we have a dilation by ratio -1/2, and center G, the centroid. The point H is dilated to O and O is dilated to B. The 4 points are on the line HO.

• Draw the line and using the ratios, explain how the points are arranged on the line.
• In particular, the four points lie in a segment, with 2 of the points as endpoints. As it happens, one of the remaining points is the midpoint.
• Draw and label a sketch of of the segment on a number line.

This line is called the Euler line. Observe the ratio relationship in your triangle ABC.

The circumcircle of A'B'C' is called the nine-point circle of ABC or the Euler circle. Notice that is seems to pass through the feet of the altitudes of ABC as well as the midpoints.

A Second Dilation to the Nine Point Circle

• Now draw another point Q in the plane and dilate triangle ABC by ratio +1/2 and center Q to get A''B''C''.
• Also dilate the point O and the circumcircle of ABC by the same dilation..
• This image of the circumcircle and the circle through A'B'C' are congruent. Why?
• Where is Q located when the circles coincide?
• Move Q to make the two congruent circles coincide. Then observe that the new figure A''B''C'' is a half-turn image of A'B'C'. What point is the center of the turn?
• When Q is located at H, then A'', B'', C'' are the midpoints of HA, HB, HC. These are the other 3 special points of the nine-point circle.

Proof

Why does Q = H work? The dilation that takes the circumcircle of ABC to the nine-point circle with ratio +1/2 is the composition of the dilation with center G and ratio -1/2 and the dilation with center B and ratio -1. You can check that H is a point that does not move under this composition, so it must be the center.