# Equal Circles, Medial Triangle and Orthocenter

The applet below illustrates the following problem [Altshiller-Court, Theorem 181]:

Any three equal circles having for centers the vertices of a given triangle cut the respective sides of the medial triangle in six points equidistant from the orthocenter of the given triangle.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

Solution

### References

1. N. Altshiller-Court, College Geometry, Dover, 1980

### Solution

Any three equal circles having for centers the vertices of a given triangle cut the respective sides of the medial triangle in six points equidistant from the orthocenter of the given triangle.

Note: The problem is equivalent to saying that the six points in question are concyclic; thus it claims the existence of a circle. One can easily surmise that the circle exists even if only four of the six points are available.

Let ΔABC be given, with Ma, Mb, Mc the midpoints and the orthocenter H.

### This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

 What if applet does not run?

Assume circle centered at A meets MbMc in U and U'. Let B be the intersection of the altitude from A and MbMc. The two right triangles APU and APU' are equal. Applying the Pythagorean identity gives

PU² = PU'² = AU² - PU² = HU² - HP².

Hence,

 AU² - HU² = AP² - HP² = (AP + HP)(AP - HP) = AH (DP - HP) = AH·DH,

implying

HU² = AU² - AH·DH.

Now, by assumption, AU = BV = CW. So it follows from a property of the orthocenter that HU = HV = HW.