Ha, I was right.

It is a pokemon.

Anyway, on to the nine-point center. What is it? It's the center of a circle containing nine important points. (Oh, God, I'm starting to sound like my math teacher with the rhetorical questions...) Because it was discovered by Feuerbach, and then studied by Euler, it's sometimes called the Feuerbach circle, or the Euler circle. But let's be honest, here - what name is better? Feuerbach or Euler? Yup, that's what I thought.

Now, it's no fun just knowing about it. Let's prove it!

First, a lemma. On that handy dandy scrap paper that I know you keep solely for the purpose of this blog, draw yourself a nice little circle. Make the diameter a side of a triangle, and then choose of the endpoints of the diameter. Any endpoint will do. Chosen one yet? Good. Now draw a line connecting that endpoint to the point on the opposite side of the triangle where it intersects the circle. A diagram will illustrate this better than I can:



Or, if Geogebra hates you, here's a picture shamelessly ripped off from some site:



Right. Now that the whole diagram thing is settled, let's move right along.

I still can't figure out how to label my vertices on Geogebra (my computer crashes every time I try it), so if you adored my gorgeous Geogebra diagram, you'll have to refer to the stolen diagrams for now, I'm sorry to say. Note that is an angle inscribed in a semi-circle, so it's , meaning the triangle is right, and meaning that is an altitude, as is . Our lemma then is that a circle that has as the diameter thus has the feet of the altitudes from and .

So far, so amazingly spectacularly wonderful?

Amazingly spectacularly wonderful.

(c wut i did thar?)

Now take a look at this next ripped off diagram:



Hopefully, this looks familiar is the altitude we were just talking about, and are the midpoints of the triangle sides. It's easy to show that the red quadrilateral is a trapezoid, because is a midline of the triangle and therefore parallel to .

is also a midline, so it's half the length of . Since is clearly the center of our lovely circle, is a radius, so it's also half the length of - making , and we have an isosceles trapezoid. We know that and are supplementary, and , so our trapezoid is not just an isosceles trapezoid, it's an awesome isosceles trapezoid. And it's not just an awesome isosceles trapezoid - it's an awesome CYCLIC isosceles trapezoid! (Yeah, that's right!)

Therefore, the circle that has all three of the midpoints of the sides has . And, because of the altitude feet stuff, this makes a total of 6 points. This is exactly the circle we want! Er, though, I did say nine points, and 9 is not 6 (I think.) So, let's go find the last three.

Here is another diagram:



Note that is the orthocenter. Since three points determine a circle, the three feet of the altitudes of any triangle whatsoever must be on its nine point circle, which therefore has the three midpoints of the sides. We did prove this is true for , but that's not the only triangle this works for - it's also true for ! This triangle has altitudes , , and , meaning the feet are - same as 's altitudes' feet!

Do you see where we're going here? That nine-point circle we have also has the three midpoints of - that makes a total of nine! In conclusion, every triangle has a nine point circle that contains the three midpoints of the sides, the three feet of the altitudes, and the three midpoints of the segments connecting each vertex to the orthocenter. The center is called the nine-point center, and it's very prettyful in all its glory:



Kudos!

Ha, I was right.

It is a pokemon.

Anyway, on to the nine-point center. What is it? It's the center of a circle containing nine important points. (Oh, God, I'm starting to sound like my math teacher with the rhetorical questions...) Because it was discovered by Feuerbach, and then studied by Euler, it's sometimes called the Feuerbach circle, or the Euler circle. But let's be honest, here - what name is better? Feuerbach or Euler? Yup, that's what I thought.

Now, it's no fun just knowing about it. Let's prove it!

First, a lemma. On that handy dandy scrap paper that I know you keep solely for the purpose of this blog, draw yourself a nice little circle. Make the diameter a side of a triangle, and then choose of the endpoints of the diameter. Any endpoint will do. Chosen one yet? Good. Now draw a line connecting that endpoint to the point on the opposite side of the triangle where it intersects the circle. A diagram will illustrate this better than I can:



Or, if Geogebra hates you, here's a picture shamelessly ripped off from some site:



Right. Now that the whole diagram thing is settled, let's move right along.

I still can't figure out how to label my vertices on Geogebra (my computer crashes every time I try it), so if you adored my gorgeous Geogebra diagram, you'll have to refer to the stolen diagrams for now, I'm sorry to say. Note that is an angle inscribed in a semi-circle, so it's , meaning the triangle is right, and meaning that is an altitude, as is . Our lemma then is that a circle that has as the diameter thus has the feet of the altitudes from and .

So far, so amazingly spectacularly wonderful?

Amazingly spectacularly wonderful.

(c wut i did thar?)

Now take a look at this next ripped off diagram:



Hopefully, this looks familiar is the altitude we were just talking about, and are the midpoints of the triangle sides. It's easy to show that the red quadrilateral is a trapezoid, because is a midline of the triangle and therefore parallel to .

is also a midline, so it's half the length of . Since is clearly the center of our lovely circle, is a radius, so it's also half the length of - making , and we have an isosceles trapezoid. We know that and are supplementary, and , so our trapezoid is not just an isosceles trapezoid, it's an awesome isosceles trapezoid. And it's not just an awesome isosceles trapezoid - it's an awesome CYCLIC isosceles trapezoid! (Yeah, that's right!)

Therefore, the circle that has all three of the midpoints of the sides has . And, because of the altitude feet stuff, this makes a total of 6 points. This is exactly the circle we want! Er, though, I did say nine points, and 9 is not 6 (I think.) So, let's go find the last three.

Here is another diagram:



Note that is the orthocenter. Since three points determine a circle, the three feet of the altitudes of any triangle whatsoever must be on its nine point circle, which therefore has the three midpoints of the sides. We did prove this is true for , but that's not the only triangle this works for - it's also true for ! This triangle has altitudes , , and , meaning the feet are - same as 's altitudes' feet!

Do you see where we're going here? That nine-point circle we have also has the three midpoints of - that makes a total of nine! In conclusion, every triangle has a nine point circle that contains the three midpoints of the sides, the three feet of the altitudes, and the three midpoints of the segments connecting each vertex to the orthocenter. The center is called the nine-point center, and it's very prettyful in all its glory:



Kudos!

A wooden cube units on a side is painted red on all six faces and then cut into unit cubes. Exactly one-fourth of the total number of faces of the unit cubes are red. What is ?

Solution