Особливі лінії та центри в трикутнику

Synthetic Geometry : First Triangle Centers

What would be better than solving a challenging geometry problem after spending hours and hours on it, with a completely synthetic solution. If you solve the problem by bashing coordinate geometry, complex numbers, or use barycentric coordinates, that would be more unsatisfying.

As such, we'll look into solving a problem syntheticly.

I assume that everyone is familiar with the basics of geometry (e.g. inscribed angle theorem, equal tangents, properties of cyclic quadrilaterals, power of point theorem).

We will prove that $H, I, O, G$ exists and are unique.

The existence and uniqueness of $O$.

Now, we'll prove that they're concurrent so that the circumcenter always exists and is unique.

Let's draw two perpendicular bisectors first.

This implies that $XBC$ is isoceles and thus $X$ lies on the perpendicular bisector of $BC$. Thus, $X$ is the circumcenter of $\triangle ABC$.

Usually, we denote $X$ by $O$.

The existence and uniqueness of $I$.

We use the same idea of the proof above. Let $X$ be the intersection of the angle bisectors of $\angle B$ and $\angle C$.

So, the angle bisectors meet at $X$.

$X$ is usually denoted as $I$.

The existence and uniqueness of $H$.

This time the proof is a bit tricky.

There's no immediate way to tackle the problem.

Similarly, the other altitudes are also the other perpendicular bisectors of $A'B'C'$. Clearly, these perpendicular bisectors are concurrent at the circumcenter of $A'B'C'$, so the altitudes of $ABC$ are concurrent.

We gain a bonus fact! That is :

The circumcenter of a triangle coincides with the orthocenter of its medial triangle. (i.e. The triangle formed by joining the midpoints of the sides of the triangle.)

The existence and uniqueness of $G$.

We first prove a very crucial fact.

Let $ABC$ be a triangle and let $D$, $E$ be the midpoints of $AB$ and $AC$. Let $F$ be the intersection of $BE$ and $CD$. Then $BF = 2EF$ and $CF = 2DF$.

This proves the claim.

Now, back to proving the existence of $G$.

Let $X$ be the intersection of the $B$-median and $A$-median, and let $X'$ be the intersection of the $C$-median and $A$-median. Let $M$ be the midpoint of $BC$.

So, $AX = 2XM, AX' = 2X'M$, so $X = X'$ and we're done.

This concludes the proof.

60 degrees

Consider a triangle $ABC$, its incenter $I$, orthocenter $H$, and circumcenter $O$. I hope that you're all familiar with the definitions of the points $I$, $H$, and $O$. We consider $\angle BAC=60^{\circ}$, because interesting stuff will happen.

Note that we are given angles, and $60^{\circ}$ is a really nice angle; it is the angle measure of each angle in an equilateral triangle. Maybe that will give us equal angles, which leads to...

CYCLIC QUADRILATERALS!!!

We just have to prove it. But since I'm lazy, you be the one to do it :-P

Exercise 1: Show that $\angle BHC=120^{\circ}$. Exercise 2: Show that $\angle BIC=120^{\circ}$. Exercise 3: Show that $\angle BOC=120^{\circ}$.

Thus BOIHC lie on a circle.

Have you finished? Click 'continue' if you're done.

Okay, time for the application to an olympiad problem.

(APMO 2007 #2) Let $ABC$ be an acute-angled triangle with $\angle BAC=60^{\circ}$ and $AB>AC$. Let $I$ be the incenter, and $H$ the orthocenter of the triangle $ABC$. Prove that $2\angle AHI=3\angle ABC$.

But just note that BIHC lie on a circle. Then just angle chase.

If you want a solution, click here: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=799589&sid=9904602403fee030110cd9b7871fe6a5#p799589

Let $P$ be a Brocard point of triangle $ABC$ ($P$ satisfies $\angle PAB = \angle PBC = \angle PCA)$. Let $D$, $E$, and $F$ be the second intersections of lines $AP$, $BP$, and $CP$ with the circumcircle of triangle $ABC$. Show that $P$ is also a Brocard point of triangle $DEF$.