# Competition style problem circa 1992

We're given a triangle $ABC$.

Going clockwise, let $B_1$ and $B_2$ be distinct points on the segment $AC$ ($B_1$ is between $A$ and $B_2$), let $A_1$ and $A_2$ be distinct points on the segment $CB$ ($A_1$ is between $C$ and $A_2$), and finally let $C_1$ and $C_2$ be distinct points on the segment $AB$ ($C_1$ is between $B$ and $C_2$).

The circumcircles of triangles $CB_1A_1$ and $CB_2A_2$ intersect at $C_3 \neq C$.

The circumcircles of triangles $BA_1C_1$ and $BA_2C_2$ intersect at $B_3 \neq B$.

The circumcircles of triangles $AB_1C_1$ and $AB_2C_2$ intersect at $A_3 \neq A$.

Prove that the lines $AA_3$, $BB_3$ and $CC_3$ have a common point.

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I hid the history of this problem with the hope of trying to get more people to attempt a solution (using high-school math material only) – Meina222 Feb 11 '13 at 15:56

Consider the region near vertex $A$. (Point $B$ on ray $AC_2$ and point $C$ on ray $AB_1$ not pictured.)

$\square AB_1A_3C_1$ is cyclic, so opposite angles $\angle B_1$ and $\angle C_1$ are supplementary; consequently, $\angle B_1 \cong \angle A_3 C_1 C_2$ (the latter itself being a supplement of $\angle C_1$). Likewise, $\angle C_2 \cong \angle A_3 B_2 B_1$.

By the Angle-Angle Similarity Theorem, we have that $\triangle A_3 B_1 B_2 \sim \triangle A_3 C_1 C_2$, and we deduce that the "bases" ($B_1B_2$ and $C_1C_2$) and corresponding "altitudes" ($A_3P$ and $A_3Q$) of these triangles have proportional lengths; but note that those altitudes are directly related to the angles into which $\angle A$ has been divided. Therefore,

$$\frac{|B_1B_2|}{|C_1C_2|} = \frac{|A_3P|}{|A_3Q|} = \frac{|AA_3|\sin\angle A_3AB_1}{|AA_3|\sin\angle A_3AC_2} = \frac{\sin\angle A_3AC}{\sin\angle A_3AB}$$

Likewise for vertex $B$ and vertex $C$, so that the trigonometric form of Ceva's Theorem is satisfied:

$$\frac{\sin\angle A_3AC}{\sin\angle A_3AB} \; \frac{\sin\angle B_3BA}{\sin\angle B_3BC} \; \frac{\sin\angle C_3CB}{\sin\angle C_3CA} \; = \; \frac{|B_1B_2|}{|C_1C_2|} \; \frac{|C_1C_2|}{|A_1A_2|} \; \frac{|A_1A_2|}{|B_1B_2|} \; = \; 1$$

proving that the lines $AA_3$, $BB_3$, $CC_3$ share a common point.

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Awesome! This is slightly different than my original idea, but in the end it does go to Ceva's theorem. – Meina222 Feb 12 '13 at 17:15
Added my original solution. Thumbs up for the less algebraic one found by Blue. – Meina222 Feb 12 '13 at 19:30

Here's my original solution when creating the problem.

Focus on vertex $C$. Let $CC_3$ intersect $AB$ at $C_4$ (one can show $C_4$ is between $A$ and $B$).

Then:

$|BC_4|\vec{CA} + |AC_4|\vec{CB} = (|AC_4|+|BC_4|)\vec{CC_4} = |AB|\vec{CC_4}$

since $C_4$ is the center of mass of $A$ and $B$ with weights $|BC_4|$ and $|AC_4|$ respectively.

If you multiply both sides by the the diameter vector incident at $C$ of $CA_1B_1$'s circumcircle, and simplify by removing the cosines, you get:

$|BC_4||CA||CB_1| + |AC_4||CB||CA_1| = |AB||CC_4||CC_3|$

Similarly:

$|BC_4||CA||CB_2| + |AC_4||CB||CA_2| = |AB||CC_4||CC_3|$

Therefore:

$|BC_4||CA||CB_1| + |AC_4||CB||CA_1| = |BC_4||CA||CB_2| + |AC_4||CB||CA_2|$

Rewrite above as:

$\frac{(|CB_1|-|CB_2|)|CA|}{(|CA_2|-|CA_1|)|CB|} = \frac{|AC_4|}{|BC_4|}$

which yields (from the way points are ordered):

$\frac{|B_1B_2||CA|}{|A_1A_2||CB|} = \frac{|AC_4|}{|BC_4|}$

If we define $B_4$ and $A_4$ in similar fashion, the same reasoning yields:

$\frac{|C_1C_2||AB|}{|B_1B_2||CA|} = \frac{|BA_4|}{|CA_4|}$

$\frac{|A_1A_2||CB|}{|C_1C_2||AB|} = \frac{|CB_4|}{|AB_4|}$

From the last 3 identities, we conclude:

$\frac{|AC_4|}{|BC_4|}\frac{|BA_4|}{|CA_4|}\frac{|CB_4|}{|AB_4|} = \frac{|B_1B_2||CA|}{|A_1A_2||CB|}\frac{|C_1C_2||AB|}{|B_1B_2||CA|}\frac{|A_1A_2||CB|}{|C_1C_2||AB|} = 1$

By Ceva's theorem it follows that $AA_4, BB_4, CC_4$ have a common point.

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