0
$\begingroup$

Let $ABC$ be a triangle. An arbitrary circle $(K_a)$ passing through $B, C$ intersects $CA, AB$ again at $A_b, A_c$, respectively. Define $B_c, B_a, C_a, C_b$ similarly. Prove that the perpendicular bisector of $C_aB_a, A_bC_b, B_cA_c$ are concurrent.

Hints (from my teacher): Use Carnot Theorem: Let $ABC$ be a triangle . $M,N,P$ are points on $BC, CA, AB$,respectively. Suppose that $\Delta_1,\Delta_2,\Delta_3$ are three lines through $M,N,P$ and perpendicular to $BC,CA,AB$, respectively. Then $\Delta_1;\Delta_2;\Delta_3$ are concurrent iff $MC^2-MB^2+NC^2-NA^2+PA^2-PB^2=0$

Thank you!

$\endgroup$
  • $\begingroup$ Use the "Power of a Point with respect to a Circle" to relate various segments lengths. For instance, $$|\overline{AA_c}||\overline{AB}| = \text{power of $A$ wrt $K_a$} = |\overline{AA_b}||\overline{AC}|$$ Also, $|\overline{AA_c}| + |\overline{BA_c}| = |\overline{AB}|$, etc; and $|\overline{AC_c}| = \frac12(|\overline{AB_c}| + |\overline{AA_c}|)$, etc, where $C_c$ is the midpoint of $\overline{A_cB_c}$. $\endgroup$ – Blue Sep 17 '14 at 21:06
  • $\begingroup$ Can you please explain things more clearly? I'm all messed up because there are too many segment lengths to calculate. $\endgroup$ – primitiveroot Sep 17 '14 at 22:44
2
$\begingroup$

Here's a diagram, where I've marked the feet of the perpendicular bisectors as $A_a$, $B_b$, $C_c$.

enter image description here

As hinted by your teacher, Carnot's Theorem ---not to be confused with Carnot's Theorem :)--- guarantees the desired concurrency once we verify that $$\left(\;|AC_c|^2 - |BC_c|^2\;\right) + \left(\;|BA_a|^2 - |CA_a|^2\;\right) + \left(\;|CB_b|^2 - |AB_b|^2\;\right) \stackrel{?}{=} 0 \tag{$\star$}$$

Let's consider one part of $(\star)$ at a time: $$|AC_c|^2 - |BC_c|^2 = \left(\;|AC_c|+|BC_c|\;\right)\left(\;|AC_c|-|BC_c|\;\right) = |AB|\;\left(\;|AC_c|-|BC_c|\;\right)$$

Since $$|AC_c| = \frac12 (\; |AA_c| + |AB_c| \;) = \frac12 (\;|AA_c|+|AB|-|BB_c|\;)$$ $$|BC_c| \phantom{= \frac12 (\; |AA_c| + |AB_c| \;)} = \frac12(\;|BB_c|+|AB|-|AA_c|\;)$$ the above becomes $$|AC_c|^2 - |BC_c|^2 = |AB|\;\left(\;|AA_c| - |BB_c|\;\right)$$ Likewise, $$\begin{align} |BA_a|^2 - |CA_a|^2 &= |BC|\;\left(\;|BB_a|-|CC_a|\;\right) \\ |CB_b|^2 - |AB_b|^2 &= |CA|\;\left(\;|CC_b|-|AA_b|\;\right) \end{align}$$ Therefore, $(\star)$ becomes $$|AB||AA_c| + |BC||BB_a| + |CA||CC_b| \stackrel{?}{=} |BA||BB_c| + |CB||CC_a|+|AC||AA_b| \qquad(\star\star)$$ where I've written "$|AB|$" as "$|BA|$", etc, on the right in anticipation of the following:

The "Power of a Point" theorem says that, if a line through point $P$ meets circle $\kappa$ at points $Q$ and $R$, then $|PQ||QR|$ is a value that depends only on $P$ and $\kappa$, not on $Q$ and $R$. This value is called the power of $P$ with respect to $K$.

In $(\star\star)$, the value of term $|AB||AA_c|$ is the power of point $A$ with respect to $\kappa_A$; but so is the value of $|AC||AA_b|$. These terms cancel. Similarly, $|BC||BB_a| = |BA||BB_c|$ (the power of $B$ with respect to $\kappa_B$), and $|CA|CC_b| = |CB||CC_a|$ (the power of $C$ with respect to $\kappa_C$). All the terms cancel, so that Carnot's Theorem is satisfied: the lines in question are concurrent. $\square$

$\endgroup$
0
$\begingroup$

And here is the elegant proof and the inventor of the problem: http://www.cut-the-knot.org/m/Geometry/GarciaCircles.shtml

$\endgroup$
  • $\begingroup$ It's always nice to include the essence of the solution here since links may break $\endgroup$ – Shailesh Jul 30 '16 at 4:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.