Thursday, February 2, 2006

Little Triangles. Topic: Geometry. Level: AIME/Olympiad.

Problem: (MOP Geometry Packet) Let $A_0B_0C_0$ be a triangle and $P$ a point. Define a new triangle whose vertices $A_1,B_1,C_1$ as the feet of the perpendiculars from $P$ to $B_0C_0,C_0A_0,A_0B_0$, respectively. Similarly, define the triangles $A_2B_2C_2$ and $A_3B_3C_3$. Show that $A_3B_3C_3$ is similar to $A_0B_0C_0$ (i.e. the third pedal triangle is similar to the original triangle).


Solution: Consider the "pedal operation" on the angles of a triangle, that is, the operation that takes a point inside a triangle, drops perpendiculars to the sides, and gives the angles of the triangle formed by the feet of the perpendiculars. Let $ f(A,B,C) $ be this operation. We want to show that $ f(f(f(A,B,C))) = (A,B,C) $.

Define the angles and points as in the diagram; $ A^{\prime}, B^{\prime}, C^{\prime} $ are the feet of the perpendiculars opposite sides $ A,B,C $ respectively, and $ \angle B^{\prime}AP = \alpha_1, \angle C^{\prime}AP = \alpha_2 $ and similarly for the other two angles.

We will define $ f(A,B,C) = (B^{\prime},C^{\prime},A^{\prime}) $.

Consider quadrilaterals $ AB^{\prime}PC^{\prime}, BC^{\prime}PA^{\prime}, CA^{\prime}PB^{\prime} $. They are all cyclic because $ \angle AB^{\prime}P = \angle AC^{\prime}P = \angle BC^{\prime}P = \angle BA^{\prime}P = \angle CA^{\prime}P = \angle CB^{\prime}P = 90^{\circ} $.

Thus we have $ \angle B^{\prime} = \angle A^{\prime}B^{\prime}C^{\prime} = \angle A^{\prime}B^{\prime}P+\angle C^{\prime}B^{\prime}P = \angle A^{\prime}CP+\angle C^{\prime}AP = \gamma_1+\alpha_2 $. Similarly, $ \angle C^{\prime} = \alpha_1+\beta_2 $ and $ \angle A^{\prime} = \beta_1+\gamma_2 $.

So $ f(\alpha_1+\alpha_2, \beta_1+\beta_2, \gamma_1+\gamma_2) = (\gamma_1+\alpha_2, \alpha_1+\beta_2, \beta_1+\gamma_2) $.

Note that the second component of each angle is invariant under $ f $. Hence $ f $ simply cycles $ \alpha_1, \beta_1, \gamma_1 $. Since there are three of them, applying $ f $ three times will give the original arrangement and the angles will the be same as of the original triangle, giving similarity. QED.


Comment: The pedal triangle is a less common geometry topic for most people, but it's pretty cool. There are all sorts of neat things you can do with pedal triangles and pedal polygons in general.


Practice Problem: Prove that the fourth pedal quadrilateral is similar to the quadrilateral. The result can be generalized.

No comments:

Post a Comment