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.

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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.

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