**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.

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