Pedal's Pedal

Given a triangle $\Delta$ and a point $P$ , the pedal triangle of $\Delta$ with respect to $P$ is the triangle whose vertices are the feet of the cevians through $P$ . Thus there is an infinite number of pedal triangles for a given triangle $\Delta$ .

What is often called the pedal triangle of $\Delta$ is actually the so-called orthic triangle of $\Delta$ , which is the pedal triangle obtained using the orthocentre of $\Delta$ as the point $P$ . In other words, the vertices of the orthic triangle of $\Delta$ are the feet of the perpendiculars of $\Delta$ . Thus, if $\Delta$ is a triangle, then $\Delta_1$ is the orthic triangle of $\Delta$ , $\Delta_2$ is the orthic triangle of $\Delta_1$ , and so on. We say that $\Delta$ is $P$ -similar if it is similar to at least one of $\Delta_n$ for $n \ge 1$ .

If a triangle $\Delta$ has angles $a,b,c$ , where $a \le b \le c$ , then it is standard that the orthic triangle $\Delta_1$ has angles

• $180^\circ-2a$ , $180^\circ-2b$ , $180^\circ - 2c$ if the triangle $\Delta$ is acute-angled, so that $c < 90^\circ$ ,

• $2a$ , $2b$ , $2c-180^\circ$ if the triangle $\Delta$ is obtuse-angled, so that $c > 90^\circ$ .

Note that the orthic triangle for $\Delta$ does not exist if $\Delta$ is right-angled.

If the angles in $\Delta$ are integer-valued, then the angles in $\Delta_1$ are even integer-valued, and the angles in $\Delta_2$ (and indeed $\Delta_n$ for all $n \ge 2$ ) are integers which are multiples of $4$ . If $\Delta$ is similar to $\Delta_1$ , then it is similar to $\Delta_n$ for all $n \ge 2$ . Thus, if a triangle $\Delta$ is $P$ -similar, then it must be similar to some $\Delta_n$ for $n \ge 2$ . Thus, if $\Delta$ is $P$ -similar, and if the angles of $\Delta$ are integers, then all the angles of $\Delta$ must be multiples of $4$ .

Consider the set $\mathcal{X} \; = \; \big\{ (a,b,c) \, \big| \, a,b,c \in \mathbb{N} \; , \; a \le b \le c \,,\, a + b + c = 45 \big\}$ Thus an element $(a,b,c)$ of $\mathcal{X}$ corresponds to a triangle $\Delta$ with angles $4a,4b,4c$ which are all integers that are multiples of $4$ . Note that, if $(a,b,c) \in \mathcal{X}$ , then either all three of $a,b,c$ are odd, or else exactly one is odd and the other two are even.

Consider the function $F\,:\, \mathcal{X} \to \mathcal{X}$ given by the formula $F\big((a,b,c)\big) \; = \; \left\{ \begin{array}{lll} (45 - 2c,45 - 2b,45 - 2a) & \hspace{2cm} & c \le 22 \\ (2a,2b,2c-45) & & c > b + 22.5 \\ (2a,2c-45,2b) & & a + 22.5 < c < b + 22.5 \\ (2c-45,2a,2b) & & 23 \le c < a + 22.5 \end{array} \right.$ so that, if a triangle $\Delta$ corresponds to an ordered triple $(a,b,c) \in \mathcal{X}$ , then the orthic triangle $\Delta_1$ corresponds to the ordered triple $F((a,b,c))$ .

Suppose that $(a,b,c)\,,\,(d,e,f)\,\in\,\mathcal{X}$ are such that $F\big((a,b,c)\big) = F\big((d,e,f)\big) = (u,v,w)$ . If $u,v,w$ are all odd, then $45-2c = 45 - 2f = u \;, \; 45 - 2b = 45 - 2e = v \;,\; 45 - 2a = 45 - 2d = u$ and hence $(a,b,c) = (d,e,f)$ . Otherwise we deduce that $2c-45 = 2f-45$ must be equal to whichever of $u,v,w$ is odd, so that $c=f$ , and that $\{2a,2b\} = \{2d,2e\}$ is equal to the pair comprising the other two of $u,v,w$ . Thus we deduce that $(a,b,c) = (d,e,f)$ .

Thus $F\,:\, \mathcal{X} \to \mathcal{X}$ is an injective map, and hence is a permutation of $\mathcal{X}$ . Thus there exists a positive integer $N\ge2$ such that $F^N = \mathrm{id}$ . This means that any triangle $\Delta$ with integer-valued angles, all of whose angles are multiples of $4$ , is similar to $\Delta_N$ , and hence is $P$ -similar.

Thus it follows that the number of $P$ -similar triangles is equal to the number of elements of $\mathcal{X}$ , which is $\sum_{a=1}^{15} \left(\left\lfloor \frac{45-3a}{2}\right\rfloor + 1\right) \; = \; 169$