Reflection Triangle

To begin with, we note the result that the reflection triangle is homothetic to the pedal triangle of the nine-point centre $N$ of $ABC$ . Let $N_A,N_B,N_C$ be the feet of the perpendiculars from $N$ , and hence the vertices of the pedal triangle of $N$ . Since $N$ is the midpoint of the line segment $ON$ , where $O$ is the outcentre and $H$ is the orthocentre, it follows that $N_A$ is the midpoint of the line segment $M_AH_A$ , where $M_A$ is the midpoint of $BC$ and $H_A$ is the foot of the altitude from $A$ . The points $N_B,N_C$ are similarly related to $M_B, H_B, M_C, H_C$ . The image of $N_A$ under the homothecy $H(G,4)$ (enlargement, scale factor $4$ , with centre the centroid $G$ of $ABC$ ) is $A'$ , where $\begin{aligned} \overrightarrow{OA'} & = \; \overrightarrow{OG} + 4\overrightarrow{GN_A} \; = \; \tfrac13(\mathbf{a} + \mathbf{b} + \mathbf{c}) + 4\overrightarrow{ON_A} - \tfrac43(\mathbf{a} + \mathbf{b} + \mathbf{c}) \\ & = \; 2\overrightarrow{OH_A} + 2\overrightarrow{OM_A} - (\mathbf{a} + \mathbf{b} + \mathbf{c}) \; = \; 2\overrightarrow{OH_A} - \mathbf{a} \end{aligned}$ and hence $A'$ is the reflection of $A$ in the line $BC$ . Similar results can be shown for $B'$ and $C'$ . Thus the image of the pedal triangle $N_AN_BN_C$ under $H(G,4)$ is the reflection triangle $A'B'C'$ . Let us find which triangles are similar to the pedal triangle of their nine-point centre. Calculations that I shall not go into show that the only triangles $ABC$ that are similar to $N_AN_BN_C$ are equilateral triangles, and that the only triangles $ABC$ that are similar to $N_AN_CN_B$ are also equilateral triangles. The only case of interest is the one where $ABC$ is similar to $N_CN_AN_B$ . If the sides of $N_AN_BN_C$ have lengths $\alpha,\beta,\gamma$ (with $\alpha$ being the length of $N_BN_C$ and so on), then we want to find the triangles $ABC$ such that $\frac{\alpha}{b} \; = \; \frac{\beta}{c} \; = \; \frac{\gamma}{a} \; = \; \lambda$ for some $\lambda > 0$ .

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Although this is not the main route through this problem, it is interesting to consider the special case where the points $N_A,N_B,N_C$ are internal to their respective line segments $BC,AC,AB$ . In this case, with the angles in the pedal triangle $N_AN_BN_C$ marked as shown, it follows from the Alternate Segment Theorem that $N_AN_C$ is a tangent to the circumcircle of $AN_BN_C$ at $N_C$ , which implies that $\angle AN_BN_C = \angle BN_CN_A$ . Another application of the Alternate Segment Theorem tells us that $\angle AN_BN_C = \angle BN_CN_A = \angle CN_AN_B = X$ , as marked on the diagram. Applying the Sine Rule to triangle $AN_BN_C$ , we see that $\frac{AN_C}{\sin X} = \frac{\beta}{\sin A}$ , and hence $\frac{AN_C}{b^2c} = \frac{\lambda \sin X}{2\Delta}$ , where $\Delta$ is the area of $ABC$ . Two more applications of the Sine Rule tell us that $\frac{AN_C}{b^2c} \; = \; \frac{BN_A}{c^2a} \; = \; \frac{CN_B}{a^2b}$ Since $N_C$ is halfway between $M_C$ and $H_C$ , we see that $AN_C \; = \; \tfrac12\big(b \cos A + \tfrac12c\big) \; = \; \frac{2c^2 + b^2 - a^2}{4c}$ and similar calculations evaluate $BN_A$ and $CN_B$ . It follows that $\begin{aligned} \frac{2c^2 + b^2 - a^2}{4b^2c^2} & = \; \frac{2a^2 + c^2 - b^2}{4a^2c^2} \; = \; \frac{2b^2 + a^2 - c^2}{4a^2b^2} \\ a^2(2c^2 + b^2 - a^2) & = \; b^2(2a^2 + c^2 - b^2) \; = \; c^2(2b^2 + a^2 - c^2) \\ u(2w + v - u) & = \; v(2u + w - v) \; = \; w(2v + u - w) \end{aligned}$

where we write $u = a^2$ , $v=b^2$ , $w=c^2$ . Putting $u=1$ we obtain $2w + v - 1 \; = \; 2v + vw - v^2 \; = \; 2vw + w - w^2$ so that $v \neq 2$ and $w \; = \; \frac{v^2 - v - 1}{v-2}$ and $\begin{aligned} 2\left(\frac{v^2 - v -1}{v-2}\right) + v - 1 & = \; (2v+1)\frac{v^2 - v - 1}{v - 2} - \left(\frac{v^2 - v - 1}{v-2}\right)^2 \\ f(v) & = \; v^4 - 6v^3 + 11v^2 - 7v + 1 \; = \; v(v-2)^3 - (v^2 - v - 1) \; = \; 0 \end{aligned}$ which implies that $w = v(v-2)^2$ . We also note that $f(v) = (v - 1)(v^3 - 5v^2 + 6v - 1) = (v-1)g(v)$ . Thus we obtain an acceptable triple $(u,v,w) = (1,v,v(v-2)^2)$ provided that $f(v) = 0$ . There are two cases:

• If $v=1$ , it follows that $w=1$ , and so $a=b=c$ , making $ABC$ an equilateral triangle.
• Consider the case $g(v) = 0$ . Suppose that $\zeta^7 = 1$ but that $\zeta \neq 1$ , and let $Z = \zeta + \zeta^{-1}$ . Then $0 \; = \; \zeta^3 +\zeta^2 + \zeta + 1 + \zeta^{-1} + \zeta^{-2} + \zeta^{-3} \; = \; (\zeta + \zeta^{-1})^3 + (\zeta+\zeta^{-1})^2 - 2(\zeta + \zeta^{-1}) - 1 \; = \; Z^3 + Z^2 - 2Z - 1$ which implies that $(Z^2-2)Z= 1 - Z^2$ , so that $Z^2(Z^2-2)^2 = (Z^2-1)^2$ , which implies that $g(Z^2)=0$ . Thus the three roots of $g(v)$ are $4\cos^2\tfrac{2\pi}7$ , $4\cos^2\tfrac{4\pi}7$ and $4\cos^2\tfrac{6\pi}7 = 4\cos^2\tfrac{\pi}7$ . If $v = 4\cos^2\tfrac{k\pi}{7}$ then $v-2 = 2\cos\tfrac{2k\pi}{7}$ , and so $w = 16\cos^2\tfrac{k\pi}{7}\cos^2\tfrac{2k\pi}{7}$ . This makes $(u,v,w) = (\sin^2\tfrac{k\pi}{7},\sin^2\tfrac{2k\pi}{7},\sin^2\tfrac{4k\pi}{7})$ a viable solution, and we end up with the heptagonal triangle.

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Life is more complicated in general, and we need to adopt a more algebraically intense approach. The formulae we derived above for distances such as $AN_C$ now have to be treated as signed distances, but we can still use the Cosine Rules to obtain expressions for the lengths $\alpha,\beta,\gamma$ of the sides of the pedal triangle $N_AN_BN_C$ . These can most easily be expressed in terms of $u=a^2,v=b^2,w=c^2$ , and if we write $U=\alpha^2,V=\beta^2,W=\gamma^2$ , we obtain $U \; = \; \frac{P}{16vw} \hspace{1cm} V \; = \; \frac{Q}{16uw} \hspace{1cm} W \;= \; \frac{R}{16uv}$ where $\begin{aligned} P & = \; (u-v)^3 + (u-w)^3 + vw(3u+v+w) - u^3 \\ Q & = \; (v-u)^3 + (v-w)^3 + uw(u+3v+w) - v^3 \\ R & = \; (w-u)^3 + (w-v)^3 + uv(u+v+3w) - w^3 \end{aligned}$ and we want to solve $\frac{U}{v} = \frac{V}{w} = \frac{W}{u} = \lambda^2$ . We see that $u,v,w$ must satisfy the polynomial equations $uvQ - w^2R \; = \; uwP - v^2Q \; = \; 0$ . Scaling so that $u=1$ we obtain the polynomial identities $A_1(v,w) \; = \; \Big(uvQ - w^2R\Big)\Big|_{u=1} \; = \; 0 \hspace{2cm} A_2(v,w) \; = \; \Big(uwP - v^2Q\Big)\Big|_{u=1} \; = \; 0$ As polynomials in $w$ , $A_1$ is quintic and $A_2$ quartic, and both have leading coefficient $-1$ . We can readily divide $A_2(v,w)$ into $A_1(v,w)$ . The remainder has a factor of $v$ , and we can write

$A_1(v,X) - (X + v^2 - 2v)A_2(v,X) \; \equiv \; vA_3(v,X)$ where $A_3(v,X)$ is a cubic polynomial in $X$ with leading term $-p_3(v)X^3 = -(v^3 - 4v^2 + 4v + 1)X^3$ . The polynomial $p_3(X)$ has no positive real zeroes, and so $p_3(v) \neq 0$ . We have the identity $A_3(v,w) \; = \; 0$ Continuing the polynomial division, we can write $p_3(v)^2A_2(v,X) \; \equiv \; \big[p_3(v)X - v^5 + 6v^4 - 12v^3 + 9v^2 - 4v - 6\big]A_3(v,X) - 2A_4(v,X)$ where $\begin{aligned} A_4(v,X) & \equiv \; \begin{array}{l}(3v^6 - 16v^5 + 11v^4 + 45v^3 - 50v^2 - 7v + 12)X^2\\ - (3v^7 - 18v^6 + 29v^5 - 9v^4 + 6v^3 - 17v^2 - 6v + 10)X \\ + (v^8 - 7v^7 + 14v^6 + v^5 - 34v^4 + 37v^3 - 8v^2 - 7v + 3) \end{array} \\ & \equiv \; p_4(v)X^2 + q_4(v)X + r_4(X) \end{aligned}$ and we must have $A_4(v,w) = 0$ . If $p_4(v) = 0$ then (since $p_4(X)$ and $q_4(X)$ are coprime) we see that $q_4(v) \neq 0$ , so we must have $w = -\frac{r_4(v)}{q_4(v)}$ , and hence $0 \; = \; A_3\big(v,-\tfrac{r_4(v)}{q_4(v)}\big) \; = \; \frac{P(v)}{q_4(v)^3}$ for some polynomial $P(X)$ of degree $27$ . But this is impossible, since it can be shown that $P(X)$ and $p_4(X)$ are coprime. Thus we deduce that $p_4(v) \neq 0$ . Performing one more polynomial division, we can show that the remainder when $A_3(v,X)$ is divided by $A_4(v,X)$ is $\begin{aligned} A_5(v,X) & \equiv \; \frac{(v-1)p_3(v)^2}{p_4(v)^2}\left[\begin{array}{l}-(6v^{10} - 41v^9 + 59v^8 + 70v^7 - v^6 -520v^5 + 501v^4 + 74v^3 - 129v^2 - 15v + 8)X \\+ (v-1)^3(3v^8 - 18v^7 + 12v^6 + 73v^5 - 61v^4 - 80v^3 + 48v^2 + 17v - 6)\end{array}\right] \\ & \equiv \; \frac{(v-1)p_3(v)^2}{p_4(v)^2}\big[p_5(v)X + q_5(v)\big] \end{aligned}$ where the polynomials $p_5(X)$ and $q_5(X)$ are coprime, and again we must have $A_5(v,w) = 0$ .

If $v=1$ we must have $A_4(1,w) = 2w(1-w) = 0$ , and hence $w=1$ . Thus we obtain an equilateral triangle in this case.

Assume now that $v \neq 1$ . We know that $p_3(v) \neq 0$ , so we deduce that $p_5(v)w + q_5(v) =0$ . Since $p_5$ and $q_5$ are coprime, we cannot have $p_5(v) = q_5(v) = 0$ . Thus $p_5(v) \neq 0$ , and $w \; = \;- \frac{q_5(v)}{p_5(v)} \; = \; \frac{(v-1)^3(3v^8 - 18v^7 + 12v^6 + 73v^5 - 61v^4 - 80v^3 + 48v^2 + 17v - 6}{6v^{10} - 41v^9 + 59v^8 + 70v^7 - v^6 -520v^5 + 501v^4 + 74v^3 - 129v^2 - 15v + 8} \hspace{1cm} (\star)$ But this implies that $0 \; = \; A_4\left(v,-\tfrac{q_5(v)}{p_5(v)}\right) p_5(v)^2 \; = \; (v-1)^5 p_4(v)^2(v^2 + v + 1) (v^3 - 5v^2 + 6v - 1)(v^6 - 2v^5 - 13v^4 + 32v^3 - 9v^2 - 6v + 1)$

We have excluded the case $v=1$ , and we know that $p_4(v)$ is nonzero. The polynomial $1 + v + v^2$ is never zero for real $v$ . As we have seen above, the case $v^3 - 5v^2 + 6v - 1 = 0$ leads to the heptagonal triangle. The remaining factor $v^6 - 2v^5 - 13v^4 + 32v^3 - 9v^2 - 6v + 1 \; = \; (v^3 - v^2 + 3v - 1)^2 - 20(v^2 - v)^2$ factorizes as $(v^3 - (1 + 2\sqrt{5})v^2 + (3 + 2\sqrt{5})v - 1)(v^3 + (2\sqrt{5}-1)v^2 + (3 - 2\sqrt{5})v - 1)$ Two of the zeros of the second cubic are negative, and the third zero gives a negative value of $w$ , so we are left with investigating the case when $v$ satisfies $v^3 - (1 + 2\sqrt{5})v^2 + (3 + 2\sqrt{5})v - 1 \; = \; 0$

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Another fit of polynomial division:

• If $v^3 - 5v^2 + 6v - 1 = 0$ , then equation $(\star)$ reduces to $w = (v-1)^2 = v(v-2)^2$ .
• If $v^3 - (1 + 2\sqrt{5})v^2 + (3 + 2\sqrt{5})v - 1 = 0$ , equation $(\star)$ reduces to $w = 1 - \tfrac12(5 + \sqrt{5})v + \tfrac12(1 + \sqrt{5})v^2$ .

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And finally:

• When $u=v=w=1$ we calculate $\lambda^2 = W = \tfrac14$ . Thus the area ratio $\frac{|A'B'C'|}{|ABC|} = 16\lambda^2 = 4$ for the equilateral triangle.
• When $u=1$ , $v^3 - 5v^2 + 6v - 1 = 0$ and $w = (v-1)^2$ , polynomial division shows that the ratio $\lambda^2 = W = \tfrac18$ . Thus the area ratio $\tfrac{|A'B'C'|}{|ABC|}=16\lambda^2 = 2$ for the heptagonal triangle.
• When $u=1$ , $v^3 - (1 + 2\sqrt{5})v^2 + (3 + 2\sqrt{5})v - 1 = 0$ and $w = 1 - \tfrac12(5 + \sqrt{5})v + \tfrac12(1 + \sqrt{5})v^2$ , one final bout of polynomial division gives $\lambda^2 = W = \tfrac{1}{16}(\sqrt{5}-1)$ , making the area ratio $\frac{|A'B'C'|}{|ABC|} = 16\lambda^2 = \sqrt{5}-1 = \tfrac{4}{\sqrt{5}+1} = \boxed{\tfrac{2}{\varphi}}$ .

In the diagram below, the left-hand triangle is the heptagonal triangle, and the right-hand triangle is the new one.