A Triangle's Stars Align

As in my solution to the previous problem concerning the reflection triangle, the reflection triangle $A'B'C'$ is homothetic to the pedal circle $N_AN_BN_C$ of the nine-point circle, and this latter triangle has edge lengths $\sqrt{U}$ , $\sqrt{V}$ , $\sqrt{W}$ , where $\begin{array}{rclcrcl} U & \;=\; & \frac{P}{16vw} & \hspace{1cm} & P & \; =\; & (u-v)^3 + (u-w)^3 + vw(3u+v+w) - u^3\\[1ex] V & \; =\;& \frac{Q}{16uv} & & Q & \; =\;& (v-u)^3 + (v-w)^3 + uw(u+3v+w) - v^3 \\[1ex] W & \; =\;& \frac{R}{16uv} & & R & \; =\;& (w-u)^3 + (w-v)^3 + uv(u+v+3w) - w^3 \end{array}$ where $a=\sqrt{u}$ , $b=\sqrt{v}$ , $c=\sqrt{w}$ are the sides of the original triangle. We want to find triangles $ABC$ such that $N_C$ is the midpoint of $N_AN_B$ , and hence such that $4U = 4V = W$ , so we want $4uP = 4vQ = wR$ . Define (scaling the calculations with $u=1$ ) $A_1(v,w) \; = \; \Big(wR - 4uP\Big)\Big|_{u=1} \hspace{2cm} A_2(v,w) \; = \; \Big(wR - 4vQ\Big)\Big|_{u=1}$ Then we have $A_1(v,w) = A_2(v,w) = 0$ . We can write $A_1(v,w) - A_2(v,2) \; = \; (v-1)A_3(v,w) \hspace{2cm} A_3(v,w) \; = \; -w^3 + 3(v+1)w^2 - (3v^2 + v + 3)w + v^3 - v^2 - v + 1$ The only positive solutions to the equation $A_2(1,w) = 0$ are $w=0$ and $w=4$ , both of which lead to degenerate triangles $ABC$ . Thus we can assume that $v \neq 1$ , and hence that $A_3(v,w) = 0$ . We can now write $A_2(v,w) + (w + 4v)A_3(v,w) \; = \; 2vA_4(v,w) \hspace{2cm} A_4(v,w) \; = \; 5w^2 - 8(v+1)w + 4(v-1)^2$ and we see that $A_4(v,w)=0$ . Continuing, we write $25A_3(v,w) + (5w - 7v-7)A_4(v,w) \; = \; A_5(v,w) \; = \; (v^2 + 47v + 1)w - 3(v-1)^2(v+1)$ and hence deduce that $A_5(v,w) = 0$ . Since $v^2 + 47v + 1 \neq 0$ for positive $v$ , we deduce that $w \; = \; \frac{3(v-1)^2(v+1)}{v^2 + 47v + 1}$ and hence $0 \; = \; A_4\left(v,\tfrac{3(v-1)^2(v+1)}{v^2 + 47v + 1}\right) (v^2 + 47v + 1)^2 \; = \; 25(v-1)^2(v^2 - 16v + 1)^2$ so we deduce that $v^2 - 16v + 1 = 0$ , so that $v = 8 \pm 3\sqrt{7}$ , and hence $w = 6 \pm 2\sqrt{7}$ . The triple $(8-3\sqrt{7},1,6-2\sqrt{7})$ is just a rescaled copy of $(1,8+3\sqrt{7},6+2\sqrt{7})$ , so we restrict attention to the case of the triple $(u,v,w) =(1,8+3\sqrt{7},6+2\sqrt{7})$ .

The smallest angle of this triangle is the angle at the vertex $N_A$ . If $\theta$ is this angle, then $\cos\theta \; = \; \frac{(6 + 2\sqrt{7}) + (8 + 3\sqrt{7}) - 1}{2\sqrt{(6 + 2\sqrt{7})(8 + 3\sqrt{7})}} \: =\; \sqrt{\frac{5+\sqrt{7}}{8}}$ But then $\sin\theta \; = \; \sqrt{\frac{3-\sqrt{7}}{8}}$ and hence $\tan\theta \; = \; \sqrt{\frac{3-\sqrt{7}}{5 + \sqrt{7}}} \; = \; \sqrt{\frac{11 - 4\sqrt{7}}{9}} \; = \; \frac{\sqrt{7}-2}{3}$ making the answer $7+2+3 =\boxed{12}$ .

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It is worth noting that the next smallest angle of this triangle is the angle at the vertex $C$ , which has cosine $\frac{1 + (8 + 3\sqrt{7}) - (6 + 2\sqrt{7})}{2\sqrt{8 +3\sqrt{7}}} \; = \; \frac{3 + \sqrt{7}}{\sqrt{2(16 +6\sqrt{7})}} \; = \; \frac{1}{\sqrt{2}}$ and so the angles of the triangle are $\theta$ , $\tfrac14\pi$ and $\tfrac34\pi -\theta$ .

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Since the pedal triangle of the nine-point centre is degenerate, it follows from the Wallace-Simpson Theorem that the nine-point centre $N$ lies on the circumcircle of $ABC$ .

Note that $\cos2\theta = \tfrac14(1 + \sqrt{7})$ and $\sin2\theta =\tfrac14(\sqrt{7}-1)$ , so that $\cos2\theta = \sin2\theta + \tfrac12$ . If we set the circumcentre $O$ as the origin, the position vectors of $A,B,C$ all have length $R$ , the circumradius. Since the angle $C$ is $\tfrac14\pi$ , we deduce that $\angle AOB = \tfrac12\pi$ , so that $\mathbf{a} \cdot \mathbf{b} = 0$ . Similarly we can show that $\mathbf{a}\cdot\mathbf{c} = -R^2\sin2\theta$ and $\mathbf{b}\cdot\mathbf{c} = R^2\cos2\theta$ .

Consider the point $X$ with position vector $\mathbf{x} = \tfrac14(\mathbf{a}+\mathbf{b}) + \mathbf{c} = \tfrac14(\mathbf{h} + 3\mathbf{c})$ , where $\mathbf{h}=\mathbf{a}+\mathbf{b}+\mathbf{c}$ is the position vector of the orthocentre of $ABC$ . Note that $X$ lies on $CH$ , with $HX\,:\,XC \,=\, 3\,:\,1$ . We calculate that $(\mathbf{x}-\mathbf{a})\cdot(\mathbf{x}-\mathbf{c}) = \tfrac{1}{16}(-3\mathbf{a}+\mathbf{b}+4\mathbf{c})\cdot(\mathbf{a}+\mathbf{b}) = \tfrac14R^2(\cos2\theta - \sin2\theta -\tfrac12) = 0$ so that $X$ is the position vector of $H_C$ , the foot of the altitude from $C$ . Thus the midpoint of $HC$ is the reflection of $C$ in $AB$ , namely the point $C'$ . But the nine-point circle passes through the midpoints of $AH_A$ , $BH_B$ and $CH_C$ , so (in particular) passes through the relection point $C'$ .

Other points are noteworthy. Again, the fact that $C = \tfrac14\pi$ tells us that the triangle $H_AH_BH_C$ is right-angled, with the right angle at the vertex $H_C$ . This means that the nine-point centre (with is the circumcentre of $H_AH_BH_C$ ) is the midpoint of $H_AH_B$ . Moreover the diameter of the nine-point circle passing through $C'$ is perpendicular to the diameter $H_AH_B$ . We can prove this by calculating the length $AH_B = AB\cos\theta$ , and hence obtaining the position vector of $H_B$ as a linear combination of $\mathbf{a}$ and $\mathbf{c}$ . It is then a matter of calculation to show that $\mathbf{c}$ and $\overrightarrow{NH_B}$ are perpendicular. The result now follows because $\overrightarrow{OC'} = \mathbf{n} + \tfrac12\mathbf{c}$ .

Let $\triangle ABC$ be the triangle and let it be scaled so $AC = 1$ . Place $A$ at $(0, 0)$ , $C$ at $(1, 0)$ , and $B$ in the first quadrant, and let $m = \tan A$ and $n = \tan B$ .

Then $AB$ is on $y = mx$ and $BC$ is on $y = -n(x - 1)$ , and the intersection of these two lines is at $B(\frac{n}{m + n}, \frac{mn}{m + n})$ , which makes the reflection of $B$ in side $AC$ is at $B'(\frac{n}{m + n}, -\frac{mn}{m + n})$ .

Let the reflection of $C$ in side $AB$ be $C'$ . Then $AC = AC' = 1$ , so that $C'$ is on $x^2 + y^2 = 1$ , and $C'$ is also on $y = -\frac{1}{m}(x - 1)$ , the line perpendicular to $AB$ through $C$ . These equations solve to $C'(\frac{1 - m^2}{m^2 + 1}, \frac{2m}{m^2 + 1})$ .

Let the reflection of $A$ in side $BC$ be $A'$ . Then $CA = CA' = 1$ , so that $A'$ is on $(x - 1)^2 + y^2 = 1$ , and $C'$ is also on $y = \frac{1}{n}x$ , the line perpendicular to $BC$ through $A$ . These equations solve to $A'(\frac{2n^2}{n^2 + 1}, \frac{2n}{n^2 + 1})$ .

If the reflected points are collinear and one point is the midpoint of the other two, then either:

$A'$ is the midpoint of $B'$ and $C'$ :

Then $\frac{n}{m + n} + \frac{1 - m^2}{m^2 + 1} = \frac{4n^2}{n^2 + 1}$ and $\frac{-mn}{m + n} + \frac{2m}{m^2 + 1} = \frac{4n}{n^2 + 1}$ , which has solutions $(m, n)$ of $(0, 0)$ , $(-1, \frac{\pm \sqrt{7} + 2}{3})$ , and $(1, \frac{\pm \sqrt{7} - 2}{3})$ .

$B'$ is the midpoint of $A'$ and $C'$ :

Then $\frac{2n^2}{n^2 + 1} + \frac{1 - m^2}{m^2 + 1} = \frac{2n}{m + n}$ and $\frac{2n}{n^2 + 1} + \frac{2m}{m^2 + 1} = \frac{-2mn}{m + n}$ , which has solutions $(m, n)$ of $(0, 0)$ , $(\frac{-\sqrt{7} - 2}{3}, \frac{\sqrt{7} - 2}{3})$ , $(\frac{-\sqrt{7} + 2}{3}, \frac{\sqrt{7} + 2}{3})$ , $(\frac{\sqrt{7} - 2}{3}, \frac{-\sqrt{7} - 2}{3})$ , and $(\frac{\sqrt{7} + 2}{3}, \frac{-\sqrt{7} + 2}{3})$ .

$C'$ is the midpoint of $A'$ and $B'$ :

Then $\frac{2n^2}{n^2 + 1} + \frac{n}{m + n} = \frac{2(1 - m^2)}{m^2 + 1}$ and $\frac{2n}{n^2 + 1} + \frac{-mn}{m + n} = \frac{4m}{m^2 + 1}$ , which has solutions $(m, n)$ of $(0, 0)$ , $(\frac{\pm \sqrt{7} - 2}{3}, 1)$ , and $(\frac{\pm \sqrt{7} + 2}{3}, -1)$ .

All cases result in either a degenerate triangle ( $(m, n) = (0, 0)$ ) or a reflection/rotation of a triangle with angles of approximately $12.15°$ , $45°$ , and $122.85°$ , the smallest of which has a tangent of $m = \frac{\sqrt{7} - 2}{3}$ . Therefore, $p = 7$ , $q = 2$ , $r = 3$ , and $p + q + r = \boxed{12}$ .