Simson Line Property

Michael Niland · Riemann Hypothesis Offline Joined: 29 Jul 2004 Posts: 460

Prove that the Simson LInes of triangles with the same incircles and circumcircles for a fixed point on the circumference are concurrent.

**Posted:** Sat Mar 19, 2005 8:43 pm

yetti

According to the 2nd Fontene theorem, the orthopole of a line passing through the circumcenter $(O)$ of a triangle $\triangle ABC$ lies on the circumcircle the pedal triangle $\triangle XYZ$ of any point $P$ on this line. For a proof, see Darij Grinberg, Generalization of the Feuerbach point ¹). The pedal triangle of the circumcenter $O$ is the medial triangle $\triangle A'B'C'$ and its circumcircle is the 9-point circle $(N)$ of the triangle $\triangle ABC$ . Hence, the orthopoles of all lines passing through the circumcenter lie on the 9-point circle. If the point $P$ is the intersection of a line through the circumcenter with the circuncircle $(O)$ , the pedal triangle degenerates into a line segment and its circumcircle into a Simson line with the pole $P$ , which passes through the orthopole of the line $OP$ .
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The pedal triangle of the incenter $I$ is the contact triangle and its circumcircle is the incircle $(I)$ of the triangle $\triangle ABC$ . The orthopole of the diacentral line $OI$ lies both on the 9-point circle and on the incircle, hence, it is identical with their tangency point $F \equiv (I) \cap (N)$ (Feuerbach point). Let the diacentral line $OI$ intersect the circumcircle and incircle at points $P, P'$ and $Q, Q'$ , so that the points $P, Q, I, Q', P'$ and the points $P, O, I, P'$ follow on the line $OI$ in these orders. The orthocenter $H$ is the external homothety center of the circumcircle and the 9-point circle. The Feuerbach point $F$ is the external homothety center of the incircle and the 9-point circle. Thus the line $HF$ is the external homothety axis of the circumcircle, 9-point circle, and the incircle passing through the external homothety center $E$ of the incircle and the circumcircle, the intersection of the diacentral line $OI$ with the homothety axis $HF$ . Let the line $HF$ intersect the circumcircle $(O)$ at a point $F'$ . Because of the homothety between the incircle and circumcircle, the right angle triangles $\triangle QFQ' \sim \triangle PF'P'$ are cenrally similar with homothety cemter $E$ . Because of the homothety between the 9-point circle and circumcircle with homothety center $H$ and homothety coefficient $\frac 1 2$ , the 9-point circle cuts the segments $PH, P'H$ at their midpoints $R, R'$ , the lines $PP' \equiv QQ' \parallel RR'$ are parallel and the segment $RR'$ is a diameter of the 9-point circle. Consequently, the right angle triangles $\triangle PFP' \sim \triangle RFR'$ are centrally similar with homothety center $H$ and the right angle triangles $\triangle QFQ' \sim \triangle RFR'$ are centrally similar with homothety center $F$ . The Simson lines of the poles $P, P'$ , perpendicular to each other, pass through these midpoints $R, R'$ and they meet at the orthopole $F$ of the diacentral line $PP' \equiv OI$ . Hence, the Simson lines of the poles $P, P'$ are the lines $RF, R'F$ . Because of the homothety of the triangles $\triangle QFQ' \sim \triangle RFR'$ , the Simson lines $RF, R'F$ pass through the points $Q, Q'$ . But these 2 points are completely independent of the "rotation" of the triangle $\triangle ABC$ around the fixed circumcircle $(O)$ and the incircle $(I)$ according to Poncelet's porism. As a result, the intersections $Q, Q'$ of the incircle with the diacentral line $OI$ are the fixed points of Simson lines with the poles $P, P'$ .
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Let $P, S$ be 2 points of the circumcircle with the central angle $0 < \measuredangle POR = \phi \le 180^o$ . Then the Simson lines $s_P, s_S$ of the poles $P, S$ form an angle of $\frac \phi 2$ . This easily follows from the usual construction of a Simson line in the desired direction. Let the lines parallel to the Simson lines $s_P, s_S$ and passing through one of the triangle vertices, say $A$ , itersect the circumcircle at points $K, L$ different from the vertex $A$ . Thus the angle between the Simson lines $s_P, s_S$ is equal to the angle $\measuredangle KAL$ . The poles $P, S$ are intersections of the lines throught the points $K, L$ perpendicular to the triangle side $BC$ with the circumcircle $(O)$ . Hence, the quadrilateral $KLSP$ is a cyclic trapezoid, which means that it is isosceles with $KL = PS$ . Consequently, the central angles $\measuredangle KOL = \measuredangle POS = \phi$ are equal and the angle formed by the 2 Simson lines is equal to $\measuredangle KAL = \frac \phi 2$ . As a special case, if the poles $P, S$ are diametrally opposite points of the circumcircle $(O)$ , the angle $\phi = \measuredangle POS = 180^o$ and the Simson lines of the poles $P, S$ are perpendicular to each other.
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Reflections of any 2 circumcircle diameters $PP', SS'$ in any side line of the medial triangle $\triangle A'B'C'$ meet on the 9-point circle. This is because the circumcenter $O$ of the triangle $\triangle ABC$ is the orthocenter of the medial triangle and the 9-point circle $(N)$ is its circumcircle. The reflection of the orthocenter in any side of a triangle lies on its circumcircle and the reflection of any line passing through the orthocenter in any triangle side must pass through the corresponding orthocenter reflection on the circumcircle. Let the reflections of the diameters $PP', SS'$ in the side line $B'C'$ of the medial triangle meet at a point $W$ on the 9-point circle $(N)$ . These reflected lines form the same angle $\phi = \measuredangle POS$ as the diameters $PP', SS'$ . According to the above reference ¹), the reflections of the diameters $PP', SS'$ in the sides $B'C', C'A', A'B'$ of the medial triangle concur at the orthopoles $U, V$ of these diameters. Thus the angle $\measuredangle UWV$ spanning the arc $UV$ of the 9-point circle $(N)$ is equal to $\measuredangle UWV = \measuredangle POS = \phi$ and the central angle of the orthopoles $U, V$ on the 9-point circle is equal to $\measuredangle UNV = 2 \measuredangle UWV = 2 \phi$ . As a special case, if the diameters $PP', SS'$ are perpendicular to each other, i.e. $\phi = \measuredangle POS = 90^o$ , their orthopoles $U, V$ are diametrally opposite points of the 9-point circle.
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Let $R, r$ be the fixed circumradius and inradius, $d = OI$ the fixed distance from the circumcenter to the incenter and $\triangle ABC$ an arbitrary triangle with the fixed circumcircle $(O)$ and incircle $(I)$ . The 9-point circle $(N)$ is the circumcircle of the medial triangle $\triangle A'B'C'$ , hence, its radius is $n = \frac R 2$ is also fixed. The 9-point circle is tangent to the fixed incircle at a point $F$ (Feuerbach point). Thus the locus of the 9-point circle centers $N$ of the triangles $\triangle ABC$ "rotated" according to Poncelet's porism is a circle $(I, m)$ centered at the incenter $I$ and with radius $m = n - r = \frac{R - 2r}{2}$ . Consider the circumcircle diameter $PP'$ , which coincides with the diacentral line $OI$ , and an arbitrary circumcircle diameter $SS'$ forming the central angle $\phi = \measuredangle POS \le 90^o$ with the diacentral diameter $PP'$ . If the angle $\phi = 0$ , i.e., if the diameters $SS' \equiv PP'$ are identical, the locus of orthopoles $F$ of this diameter (Feuerbach points) is the incircle $(I)$ . If the angle $\phi = 90^o$ , i.e., if the diameters $SS' \perp PP'$ are perpendicular, the orthopole $V$ is the diametrally opposite point of the Feuerbach point on the 9-point circle. Hence, the locus of the orthopoles $V$ diametrally opposite to the Feuerbach points $F$ on the 9-point circles is a circle $(I, k)$ centered at the incenter $I$ and with radius $k = m + n = \frac{R - 2r}{2} + \frac R 2 = R - r$ . If the angle $0 < \phi < 90^o$ , consider the triangle $\triangle INV$ . The side $IN = m = \frac{R - 2r}{2}$ is fixed, the side $NV = n = \frac R 2$ is fixed and the angle $\measuredangle INV \equiv \measuredangle FNV$ is also fixed, because the Feuerbach point $F$ lies on the center line $IN$ of the incircle and the 9-point circle and the 9-point circle central angle of the orthopoles $F, V$ of the circumcircle diameters $PP', SS'$ is equal to $\measuredangle FNV = 2 \measuredangle POS = 2 \phi$ . Hence, the remaining side of this triangle is fixed as well and given by the cosine theorem:

$IV = \sqrt{NV^2 + IV^2 - 2\ NV \cdot IV \cos{\widehat{INV}}} = \sqrt{\left(\frac R 2\right)^2 + \left(\frac{R - 2r}{2}\right)...$

$= \sqrt{r^2 + (R^2 - 2rR)\ \frac{1 - \cos{2 \phi}}{2}} = \sqrt{r^2 + (R^2 - 2rR) \sin^2{\phi}} = \sqrt{(R - r)^2 - (R^2 - 2rR...$

Since the incenter $I$ is fixed and the distance $IV$ is independent of the "rotation" of the triangle $\triangle ABC$ according to Poncelet's porism, the locus of orthopoles $V$ of the circumcircle diameter $SS'$ is a circle $(I, v)$ centered at the incenter $I$ and with radius $v = IV$ . From the above formula, it is clear that the radius of this circle is limited by $r \le v \le R - r$ for $0 \le \phi \le 90^o$ .
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Since the angle $\measuredangle FNV = 2 \phi$ is fixed and the triangle $\triangle INV$ just rotates around the incenter $I$ , the angle $\measuredangle FNV = 2 \psi$ is also fixed. Using the sine theorem for the triangle $\triangle INV$ ,

$\frac{R}{2 \sqrt{r^2 + (R^2 - 2rR) \sin^2{\phi}}} = \frac{NV}{IV} = \frac{\sin{\widehat{NIV}}}{\sin{\widehat{INV}}} = \frac{\...$

$\sin{2 \psi} = \frac{R \sin{2 \phi}}{2 \sqrt{r^2 + (R^2 - 2rR) \sin^2{\phi}}}$

From the above formula, it is clear that $0 \le 2 \psi < 180^o$ for $0 \le 2 \phi \le 180^o$ and that $\sin{2 \psi} = 1,\ 2 \psi = 90^o$ for

$4R^2 \sin^2{\phi}\ \cos^2{\phi} = 4\left[r^2 + (R^2 - 2rR) \sin^2{\phi}\right]$

$R^2 \sin^4{\phi} - 2rR\ \sin^2{\phi} + r^2 = (R\ \sin^2{\phi} - r)^2 = 0$

$\sin^2{\phi} = \frac r R,\ \ \cos{2 \phi} = \frac{R - 2r}{R}$

The Simson lines $s_P \equiv QF \perp s_P' \equiv Q'F$ with the poles $P, P'$ pass through the orthopole $F \in (I, r)$ (Feuerbach point) of the diacentral line $OI \equiv PP'$ and they are perpendicular to each other. Likewise, the Simson lines $s_S \perp s_S'$ with the poles $S, S'$ pass through the orthopole $V \in (I, v)$ of the line $SS'$ and they are also perpendicular to each other. In addition, the Simson line pairs $s_P, s_S$ and $s_P', s_S'$ both form a constant angle equal to $\frac \phi 2$ . Let $V' \equiv IV \cap (I)$ be the intersection of the line $IV$ with the incircle $(I)$ and $p \parallel s_S, p' \parallel s_S'$ 2 lines through the point $V'$ parallel to the Simson lines $s_S, s_S'$ . The concentric circles $(I, v), (I, r)$ are centrally similar with homothety center $I$ and so are the parallel lines $s_S \parallel p, s_S' \parallel p'$ passing through the corresponding points $V, V'$ in this homothety. It is clear that the mutually perpendicular Simson lines $s_S \perp s_S'$ pass through fixed points of the circle $(I, v)$ iff their mutually perpendicular parallels $p \perp p'$ pass through fixed points of the circle $(I, r)$ in a rotation of the central angle $\measuredangle FIV \equiv \measuredangle FIV'$ around the incenter $I$ by an arbitrary angle $\theta$ . Therefore, to complete the solution of the problem, we have to prove the following lemma:
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Let $(I)$ be an arbitrary circle with a fixed diameter $Q, Q'$ and $F$ an arbitrary point on this circle, forming a right angle triangle $\triangle QFQ'$ with the fixed hypotenuse $QQ'$ . Let $0 \le \measuredangle FIV = 2 \psi \le 180^o$ be an arbitrary central angle of this circle with a fixed measure, the point $V$ also lying on the circle $(I)$ . Let $p \perp p'$ be 2 perpendicular lines through the point $V$ , both forming the same arbitrary fixed angle $\frac \phi 2 \le 90^o$ with the perpendicular lines $QF \perp Q'F$ . In a rotation of the point $F$ around the circle center $I$ by an arbitrary angle $\theta$ , the perpendicular lines $p \perp p'$ pass through fixed points $X, X'$ of the circle $(I)$ , forming a right angle triangle $\triangle XVX'$ with the fixed hypotenuse $XX'$ .
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The proof of this lemma is easy: Let the perpendicular lines $p \perp p'$ through the point $V$ intersect the circle $(I)$ at points $X, X'$ . Since the angle $\measuredangle XVX' = 90^o$ is right, $XX'$ is a diameter of the circle $(I)$ . The triangles $\triangle FIQ, \triangle VIX$ are both isosceles with $IF = IQ = IV = IX = r$ , where $r$ is radius of the circle $(I)$ . Hence, the angles $\measuredangle IFQ = \measuredangle IQF$ are equal and the angles $\measuredangle IVX = \measuredangle IXV$ are also equal. Let the lines $FQ, VX$ intersect at a point $Z$ . From the concave quadrilateral $IFZV$ , sum of the angles $\measuredangle IFQ + \measuredangle IVX$ is equal to

$\measuredangle IFQ + \measuredangle IVX = 360^o - (\measuredangle FZV + 360^o - \measuredangle FIV) = \measuredangle FIV -\me...$

From the convex quadrilateral $QIXZ$ , the angle $\measuredangle QIX$ is equal to

$\measuredangle QIX = 360^o - (\measuredangle QZX + 180^o - \measuredangle IQF + 180^o - \measuredangle IXV) =$

$= \measuredangle IQF + \measuredangle IXV - \measuredangle QZX = 2 \psi - \frac \phi 2 - \frac \phi 2 = 2 \psi - \phi$

Since the measures of the angles $2 \psi, \frac \phi 2$ are both fixed and the point $Q$ is also fixed, the point $X$ is necessarily fixed as well. Since $XX'$ is a diameter of the circle $(I)$ , both the points $X, X'$ are fixed and they do not depend on the arbitrary angle of rotation $\theta$ .
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This concludes the proof of the problem proposition. Wow !!! Cool

And look mom, no cubics !

Related problem: Instead of "rotating" the triangle $\triangle ABC$ around he fixed circumcircle $(O)$ and incircle $(I)$ according to Poncelet's porism, we can keep the triangle fixed and rotate the diametrally opposite poles $S, S' \in (O)$ around the circumcenter. When the poles $S, S'$ are identical with the intersections $P, P'$ of the diacentral line $OI$ with the circumcircle, i.e., $\phi = \measuredangle POS = 0$ , the Simson line pivot points $X, X'$ are identical with the intersections $Q, Q'$ of the incircle with the diacentral line $OI$ , i.e., $IQ = IQ' = r$ . When the diameter $SS'$ is perpendicular to the diacentral line $OI$ , i.e., $\phi = \measuredangle POS = 90^o$ , the Simson line pivot points $X, X'$ are on a line perpendicular to the diacentral line $OI$ and $IX = IX' = R - r$ . From this, we may conjecture that the locus of the Simson line pivot points is an ellipse centered at the incenter $I$ , with the semiminor axis $IQ = r$ , semimajor axis $IX = R - r$ and the focal length (using Euler's formula for $R, r, d = OI)$

$\sqrt{(R - r)^2 - r^2} = \sqrt{R^2 - 2rR} = d = OI$

I am 100 % sure that this is true, even though I have not performed the calculation yet.

Yetti