The Simson Triangle and Its Properties

Transcription

1 Forum Geometricorum Volume FORUM GEOM ISSN The Simson Triangle and Its Properties Todor Zaharinov bstract Let be a triangle and P P P 3 points on its circumscribed circle The Simson triangle for P P P 3 is the triangle bounded by their Simson lines with respect to triangle We study some interesting properties of the Simson triangle Introduction The following theorem is often called Simson s theorem see [ p 37 Theorem 9] Theorem Wallace-Simson line The feet of the perpendiculars to the sides of triangle from a point are collinear if and only if the point is on the circumscribed circle of the triangle This is Simson line or Wallace-Simson line Definition Simson triangle The Simson triangle is the triangle bounded by the Simson lines of three points on the circumscribed circle of a fixed triangle It is degenerate if the three Simson lines are concurrent l 3 N P P 3 N 3 l N l P Figure Theorem Orthopole see [ p 47 Theorem 406] If perpendiculars are dropped on a line from the vertices of a triangle the perpendiculars to the opposite sides from their feet are concurrent at a point called the orthopole of the line Publication Date: June 8 07 ommunicating Editor: Paul Yiu

2 374 T Zaharinov Notations is a triangle of reference The circumcircleof has center O and radius R P P P 3 l l l 3 are the Simson lines of P P P 3 with respect to N N N 3 is the Simson triangle bounded by these Simson lines: N = l l 3 N = l 3 l N 3 = l l The circumcircle of N N N 3 is with center O and radius R The orthocenter of is H The orthocenter of P P P 3 is H The orthocenter of N N N 3 is H The center of nine-point circle for ise We will use complex numbers in the proofs y u we shall denote the complex number corresponding to point U Without loss of generality we take the circumcircleto be the unit circle Then R = ando = 0 a ā = b b = c c = p p = p p = p 3 p 3 = h = abc;e = /abc;h = p p p 3 Lemma 3 LetV andw be points on the unit circle The orthogonal projection of a point P onto the linel = VW is given by p l = v w p vw p In particular if P is also on the unit circle then p l = v w p vw p Proof Write the orthogonal projection asp l = tvtw for some real number t The vector p tv tw is perpendicular to This means that p tv tw v w p t v t wv w = 0 Sincev andw are on the unit circle v = v w = w We have p tv tw v p t t v w = 0 w v w From this t = v w pvw p v w and the orthogonal projection is p l = tv tw = v w p vw p IfP is also on the unit circle then p = p and p l = v w p bc p

3 The Simson triangle and its properties 375 Proposition 4 Let P be a point on the unit circumcircle of triangle The equation of its Simson line is abc z pz p abcp bccaab abc p = 0 Proof Let P be a point on the unit circumcircle Its projections on the side lines of triangle are by Lemma 3 p a = bcp bc p p b = cap ca p p c = abp ab p The line joining p b andp c has equation z p b p c z 0 = z p b p c = cap ca p abp ab p z c a p ca p a b p p ab = b cp aabcp z p z p 3 abcp bccaabp abc 4abcp = b cp aabc z pz p abcp bccaab abc 4abcp Therefore the equation of the Simson line of P is given by above Proposition 5 The intersection of the Simson lines of two points P Q is the point with coordinates pq abc abc pq Proof Let l p l q be the Simson lines of two points P Q on the unit circumcircle of Their intersection is given by the solution of abc z pz p abcp bccaab abc p abc z qz q abcq bccaab abc q Subtracting 4 from 3 we obtain p qz p q abcp q abc Dividing byp q we obtainz as given in above p = 0 3 = 0 4 p = 0 q

4 376 T Zaharinov 3 Simson triangle Theorem 6 The Simson triangle N N N 3 is directly similar to triangle P P P 3 see Figure Proof For three points P P P 3 on by Proposition 5 the pairwise intersections of their Simson lines are n = p p 3 abc abc p p 3 n = p 3 p abc abc p 3 p n 3 = p p abc abc p p the vertices of the Simson triangle Note that n n 3 = p 3 p abc abc p 3 p = p 3 p abcp p 3 p p p 3 = abc p p p 3 p p p 3 p p 3 p p abc abc p p Since the factork := abc p p p 3 p p p 3 is symmetric inp p p 3 we conclude that N N 3 = k P P 3 N 3 N = k P 3 P N N = k P P and the triangles N N N 3 andp P P 3 are directly similar orollary 7 The Simson triangle N N N 3 has circumradius abc p p p 3 and circumcenter at the midpoint of the segment joining the orthocenters H of andh ofp P P 3 see Figure Proof Since the Simson triangle is similar to P P P 3 with similarity factor k = abc p p p 3 it clearly has circumradius k The orthocenters of triangles andp P P 3 are the points h = abc and h = p p p 3 With these we rewrite n = h h abc p = h h abc p p p 3 p p 3 p p 3 Therefore n h h = abc p p p 3 p p 3 = k since p = p 3 = The same relation holds ifn is replaced byn andn 3 This shows that the Simson triangle has circumcenter h h which is the midpoint of H andh It also confirms independently that the circumradius isk

5 The Simson triangle and its properties 377 l 3 N P P 3 H H O N 3 l N l P Figure 4 The Simson triangle and orthopoles Lemma 8 Let V and W be points on the unit circumcircle of triangle the orthogonal projection of onto VW The perpendicular from to has equation z z bc a v w vw a bcav w vw abc = 0 5 Proof y Lemma 3 the coordinatesa of and its complex conjugate are a = v w a vw a ā = a av wvw avw y Lemma 3 again the coordinates a of the orthogonal projection of onto together with its complex conjugate are a = bca bcā = a bc abcv wa bccaabvw avwv w v w 4avw ā = a bcabcv w a bc ca abvw avwv wv w 4abcvw

6 378 T Zaharinov The line contains a point with coordinates z if and only if z a a 0 = z ā ā where = fabcvwgabcvwz 8a bcv w fabcvw := a bc abcv w a bc ca abvw avwv wv w gabcvwz := abcvw z avwz a bc abcv w a bcvw avwv w v w Therefore the equation of the perpendicular is gabcvwz = 0 Dividing by abcvw we obtain the equation 5 Now we consider the construction in Lemma 8 beginning with all three vertices of This results in the three lines z z bc a v w vw a bcav w vw abc z z b v w b cabv w vw ca vw abc z z ab c v w vw c abcv w vw abc = 0 = 0 = 0 The intersection of the last two lines is given by z ca z b v w c v w ab vw vw b cabv w vw abc b cz abc b c vw c abcv w vw abc b cabcv w abc = 0 = 0 Multiplying by abc b c we obtain z = abcv w abc vw Note that this is symmetric in a b c This means that the three perpendiculars form to to and to are concurrent The point of concurrency is the orthopole N of the line VW y Proposition 5 this is also the same as the intersection of the Simson lines of V and W with respect to see Figure 3

7 The Simson triangle and its properties 379 and [ p89 Theorem 697] pplying this to the three side lines of the triangle P P P 3 for three points P P P 3 we obtain the following theorem l w l v N V W Figure 3 Theorem 9 For three points P P P 3 on the circumcircle of the orthopoles of the lines P P 3 P 3 P P P coincide with the vertices N N N 3 of the Simson triangle bounded by the Simson lines of P P P 3 with respect to 5 Examples Example Let P P P 3 be an equilateral triangle The circumcenter of the Simson triangle coincides with the center of nine-point circle for Proof If P P P 3 is equilateral its orthocenter coincides with the circumcenter This means that p p p 3 = 0 The circumcenter of the Simson triangle is abcp p p 3 = abc the centere of the nine-point circle of Example LetP and letp E meet the circleagain inp 3 E is the center of nine-point circle for Let P and EP P P 3 Then the circumcenter of the Simson triangle lies on the circle with centerh the orthocenter and radius R see Figure 4 Proof LetP 4 be another point of on the linep E P P 3 P P 4 p p 3 p p 4 p p 3 p p 4 = 0 p p 3 p p 4 p p 3 P p 4 = 0

8 380 T Zaharinov P 3 l l N O P 4 E H O N N 3 P P l 3 Figure 4 Thereforep p 3 p p 4 = 0 see [3 p 45] y Lemma 3 or [3 p 45] Therefore E P P 3 p p 3 = ep p 3 ē E P P 4 p p 4 = ep p 4 ē p p p 3 p 4 = ep p 3 p p 4 = e = abc y Theorem 7 the circumcenter O of the Simson triangle of P P P has coordinates o = abcp p p 3 = abc p 4 and o h = p 4 = Hence O lies on a circle with radius R and center H Example 3 Let be points on a circle onstruct the Simson triangle for with respect to and the Simson triangle for with respect to The six vertices of these two Simson triangles lie on a circle see Figure 5 Proof LetN N N 3 be the Simson triangle for with respect to and N N N 3 that of with respect to y Theorem 7 the circumcircles of N N N 3 and N N N 3 both have radius abc a b c and center abca b c Therefore the two circumcircles coincide

9 The Simson triangle and its properties 38 l 3 l N l 3 N 3 N l = l N 3 N N l Figure 5 References [] N ourt ollege Geometry arnes & Noble New York 957 [] R Johnson dvanced Euclidean Geometry New York 960 [3] R Malcheski S Grozdev and nevska Geometry Of omplex Numbers 05 Sofia ulgaria Todor Zaharinov: Darvenitsa jk bl5 vhd ap Sofia ulgaria address: zatrat@abvbg