July 10, 2020. Math 9+. Geometry Revisited and more. Collinearity and Concurrence Degenerate pedal triangle. Simson lines. Definition. The feet of the perpendiculars from any C1 point 푃 of the plane, inside or outside a triangle (pedal P point) to the three sides of that triangle are the vertices of the pedal triangle. Note: this includes a special case of 푃 on a circumcircle considered below, which corresponds to a degenerate triangle. B Theorem. If the distances from a pedal point 푃 of a triangle 퐴퐵퐶 to its vertices are |퐴푃| = 푥, |퐵푃| = 푦, A1 and |퐶푃| = 푧 and the sides of the triangle are O |퐴퐵| = 푐, |퐵퐶| = 푎, and |퐴퐶| = 푏, then the sides of B the pedal triangle are, C 1 A 푎푥 푏푦 푐푧 |퐵 퐶 | = , |퐶 퐴 | = , and |퐴 퐵 | = , 1 1 2푅 1 1 2푅 1 1 2푅 where 푅 is the circumradius of the triangle 퐴퐵퐶. Exercise. Prove the above theorem for P configurations of pedal triangle shown in the figures, with pedal point outside of triangle 퐴퐵퐶. Is C 1 O there any difference between the cases of obtuse B A1 and acute triangle 퐴퐵퐶? C B1 A Proof. Consider the circumcircles of the triangles 퐴퐶1퐵1, 퐶1퐵퐴1and 퐵1퐴1퐶. Using the extended sine |퐵 퐶 | 푎 |퐴푃| theorem for ∆퐴퐶 퐵 , 1 1 = |퐴푃| and = 2푅, so |퐵 퐶 | = 푎 . For 1 1 sin 퐴 sin 퐴 1 1 2푅 |퐴 퐶 | 푏 |퐵푃| ∆퐶 퐵퐴 , 1 1 = |퐵푃|; using = 2푅, |퐴 퐶 | = 푏 . 1 1 sin(180−퐵) sin 퐵 1 1 2푅 Theorem. The feet of the perpendiculars from a point to the sides of a triangle are collinear if and only if the point lies on the circumcircle. The line they lie on is called Simson line. P C1 P C1 B A1 O B O A 1 C C B1 B1 A A Proof. Consider the figure. Note that ∠퐶1푃퐵1 = ∠퐵푃퐶 = 180° − ∠퐵퐴퐶, wherefrom follows that ∠퐵푃퐶1 = ∠퐵1푃퐶. Furthermore, ∠퐵퐴1퐶1 = ∠퐵푃퐶1 and ∠퐵1퐴1퐶 = ∠퐵1푃퐶 by the inscribed angle theorem. The collinearity of 퐶1, 퐴1, and 퐵1then follows from the congruence of angles ∠퐵퐴1퐶1 and ∠퐵1퐴1퐶. Exercise. Does the proof of the above theorem require any modification when Δ 퐴퐵퐶 is obtuse? Exercise. Is there a point on the circle that has side 퐵퐶 as its Simson line? Theorem. If the perpendicular 푃퐵1 from a pedal point 푃 to the side 퐴퐶 of ∆퐴퐵퐶 is extended to meet the circumcircle at point 퐵2, then 퐵퐵2 is parallel to the Simson line, 퐶1퐵1. C1 P Proof. Consider the figure on the right. By the B inscribed angle theorem, ∠퐵퐵2푃 = ∠퐵퐶푃 = A1 ∠푃퐵1퐴1, so that 퐶1퐵1 and 퐵퐵2 make equal O angles with 푃퐵2 and are therefore parallel. C B1 A B2 C Theorem. The angle between the Simson lines of 1 C’ P the two points, 푃 and 푃′ on the circumcircle of 1 B P’ ∆퐴퐵퐶 is half the angular measure of the arc 푃푃′. A1 Proof. Consider the figure. Note that the arcs 푃푃′ O A’ 1 and 퐵2퐵′2 are equal because 푃퐵2 and 푃′퐵′2 are C B’1 parallel. Using the above theorem, we conclude B1 B’2 that the angle between the Simson lines of points, A 푃 and 푃′ equals the angle ∠퐵2퐵퐵′2 and therefore B2 half the angular measure of the arc 푃푃′. Exercise. Are there points that lie on their own Simson lines? What are these points? Exercise. Prove the following theorem. Theorem. The Simson line of a points, 푃 on the circumcircle of ∆퐴퐵퐶 bisects the segment joining that point to the orthocenter of ∆퐴퐵퐶. Ptolemy’s theorem D Theorem. If a quadrilateral 퐴퐵퐶퐷 is inscribed in a B circle, the sum of the products of the two pairs of opposite sides is equal to the product of diagonals, O |퐴퐵| ∙ |퐶퐷| + |퐵퐶| ∙ |퐴퐷| = |퐴퐶| ∙ |퐵퐷| C Proof. Apply theorem expressing the length of the A sides if a pedal triangle in the degenerate case when pedal triangle is a Simson line. Theorem. If 퐴퐵퐶퐷 is not an inscribed quadrilateral, the sum of the products of the two pairs of opposite sides is larger than the product of diagonals, |퐴퐵| ∙ |퐶퐷| + |퐵퐶| ∙ |퐴퐷| > |퐴퐶| ∙ |퐵퐷| Proof. Apply theorem expressing the length of the sides if a pedal triangle in the case when pedal triangle is not a Simson line and then use a triangle inequality. C Exercise. Prove that if a circle cuts two sides and a diagonal of a parallelogram D ABCD at points P, Q, R, as shown in the Q figure, then, R |퐴푃| ∙ |퐴퐵| + |퐴푅| ∙ |퐴퐷| = |퐴푄| ∙ |퐴퐶| O P B A The nine-points circle problem. Theorem. The feet of the three altitudes of any triangle, the midpoints of the three sides, and the midpoints of the segments from the three vertices lo the orthocenter, all lie on the same circle, of radius ½푅. This theorem is usually credited to a German geometer Karl Wilhelm von Feuerbach, who actually rediscovered the theorem. The first complete proof appears to be that of Jean-Victor Poncelet, published in 1821, and Charles Brianson also claimed proving the same theorem prior to Feuerbach. The theorem also sometimes mistakenly attributed to Euler, who proved, as early as 1765, that the orthic triangle and the medial triangle have the same circumcircle, which is why this B circle is sometimes called "the H C H Euler circle". Feuerbach A H rediscovered Euler's partial result MC even later, and added a further MA property which is so remarkable N that it has induced many authors A O to call the nine-point circle "the Feuerbach circle". HB MB C Proof. Consider rectangles formed by the mid-lines of triangle ABC and of triangles 퐴퐵퐻, 퐵퐶퐻 and 퐴퐶퐻. Theorem. The orthocenter, 퐻, centroid, 푀, and the circumcenter, 푂, of any triangle are collinear: all these three points lie on the same line, 푂퐻, which is called the Euler line of the triangle. The orthocenter divides the distance from the centroid to the circumcenter in 2: 1 ratio. B HC Proof. Note that the altitudes of the medial HA H triangle푀퐴푀퐵푀퐶 are the perpendicular MC bisectors of the triangle 퐴퐵퐶, so the M orthocenter of Δ 푀 푀 푀 is the M A 퐴 퐵 퐶 A circumcenter, 푂, of Δ 퐴퐵퐶. Now, using the O property that centroid divides medians of a HB MB triangle in a 2:1 ratio, we note that triangles C 퐵푀퐻 and 푀퐵푀푂 are similar, and homothetic with respect to point 푀, with the homothety coefficient 2. Theorem. The center of the nine-point-circle lies on the (Euler’s) line passing through orthocenter, centroid, and circumcenter, midway between the orthocenter and the circumcenter. B HC B1 Proof. Consider the figure. Note the colored HA triangle 퐴1퐵1퐶1, which is formed by MC H medians of triangles 퐴퐵퐻, 퐵퐻퐶 and 퐶퐻퐴, A O9 1 M and is therefore congruent to the medial M A A C1 triangle 푀퐴푀퐵푀퐶, but rotated 180 degrees. O The 9 points circle is the circumcircle for HB MB both triangles, which means that rotation C by 180 degrees about the center 푂9 of the 9 point circle moves Δ 푀퐴푀퐵푀퐶 onto Δ 퐴1퐵1퐶1, and the orthocenter, 푂, of the Δ 푀퐴푀퐵푀퐶 onto the orthocenter, 퐻, of the Δ 퐴1퐵1퐶1. More problems. Problem. Rectangle DEFG is inscribed in triangle ABC such that the side DE belongs to the base AB of the triangle, while points F and G belong to sides BC abd CA, respectively. What is the largest area of rectangle DEFG? C Solution. Notice similar triangles, 퐶퐷퐸~퐴퐵퐶, (a) wherefrom the vertical side of the rectangle is, x|CH| |퐷퐸| D H’ E |퐷퐺| = |퐸퐹| = |퐶퐻| − |퐶퐻′| = (1 − ) |퐶퐻|, |퐴퐵| (1-x)|CH| A B so that the area of the rectangle is, 푆퐷퐸퐹퐺 = G G’ H F |퐷퐸| S +S +S = S +S = x2 S +(1-x)2S >= ½ S |퐷퐸||퐷퐺| = |퐷퐸| (1 − ) |퐶퐻| = AGD EFB DEC AG’D DEC AB C ABC ABC |퐴퐵| x22 +(1-x) = 1-x+2x22 = ½+2(x-½) ) >= ½ |퐷퐸| |퐷퐸| |퐷퐸| |퐷퐸| (1 − ) |퐴퐵||퐶퐻| = (1 − ) 2푆퐴퐵퐶. |퐴퐵| |퐴퐵| |퐴퐵| |퐴퐵| (b) (c) Using the geometric-arithmetic mean C C |퐷퐸| |퐷퐸| 2 |퐷퐸| |퐷퐸| +1− 1 inequality, (1 − ) ≤ (|퐴퐵| |퐴퐵|) = , |퐴퐵| |퐴퐵| 2 4 D E D E where the largest value of the left side is A B A B |퐷퐸| |퐷퐸| G C’ F G F achieved when = 1 − , and therefore C’ |퐴퐵| |퐴퐵| DC’ || CB, EC’ || AC, SDEC = S DEC’ 1 S +S +S = sum of the areas of shaded triangles >= ½S 푆 = 푆 . There are a number of other AGD EFB DEC ABC 퐷퐸퐹퐺 2 퐴퐵퐶 possible solutions, some of which are shown in the figures. Problem. Prove that for any triangle 퐴퐵퐶 with sides 푎, 푏 and 푐, the area, 푆 ≤ 1 (푏2 + 푐2). 4 C’ Solution. Notice that of all triangles with given C two sides, 푏 and 푐, the largest area has triangle 퐴퐵퐶′, where the sides with the given lengths, b a |퐴퐵| = 푐 and |퐴퐶| = 푏 form a right angle, b 퐵퐴퐶̂ = 90° (푏 is the largest possible altitude to side 푐). Therefore, ∀∆퐴퐵퐶, 푆퐴퐵퐶 ≤ 푆퐴퐵퐶′ = A c B 1 1 푏2+푐2 푏푐 ≤ , where the last inequality follows from the arithmetic-geometric 2 2 2 푏2+푐2 mean inequality, 푏푐 ≤ (or, alternatively, follows from 푏2 + 푐2 − 2푏푐 = 2 (푏 − 푐)2 ≥ 0. Problem. In an isosceles triangle 퐴퐵퐶 with the side |퐴퐵| = |퐵퐶| = 푏, the segment |퐴′퐶′| = 푚 connects the intersection points of the bisectors, 퐴퐴′ and 퐶퐶′ of the angles at the base, 퐴퐶, with the corresponding opposite sides, 퐴′ ∈ 퐵퐶 and 퐶′ ∈ 퐴퐵.