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A B C P A1 C1 B1 Z July 3, 2020. Math 9+. Geometry Revisited and more. Circles and Triangles Pedal triangles. Exercise. Prove the Viviani’s theorem: the sum of the distances from a point inside an equilateral triangle to its sides equals the altitude of that triangle. Definition. The feet of the perpendiculars from any point 푃 inside a triangle (pedal point) to the three sides of that triangle are the vertices of the pedal triangle. Theorem. If the distances from a pedal point 푃 of a triangle 퐴퐵퐶 to its vertices are |퐴푃| = 푥, |퐵푃| = 푦, and |퐶푃| = 푧 and the sides of the triangle are |퐴퐵| = 푐, |퐵퐶| = 푎, and |퐴퐶| = 푏, then the sides of the pedal triangle are, 푎푥 푏푦 푐푧 퐵 퐶 = , 퐶 퐴 = , and 퐴 퐵 = , B 1 1 2푅 1 1 2푅 1 1 2푅 where 푅 is the circumradius of the y A1 triangle 퐴퐵퐶. P C1 Proof. Consider the circumcircles of z the triangles 퐴퐶1퐵1, 퐶1퐵퐴1and x 퐵1퐴1퐶. Using the extended sine C 퐵 퐶 푎 theorem, 1 1 = |퐴푃| and = 2푅, sin 퐴 sin 퐴 A B1 |퐴푃| so 퐵 퐶 = 푎 . Similarly, for the 1 1 2푅 other sides. Theorem. The third pedal triangle is similar to the original triangle. Proof. Consider the figure. Right triangles A 퐵퐶1푃 and 퐵퐴1푃 share a common circumcircle with diameter 퐵푃, which is the same as the circumcircle of the inscribed rectangle, 푃퐶1퐵퐴1. Hence, ∠퐴1퐵푃 = ∠퐴1퐶1푃 and ∠퐶1퐵푃 = ∠퐶1퐴1푃. Considering now the B circumcircle of the rectangle 푃퐵2퐴1퐴2 we 1 similarly conclude that ∠퐶 퐴 푃 = ∠퐵 퐴 푃 = 1 1 2 1 B3 A2 C 2 C ∠퐵2퐶2푃. By a similar reasoning, ∠퐴1퐶1푃 = C P 3 ∠퐵2퐴2푃. We observe that angle 퐴퐵퐶 is split in A3 C1 two angles, on of which “travels” clockwise in B2 A1 each subsequent pedal triangle, while the other “travels” counter-clockwise. Considering the circumcircle of the rectangle B 푃퐵3퐶2퐴3, we conclude that B ∠퐵2퐶2푃 = ∠퐴3퐵3푃 and, similarly, ∠퐵 퐴 푃 = ∠퐶 퐵 푃. 2 2 3 3 B2 A1 Hence, angle ∠퐴3퐵3퐶3 = A3 C3 C1 ∠퐴3퐵3푃 ∪ ∠퐶3퐵3푃 consists of the C2 P B3 same two angles as ∠퐴퐵퐶, which proves that ∠퐴3퐵3퐶3 = ∠퐴퐵퐶. A2 The equality of two other corresponding angles is proven C in a similar manner. This proves B1 A A that the first and the third pedal 1 B triangles are similar, ∆퐴3퐵3퐶3~∆퐴퐵퐶. Ha The orthocenter and the circumcenter. C1 O Exercise. Prove that if 퐻 is an orthocenter of Hc H triangle 퐴퐵퐶, then 퐴 is the orthocenter of ∆퐻퐵퐶, C A Hb B1 퐵 is the orthocenter of ∆퐻퐶퐴, and 퐶 is the orthocenter of ∆퐻퐴퐵. Exercise. In a triangle 퐴퐵퐶, altitudes, 퐴퐻퐴, 퐵퐻퐵, and 퐶퐻퐶 are extended to meet the circumcircle at points 퐴1, 퐵1 and 퐶1, respectively. Prove that triangle 퐴1퐵1퐶1 is similar to the orthic triangle, ∆퐴1퐵1퐶1~∆퐻퐴퐻퐵퐻퐶. N B A 1 H a AO B M Ha a Mc O Hc C1 O H Hc C H M C H b A b A Hb B1 Exercise. Prove that for a triangle 퐴퐵퐶 with sides 푎, 푏 and 푐, the length of the 푏푐 altitude from the vertex 퐴 (to the side of length 푎), ℎ , equals, ℎ = , where 푎 푎 2푅 푅 is the circumradius (the radius of the circumscribed circle). Exercise. Prove that for a triangle 퐴퐵퐶, ∠퐻퐴퐴퐴푂 = |∠퐵 − ∠퐶|, where 퐴퐻퐴 is the altitude and 퐴퐴푂 is the diameter of the circumcircle (see figure; consider cases ∠퐵 − ∠퐶 > 0 and ∠퐵 − ∠퐶 < 0 separately). Definition. The power of a point 푃 with respect to a given circle is the product, |푃퐴| × |푃퐵|, where 퐴퐵 is the chord of the circle on the line 퐴퐵 passing through point 푃 (why is this product the same for all lines passing through point 푃?). Theorem. If 퐻 is an orthocenter of a triangle 퐴퐵퐶 with altitudes, 퐴퐻퐴, 퐵퐻퐵, and 퐶퐻퐶, then |퐴퐻| × |퐻퐻퐴| = |퐵퐻| × |퐻퐻퐵| = |퐶퐻| × |퐻퐻퐶|. Proof. Consider the above figure. Find the congruent right triangles and use the theorem on chords in a circle to obtain the needed set of equalities. Definition. The radical axis of the two circles is the locus of points having the same power of the point with respect to both circles. Observation. If a circle is constructed on a cevian 퐴퐴1 of a triangle 퐴퐵퐶 as diameter, it passes through a foot of the corresponding altitude, 퐻퐻퐴. B Theorem. If circles are constructed on two Ha cevians as diameters, their radical axis passes A through an orthocenter 퐻 of the triangle. 1 Hc Proof. Consider the figure on the right and use H C the theorem given above. B1 A Hb Theorem. For any three non-coaxial circles having cevians for diameters, the orthocenter, 퐻, is their radical center. Exercise. Prove that altitudes in a triangle are concurrent (Hint: in the above theorems, consider cevians coincident with the sides of the triangle, i.e. consider circles built on sides of a triangle for diameters). 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.
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