Geometry. Class Examples (July 1) Paul Yiu. Department of Mathematics Florida Atlantic University. Summer 2014

Size: px

Start display at page:

Download "Geometry. Class Examples (July 1) Paul Yiu. Department of Mathematics Florida Atlantic University. Summer 2014"

Erick Clifford Park
5 years ago
Views:

1 Geometry lass Examples (July 1) Paul Yiu Department of Mathematics Florida tlantic University c b a Summer 2014

2 1 Example 1(a). Given a triangle, the intersection P of the perpendicular bisector of and a trisector of angle is such that P =. Show that 3 + P =30. M P D

3 2 Solution to Example 1(a). Given a triangle, the intersection P of the perpendicular bisector of and a trisector of angle is such that P =. Show that 3 + P =30. M P D Solution. Let M be the midpoint of P. M is the bisector of angle P, hence a trisector of angle. M = sin 3. In triangle P, P = P =2 M =2 sin 3. y the law of sines, P sin P = P sin 3 Therefore, P =30. = sin P = 2 sin 3 sin 3 = sin P = 1 2. Since P =90 and P = P,wehave 3 ( + 90 ) +2 P +30 = From this, 3 + P =30. Exercise Find the angles of triangle if it is isosceles. onsider all possibilities.

4 3 Example 2. Given an isosceles triangle, the intersection P of the perpendicular bisector of and a trisector of angle is such that P =. Find the angles of triangle. M P D

5 4 Solution to Example 2. Given an isosceles triangle, the intersection P of the perpendicular bisector of and a trisector of angle is such that P =. Find the angles of triangle. P P nswer. (,, ) =(45, 90, 45 ) or (72, 72, 36 ). Solution. Let P = P = θ. We have known that + θ 3 =30. (i) Suppose =. In this case, = = P = P = 90 3 =30. From this, 3 =60, an impossibility. (ii) Suppose =. In this case, =30 + θ = = =45. Therefore, = =45 and =90. (iii) Suppose =. In this case, = θ = = 5 3 = 120 = =72. Therefore, = =72 and =36.

6 5 Example 3. coincide. triangle is equilateral if and only if its circumcenter and centroid

7 6 Example 4. Let H be the orthocenter of triangle. Show that (i) is the orthocenter of triangle H; (ii) the triangles H, H, H and have the same circumradius.

8 Example 5. In triangle with circumcenter O, orthocenter H, midpoint D of, and perpendicular foot X of on, OHXD is a rectangle of dimensions alculate the length of the side. 7 H 11 O 5 X D

9 8 Example 6. Given triangle with circumcenter O and orthocenter H, prove that the altitude H and the circumradius O are equally inclined to the sides and. H O

10 9 Solution to Example 6. Given triangle with circumcenter O and orthocenter H, prove that the altitude H and the circumradius O are equally inclined to the sides and. H O

11 10 Example 6. In triangle, one pair of trisectors of the angles and meet at the orthocenter. (a) Show that the other pair of trisectors of these angles meet at the circumcenter. (b) Find all possible values of the angles of the triangle. H O O H O H

12 11 Example 7. is a triangle with a =9, b =11, c =12. Z is a point on such that Z =9and Z =3. alculate the length of Z Z

13 12 Example 8: The law of cosines (proof without words).

14 Example 9: (3, 5, 7)-triangle. is a triangle with = 3, = 5 and =7. (a) Show that = 120. (b) n equilateral triangle Z is constructed externally on the side. alculate the length of the segment Z Z

15 14 Solution to Example 10: (3, 5, 7)-triangle. is a triangle with =3, =5and =7. (a) Show that = 120. (b) n equilateral triangle Z is constructed externally on the side. alculate the length of the segment Z Z Solution. (a) cos = = 15 = 1 = = (b) Since + Z = 180,,,, Z are concyclic. y Ptolemy s theorem, Z + Z = Z = Z = + =5+3=8.

16 15 Example 11. is a triangle with = 7, = 7+1, and = 7 1. Show that α =90 + β and find γ nswer. cos α = 1(1 7), cos β = 1(1 + 7), cos γ = Solution. cos α = b2 + c 2 a 2 2bc = (8 2 7) + 7 ( ) 2 7( 7 1) = (8 2 7) 4( 7 1) cos β = c2 + a 2 b 2 2ca = ( 7 1) 2 +( 7) 2 ( 7+1) 2 2( 7 1) 7 = ( 7 1) 2 4( 7 1) = = ( 7 1) = 7 1 ; 4 = ( 7) 2 +( 7+1) 2 ( 7 1) 2 2 7( 7+1) = 7+(8+2 7) (8 2 7) 2 7( = ) 2 7( 7+1) = = ( 7+1) = ( 7+1) ( 7+1) = ; 4 cos γ = a2 + b 2 c 2 2ab = ( ) + (8 2 7) 7 2(7 1) = ( 7+1) 2 +( 7 1) 2 ( 7) 2 2( 7+1)( 7 1) = 9 12 = ( 7 4) 2 7( 7 1) 7( 7+4) 2 7( 7+1) Note that cos 2 α +cos 2 β = ( 7 1) 2 +( 7+1) 2 = (8 2 7) + ( ) = Since cos α<0,wehavecos α = sin β = cos(90 + β). It follows that α =90 + β.

17 16 Example :. is an equilateral triangle inscribed in a circle. P is a point on the minor arc. Show that P 2 + P 2 + P 2 does not depend on the choice of P. O P

18 17 Solution to Example :. is an equilateral triangle inscribed in a circle. P is a point on the minor arc. Show that P 2 + P 2 + P 2 does not depend on the choice of P. O P Solution. Note that P and P are each 60. pplying the law of cosines to triangles P and P,wehave From these, 2 = P 2 + P 2 P P, 2 = P 2 + P 2 P P. P 2 +P 2 +P 2 = P (P+P) P 2 =2 2 +P (P P P )=2 2.

19 18 Example 12(a): Fermat point. Given triangle, construct externally similar isosceles triangles X, Y, and Z with base angles θ. Z Y θ θ (a) pplying the law of cosines to triangle X, X X 2 = 2 + X 2 2 X cos X ( a ) 2 = c 2 ca 2cosθ cos θ cos( +60 ) = c 2 + a2 4cos 2 θ ca(cos cos 60 sin sin 60 ) cos θ 1 ( = 4c 2 cos 2 θ + a 2 2ca cos cos θ +4 ) 3Δ cos θ 4cos 2 θ 1 ( = 4c 2 cos 2 θ + a 2 (c 2 + a 2 b 2 )cosθ +4 ) 3Δ cos θ 4cos 2 θ 1 ( = a 2 (1 cos θ)+b 2 cos θ + c 2 (4 cos 2 θ cos θ)+4 ) 3Δ cos θ. 4cos 2 θ (b) This expression is symmetric in a, b, c if and only if cos θ = 1 2. With this choice of θ,wehave X 2 = 1 2 (a2 + b 2 + c Δ). This means that if equilateral triangles X, Y, and Z are constructed externally of triangle, the segments X, Y, Z have equal lengths.

20 19 Example 12(b): The Fermat point. Given triangle, construct equilateral triangles X, Y, and Z externally on the sides. Let the circumcircle of X intersect the line X at F. Y Z F X (a) Prove that FX = XF =60. (b) Prove that, F,, Y are concyclic. Similarly,, F,, Z are also concyclic. (c) Prove that, F, Y are collinear. Similarly,, F, Z are also collinear. The point F is called the Fermat point of triangle. It is the point of concurrency of the lines X, Y, Z. It is also the common point of the circumcircles of the equilateral triangles X, Y, Z.

21 20 Example 13. is a triangle with a =8, b =9, c =11. Two of its medians have rational lengths. What are these? 8 D 9 E 1 11 F 1 mb = 17 2 ; m c = 13 2.

22 21 Example 14(a). In the diagram below, show that S 1 + T 1 =2(S 2 + S 3 ). T 1 S 3 S 2 S 1

23 22 Example 14(b). In the diagram below, show that T 1 + T 2 + T 3 =3(S 1 + S 2 + S 3 ). T 1 S 3 S 2 T 3 S 1 T 2

24 23 Example 15. The lengths of the sides of a triangle are 136, 170, and 174. alculate the lengths of its medians.

25 24 Example 16: Triangles with perpendicular medians. Suppose the medians E and F of triangle are perpendicular. This means that G 2 + G 2 = 2, where G is the centroid of the triangle. In terms of the lengths, we have 4 9 m2 b m2 c = a 2 ; 4(m 2 b + m2 c)=9a 2 ; (2c 2 +2a 2 b 2 )+(2a 2 +2b 2 c 2 )=9a 2 ; b 2 + c 2 =5a 2. This relation is enough to describe, given points and, the locus of for which the medians E and F of triangle are perpendicular. Here, however, is a very easy construction: From b 2 + c 2 =5a 2,wehavem 2 a = 1 4 (2b2 +2c 2 a 2 )= 9 4 a2 ; m a = 3 a. The 2 locus of is the circle with center at the midpoint of, and radius 3. 2

26 25 Example 17: utomedian triangles. triangle (a, b, c) is automedian if the squares of the sides are in arithmetic progression. Let be an automedian triangle (satisfying a 2 + c 2 =2b 2 ). Prove that the points, F, G, D are concyclic. F D G E Proof. G D = 2 3 m a m a = 2 3 m2 a = (2b2 +2c 2 a 2 )= 1 6 (a2 + c 2 +2c 2 a 2 ) = 1 2 c2 = 1 2 c c = F. y (the converse of) the intersecting chords theorem,, F, G, D are concyclic.

27 26 Example 18: utomedian triangles. Let be an automedian triangle (satisfying a 2 + c 2 =2b 2 ). The median E is extended to intersect the circumcircle at P. Prove that the centroid G is the midpoint of P. F D G O E P Proof. Since is automedian with a 2 + c 2 =2b 2, m b = 3 2 chord theorem, b. y the intersecting E EP = E E = m b EP = b2 4 = 1 3 m2 b = EP = 1 3 m b. Therefore, GP = GE + EP = m b G is the midpoint of P. 3 + m b 3 = 2 3 m b = G.

Geometry. Class Examples (July 1) Paul Yiu. Department of Mathematics Florida Atlantic University. Summer 2014

Geometry. Class Examples (July 1) Paul Yiu. Department of Mathematics Florida Atlantic University. Summer 2014 Geometry lass Examples (July 1) Paul Yiu Department of Mathematics Florida tlantic University c b a Summer 2014 21 Example 11: Three congruent circles in a circle. The three small circles are congruent.