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

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

2 21 Example 11: Three congruent circles in a circle. The three small circles are congruent. If the radius of the large circle is R, calculate the common radius is of the small circles. P O

3 22 Solution to Example 11: Three congruent circles in a circle. The three small circles are congruent. If the radius of the large circle is R, calculate the common radius is of the small circles. P M O T Q Solution. Let r be the common radius of the congruent circles. In the right triangle OT, O = R r, T = r, OT = MT OM = MT (OP MP)=r (R 2r) =3r R. y the Pythagorean theorem, (R r) 2 = r 2 +(3r R) 2, R 2 2Rr + r 2 = r 2 +9r 2 6Rr + R 2, 4Rr =9r 2, r = 4R 9.

4 Example 12. The large circle has radius R. alculate the common radius of the three congruent circles are congruent. 23

5 24 Solution to Example 12. The large circle has radius R. alculate the common radius of the three congruent circles are congruent. X Y T R 2ρ O ρ Solution. Let ρ be the common radius of the small circles. Since OT and X are parallel, R 2ρ R = ρ 2R ρ, (R 2ρ)(2R ρ) = Rρ, 2ρ 2 6Rρ +2R 2 = 0, ρ 2 3Rρ + R 2 = 0, ρ = 3 5 R. 2

6 Example 12: Geogebra exercise. onstruct the following diagram in which the three small circles are congruent. 25

7 26 Solution to Example 12: Geogebra exercise. onstruct the following diagram in which the three small circles are congruent. O D Solution. Since the common radius of the small circles is ρ = R, OD = R 2ρ = R (3 5)R =( 5 2)R. P E O D Let OP be a radius perpendicular to the diameter, and E the reflection of the center O in. onstruct the circle E(P ) to intersect at D. The midpoint of D is the center of one of the small circles. Remark. divides O in the golden ratio.

8 The law of sines Theorem (The law of sines). Let R denote the circumradius of a triangle. 2R = a sin α = b sin β = c sin γ. α F O E O α D D Since the area of a triangle is given by Δ= 1 2 bc sin α, the circumradius can be written as R = abc 4Δ.

9 2 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

10 3 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 = sin P = 2 sin 3 sin 3 = sin P = 1 2. Therefore, P =30. Since P =90 3 and P = P,wehave ( + 90 ) +2 P +30 = From this, 3 + P =30. Exercise Find the angles of triangle if it is isosceles. onsider all possibilities.

11 4 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

12 5 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 3 + θ =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.

13 8 The orthocenter Why are the three altitudes of a triangle concurrent? Let be a given triangle. Through each vertex of the triangle we construct a line parallel to its opposite side. These three parallel lines bound a larger triangle. Note that and are both parallelograms since each has two pairs of parallel sides. It follows that = = and is the midpoint of. Y Z H X onsider the altitude X of triangle. Seen in triangle, this line is the perpendicular bisector of since it is perpendicular to through its midpoint. Similarly, the altitudes Y and Z of triangle are perpendicular bisectors of and. s such, the three lines X, Y, Z concur at a point H. This is called the orthocenter of triangle.

14 9 The orthocenter The superior triangle is similar to. The ratio of similarity is 2. If D is the midpoint of, then H =2 OD. O H D

15 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. 9

16 10 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. H 11 O 5 X D

17 11 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

18 12 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

19 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. 13 H O O H O H

20 The law of cosines Given a triangle, we denote by a, b, c the lengths of the sides,, respectively. Theorem (The law of cosines). c 2 = a 2 + b 2 2ab cos γ. c b c b X a a X Proof. Let X be the altitude on. c 2 = X 2 + X 2 =(a bcos γ) 2 +(bsin γ) 2 = a 2 2ab cos γ + b 2 (cos 2 γ +sin 2 γ) = a 2 + b 2 2ab cos γ.

21 18 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

22 Example 8: (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

23 20 Solution to Example 8: (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 = = = 1 2 = = 120. (b) Since + Z = 180,,,, Z are concyclic. y Ptolemy s theorem, Z + Z = Z = Z = + =5+3=8.

24 21 Example 9. is a triangle with = 7, = 7+1, and = 7 1. Show that α =90 + β and find γ nswer. cos α = 1 4 (1 7), cos β = 1 4 (1 + 7), cos γ = 3 4. 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) = 7 1) ( 7 1) = ; 4 = ( 7) 2 +( 7+1) 2 ( 7 1) 2 2 7( 7+1) 7( 7 4) 2 7( 7 1) = 7+(8+2 7) (8 2 7) 2 7( = ( 7+4) 7+1) 2 7( 7+1) = 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) Note that = ( 7+1) 2 +( 7 1) 2 ( 7) 2 2( 7+1)( 7 1) = 9 12 = 3 4. 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.

25 24 Napoleon s theorem If similar isosceles triangles X, Yand Z (of base angle θ) are constructed externally on the sides of triangle, the lengths of the segments YZ, ZX, XZ can be computed easily. For example, in triangle Y Z, Y = b 2 sec θ, Z = c 2 sec θ and YZ = α +2θ. θ θ Y Z θ θ θ θ X y the law of cosines, YZ 2 = Y 2 + Z 2 2Y Z cos YZ = sec2 θ (b 2 + c 2 2bc cos(α +2θ)) 4 = sec2 θ (b 2 + c 2 2bc cos α cos 2θ +2bc sin α sin 2θ) 4 = sec2 θ (b 2 + c 2 (b 2 + c 2 a 2 )cos2θ +4Δsin2θ) 4 = sec2 θ (a 2 cos 2θ +(b 2 + c 2 )(1 cos 2θ)+4Δsin2θ). 4 Likewise, we have ZX 2 = sec2 θ (b 2 cos 2θ +(c 2 + a 2 )(1 cos 2θ)+4Δsin2θ), 4 XY 2 = sec2 θ (c 2 cos 2θ +(a 2 + b 2 )(1 cos 2θ)+4Δsin2θ). 4 It is easy to note that YZ = ZX = XY if and only if cos 2θ = 1 2, i.e., θ =30. In this case, the points X, Y, Z are the centers of equilateral triangles erected externally on,, respectively. The same conclusion holds if the equilateral triangles are constructed internally on the sides. This is the famous Napoleon theorem.

26 Theorem (Napoleon). If equilateral triangles are constructed on the sides of a triangle, either all externally or all internally, then their centers are the vertices of an equilateral triangle. 25 Y Z Z X Y X

27 28 pollonius Theorem Theorem. Given triangle, let D be the midpoint of. The length of the median D is given by =2(D 2 + D 2 ). D Proof. pplying the law of cosines to triangles D and D, and noting that cos D = cos D,wehave 2 = D 2 + D 2 2D D cos D; 2 = D 2 + D 2 2D D cos D, 2 = D 2 + D 2 +2D D cos D. The result follows by adding the first and the third lines. If m a denotes the length of the median on the side, m 2 a = 1 4 (2b2 +2c 2 a 2 ).

28 29 Example 12. is a triangle with a =8, b =9, c =11. Two of its medians have rational lengths. What are these? 8 D 9 E F mb = 17 2 ; m c = 13 2.

29 30 Example 13. The lengths of the sides of a triangle are 136, 170, and 174. alculate the lengths of its medians.

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 1 Example 1(a). Given a triangle, the intersection P of the perpendicular bisector of and