Geometry lass Examples (July 8) Paul Yiu Department of Mathematics Florida tlantic University c b a Summer 2014
1 The incircle The internal angle bisectors of a triangle are concurrent at the incenter of the triangle. This is the center of the incircle, the circle tangent to the three sides of the triangle. Let the bisectors of angles and intersect at I. onsider the pedals of I on the three sides. Since I is on the bisector of angle, IX = IZ. Since I is also on the bisector of angle, IX = IY. It follows IX = IY = IZ, and the circle, center I, constructed through X, also passes through Y and Z, and is tangent to the three sides of the triangle. s a s a Y Y Z I Z I s c s b X s b X s c This is called the incircle of triangle, and I the incenter. Let s be the semiperimeter of triangle. The incircle of triangle touches its sides,, at X, Y, Z such that Y =Z = s a, Z =X = s b, X =Y = s c. The inradius of triangle is the radius of its incircle. It is given by r = 2Δ a + b + c = Δ s.
2 Example E1. If triangle has a right angle at, then the inradius r = s c. s b s b r I r s a s c r s c s a It follows that if d is the diameter of the incircle, then a + b = c + d.
Example E1. square of side c is partitioned into 4 congruent right triangles and a small square, all with equal diameter d. alculate d. 3
4 Solution to Example E1. square of side c is partitioned into 4 congruent right triangles and a small square, all with equal diameter d. alculate d. c a b Solution. The diameter of each circle inscribed in a right triangle = a+b c. Diameter of circle at the center = b a. a + b c = b a = c =2a. a = c 2, b = 3 2 c, and d = b a = ( 3 1) 2 c.
Solution to Example E1: onstruction. 5
6 Example E2. The incenter of a right triangle is equidistant from the midpoint of the hypotenuse and the vertex of the right angle. Show that the triangle contains a 30 angle. D I E M F
7 Solution to Example E2. The incenter of a right triangle is equidistant from the midpoint of the hypotenuse and the vertex of the right angle. Show that the triangle contains a 30 angle. D I E Solution. M be the midpoint of, and D, E, F the points of tangency of the incircle with the sides. If I = IM, then the triangles EI and FMI are congruent. E = MF = s c = c 2 (s a), a+b c 2 = a b 2, c =2b. From this, =30. M F (RHS)
8 Example E3. is an isosceles triangle with = =25and =14. Y is a point on such that Y =, and X is the midpoint of Y. alculate the inradii of the triangles X, XY, and Y. 11 25 Y X 14 14
9 Solution to Example E3. is an isosceles triangle with = =25and =14. Y is a point on such that Y =, and X is the midpoint of Y. alculate the inradii of the triangles X, XY, and Y. 11 25 Y X 14 y Stewart s theorem, Y = 84 5. Each of the triangles X and Y X is 14 5 Their inradii are each equal to 14 Now, triangle Y is 1 5 times a (84, 55, 125) triangle. This has inradius 1 5 1 2 The three incircles are congruent. 5. 14 times a (3, 4, 5) triangle. ( 84+55+125)(84 55+125)(84+55 125) 84+55+125 = 14 5.
10 Example E4. The medians E and F of triangle intersect at the centroid G. If the inradii of triangles GF and GE are equal, prove that the triangle is isosceles. F E I G J P Q
11 Solution to Example E4. The medians E and F of triangle intersect at the centroid G. If the inradii of triangles GF and GE are equal, prove that the triangle is isosceles. F E I G J P Q Proof. The triangles GF and GE have equal areas. If they have equal inradii, then they have the same semiperimeter s. Now, if I and J are the incenters of the triangles, and P, Q points their points of tangency with the medians E and F, then triangles GIP and GJQ are congruent. It follows that GP = GQ. Since GP = s c 2 and GQ = s b 2, it is clear that b = c, and triangle is isosceles.
12 Example E5: (5, 6, 7) triangle. is a triangle with =5, =6and =7. The incircle touches at X. Prove that X =5. 5 7 Y X 6
13 Solution to Example E5: (5, 6, 7) triangle. is a triangle with =5, =6and =7. The incircle touches at X. Prove that X =5. 5 7 Y X 6 Solution. Let Y with Y on. If Y = y, then Y = 6 y. y the Pythagorean theorem, 5 2 y 2 =7 2 (6 y) 2 = y =1. Since X = 1 2 (5 + 6 7) = 2, Y is the perpendicular bisector of X, and X = =5.
14 Example E6: (5, 6, 7) triangle. is a triangle with =5, =6and =7. The incircle touches at X. Find the ratio of the inradii of the triangles X, X,. 5 7 r r 1 r 2 X 6
15 Solution to Example E6: (5, 6, 7) triangle. is a triangle with =5, =6and =7. The incircle touches at X. Find the ratio of the inradii of the triangles X, X,. 5 7 r 1 r 2 r X 6 Solution. Let Δ be the area of triangle. For the inradii, Δ(X) :Δ(X) :Δ() =2:4:6=1:2:3. 1 r 1 : r 2 : r = + X + X : 2 + X + X : 3 + + = 1 12 : 2 16 : 3 18 = 2 : 3 : 4. Exercise Show that the incircles of triangles X and X touch X at the same point.
16 Example E7: (21, 22, 23) triangle. is a triangle with a =21, b =22, c =23. The incircle touches at Y. alculate the length of Y. 23 21 Y 22
17 Solution to Example E7: (21, 22, 23) triangle. is a triangle with a =21, b =22, c =23. The incircle touches at Y. alculate the length of Y. 23 21 Y 22 Solution. (1) cos = b2 +c 2 a 2 2bc = 222 +23 2 21 2 2 22 23 = 123 23. (2) Y = 1 2 (22 + 23 21) = 12. (3) pplying the law of cosines to triangle Y : Therefore, Y =19. Y 2 = 2 + Y 2 2 Y cos = 23 2 +12 2 2 23 12 13 23 = 361.
32 Example E16. n equilateral triangle of side 2 is partitioned symmetrically into a quadrilateral, an isosceles triangle, and two other congruent triangles. Suppose the quadrilateral and the isosceles triangle have congruent incircles of radii r. X r Y r θ Let X = θ. 2 (a) Show that r = cot 30 +cot θ by considering triangle X. (b) Show that Y =30 θ. cot θ cot 30 (c) Show that r = cot 30 cot θ+1. (d) y noting that cot 30 = 3, make of use (a) and (b) to show that cot θ = 3+2 2. (e) Deduce that r = 3 2.
33 Solution to Example E16. n equilateral triangle of side 2 is partitioned symmetrically into a quadrilateral, an isosceles triangle, and two other congruent triangles. Suppose the quadrilateral and the isosceles triangle have congruent incircles of radii r. r Z X θ Y r T Let X = θ. 2 (a) Show that r = cot 30 +cot θ by considering triangle X. (b) Show that Y =30 θ. (c)... Solution. (a) Z + Z = = r cot 30 + r cot θ =2 = r = 2 cot 30 +cot θ. (b) Y = 1 2 (60 2θ) =30 θ.
34 Solution to Example E16 (continued). n equilateral triangle of side 2 is partitioned symmetrically into a quadrilateral, an isosceles triangle, and two other congruent triangles. Suppose the quadrilateral and the isosceles triangle have congruent incircles of radii r. r Z X θ Y r T Let X = θ. 2 (a) Show that r = cot 30 +cot θ by considering triangle X. (b) Show that Y =30 θ. cot θ cot 30 (c) Show that r = cot 30 cot θ+1. (d) y noting that cot 30 = 3, make of use (a) and (b) to show that cot θ = 3+2 2. (e) Deduce that r = 3 2. Solution. (c) In triangle YT, r =tan(30 θ) = tan 30 tan θ 1+tan30 tan θ = cot θ cot 30 cot 30 cot θ +1. (d) Note that cot 30 = 3. Putting x =cotθin the expressions for r, 2 we have 3+x = x 3 3x+1 = x 2 3= 2 3x+2 = x 2 2 3x 5 =0, and x = 3+2 2, rejecting the negative root. (e) From (a), r = 2 3+x = 1 3+ 2 = 3 2.
Example E17 (Geogebra exercise). onstruct the following in which an equilateral triangle is partitioned symmetrically so that the quadrilateral and the isosceles triangle have congruent incircles. 35
36 Solution to Example E17 (Geogebra exercise). onstruct the following in which an equilateral triangle is partitioned symmetrically so that the quadrilateral and the isosceles triangle have congruent incircles.
37 The excircle b I b c I c c b c a b a a I a
38 The excircles The internal bisector of each angle and the external bisectors of the remaining two angles are concurrent at an excenter of the triangle. n excircle can be constructed with this as center, tangent to the lines containing the three sides of the triangle. I b I c X Y r a Z r a r a The exradii of a triangle with sides a, b, c are given by r a = Δ s a, r b = Δ s b, r c = Δ s c. I a The areas of the triangles I a, I a, and I a are 1 2 ar a, 1 2 br a, and 1 2 cr a respectively. Since Δ= ΔI a +ΔI a +ΔI a, we have Δ= 1 2 r a( a + b + c) =r a (s a), from which r a = Δ s a.
39 Heron s formula for the area of a triangle onsider a triangle with area Δ. Denote by r the inradius, and r a the radius of the excircle on the side of triangle. It is convenient to introduce the semiperimeter s = 1 2 (a + b + c). I a I r a r Y Y (1) From the similarity of triangles IY and I Y, r = s a. r a s (2) From the similarity of triangles IY and I Y, r r a =(s b)(s c). (3) From these, (s a)(s b)(s c) r =, s s(s b)(s c) r a =. s a Theorem (Heron s formula). Δ= s(s a)(s b)(s c). Proof. Δ=rs.
Example E18. Find the inradius and the exradii of the (13,14,15) triangle. 1
2 Example E19. If one of the ex-radii of a triangle is equal to its semiperimeter, then the triangle contains a right angle.
4 Example E21. The length of each side of the square is 6a, and the radius of each of the top and bottom circles is a. alculate the radii of the other two circles. 1 1 a and 3 4 a.
5 Solution to Example E21. The length of each side of the square is 6a, and the radius of each of the top and bottom circles is a. alculate the radii of the other two circles. D P Q Suppose each side of the square has length 6. Let x be the length of the tangents from P to the top circle. (x +5) 2 =(x +1) 2 +6 2 = x = 3 2. Therefore, PD = 5 2 and P =5+3 2 = 13 2. The length of the tangents from to the bottom circle = 5 2 +5 2 1 2 = 7. Let y be the length of the tangents from Q to the bottom circle. From (y +7) 2 =(5 y) 2 +6 2, we have y = 1 2. Therefore, Q =1+ 1 2 = 3 2 and Q =7+ 1 2 = 15 2. Triangle PQ has sides P = 13 2, Q = 15 2 and PQ =6 3 2 5 2 =2. It has semiperimeter 8. Its incircle has radius 8 on PQhas radius 3 2 1 2 6 =1. 3 2 1 2 6 8 = 3 4, and the excircle