Example 1. Show that the shaded triangle is a (3, 4, 5) triangle.

Example 1. Show that the shaded triangle is a (3, 4, 5) triangle.

Solution to Example 1. Show that the shaded triangle C is a (3, 4, 5)-triangle. E D t C 4 T t 4 4 Solution. Suppose each side of the square has length 4, and CD = t. Then C = T + TC = E + CD =4+t. pplying the Pythagorean theorem to triangle C,wehave (4 + t) 2 =4 2 +(4 t) 2 = t =1, and C =3. Therefore, C is a 3, 4, 5)-triangle.

Example 2. Show that the shaded triangle is a (3, 4, 5) triangle.

Solution to Example 2. Show that the shaded triangle is a (3, 4, 5)-triangle. D C O T Solution. Suppose each side of the square has length 4, and CO = t. Then O =4+t. pplying the Pythagorean theorem to triangle O,wehave (4 + t) 2 =4 2 +(4 t) 2 = t =1, and O =3, and =5. Therefore, O is a (3, 4, 5)-triangle.

Example 3. Each side of the square has length 8. Calculate the radius of the circle tangent to a side of the square and the two quadrants, both internally.

Solution to Example 3. Each side of the square has length 8. Calculate the radius of the circle tangent to a side of the square and the two quadrants, both internally. Solution. Let t be the radius of the circle. (8 t) 2 = t 2 +4 2 = 64 16t =16 = t =3.

Example 4. Each side of the square has length 6. Calculate the radius of the circle tangent to a side of the square and the two quadrants, one internally and the other externally.

Solution to Example 4. Each side of the square has length 6. Calculate the radius of the circle tangent to a side of the square and the two quadrants, one internally and the other externally. t 6+t s Solution. Let t be the radius of the circle. Computing the distance of the center from the base in two ways, we have (6 + t) 2 (6 t) 2 =(6 t) 2 t 2 = 24t =36 12t = t =1.

Example 5. Each side of the square has length 16. Calculate the radius of the circle tangent to a side of the square and the two quadrants, both externally.

Solution to Example 5. Each side of the square has length 16. Calculate the radius of the circle tangent to a side of the square and the two quadrants, both externally. Solution. Let t be the radius of the circle. (16 + t) 2 =8 2 +(16 t) 2 = 32t =64 32t = t =1.

Example 6. Two circles () and () with radii a and b are tangent to each other externally. common tangent touches the circles at X and Y. Calculate the length of XY. X Y

Solution to Example 6. Two circles () and () with radii a and b are tangent to each other externally. common tangent touches the circles at X and Y. Calculate the length of XY. Q nswer. 2 ab. X Y Solution. If the radii of the circles () and (b) are a and b respectively, XY 2 = Q 2 = 2 Q 2 = 2 (X Y ) 2 = (a + b) 2 (a b) 2 = 4ab; XY =2 ab.

Example 7. Two circles () and () with radii a and b are tangent to each other externally. common tangent touches the circles at X and Y. third circle C is tangent to both circles (), (), and to the line XY. Calculate the radius of the circle. C X Z Y

Solution to Example 7. Two circles () and () with radii a and b are tangent to each other externally. common tangent touches the circles at X and Y. third circle C is tangent to both circles (), (), and to the line XY. Calculate the radius of the circle. X Solution. Let r be the radius of this third circle, tangent to XY at Z. XY =2 ab, XZ =2 ar, ZY =2 br. Since XZ + ZY = XY, Z C Y 2 ar +2 br = 2 ab, ( a + b) r = ab, ab r = ; a + b r = ( ) 2 ab a + b ab = ( a + b) 2 ab = a + b +2 ab. Remark. This relation can also be be put into the form 1 = 1 + 1. r a b

Example 8. The two congruent circles are tangent to each other. Suppose P divides O in the golden ration, PQ O, and XQ PQ. Show that Y divides QX in the golden ratio. X Y Q O P

Solution to Example 8. The two congruent circles are tangent to each other. Suppose P divides O in the golden ration, PQ O, and XQ PQ. Show that Y divides QX in the golden ratio. X Z Y Q T O P Solution. Let P =1. Since P divides O in the golden ratio, O = ϕ. If the circles touch at T and Z is the midpoint of XY, then ZX = ZY = Q = P, and YX =2 P. Therefore, QY : YX =2 OP :2 P = OP : P. Y divides QX in the golden ratio.

Example 9. Show that = CD. C D

Solution to Example 9. Show that = CD. E F G C D H Solution. Label the remaining points of tangency. EF = E + F = + C =2 + C, GH = DH + GD = CD + D =2 CD + C. y symmetry, EF = GH. Therefore, by comparison, = CD.

Example 10. From the centers of each of two nonintersecting circles tangents are drawn to the other circle. Prove that the chords PQand XY are equal in length. P X Q Y

Solution to Example 10. From the centers of each of two nonintersecting circles tangents are drawn to the other circle. Prove that the chords PQand XY are equal in length. S T P X H K Q Y Solution. Let H and K be the midpoints of PQand XY respectively. Triangles P H and T are similarly (since they contain two pairs of equal angles). Therefore, HP P = T = PQ =2 HP = 2 P T = 2r a r b, where r a and r b are the radii of the circles. lso, from the similarity of triangles XK and S,wehave KX X = S Therefore, PQ = XY. = XY =2 KX = 2 S X = 2r a r b.

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

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.

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

Solution to Example 12. The large circle has radius R. Calculate 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