Intermediate Math Circles Wednesday October Problem Set 3

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1 The CETRE for EDUCTI in MTHEMTICS and CMPUTIG Intermediate Math Circles Wednesday ctober Problem Set 3.. Unless otherwise stated, any point labelled is assumed to represent the centre of the circle. 1. Determine the length of the chord if = and = 3. In the diagram, the radius = and = 3. This forms a right triangle. y the Pythagorean Theorem, = 2 2 = = = 1 = 4 ( > 0) y the right bisector chord property, is the midpoint of. So = and = + = = 8 follows. 2. If = 10 and =, determine the length of. y the right bisector chord property, bisects. Hence =. Since = 10, then =. bserve is right angled. y the Pythagorean Theorem, = 2 2 = = 2 2 = 144 = 12 ( > 0) 3. circle has a diameter of length 2. If a chord of the same circle has a length of 10, how far is the chord from the centre? Given chord, draw diameter CD. Then the distance from to will be the length of the perpendicular bisector of (the perpendicular bisector of a chord passes through the centre of a circle). Let M be the perpendicular bisector of. M is therefore the midpoint of. Then M = M =. ote is a radius of the circle. Since the circle has diameter 2, the radius =. 1 C 2 10 M D

2 The CETRE for EDUCTI in MTHEMTICS and CMPUTIG M by construction, and hence M M. Then M is right angled. y the Pythagorean Theorem, M 2 + M 2 = 2 M = 2 M 2 = 144 M = 12 (M > 0) Therefore the distance from the chord to the centre of the circle is 12 units. 4. Calculate the distance between the parallel chords P Q and XY if P Q =, XY = 8, and the radius of the circle is. The distance between parallel chords is the length of a perpendicular line segment which extends from one to the other. Let M be the perpendicular bisector of P Q. y the right bisector property, this passes through, and it is also the perpendicular bisector of XY. Then P M = MQ = 3, X = Y = 4 and the length of M will be the distance between the chords. bserve X and Q are radii. Therefore X = Q =. bserve that M Q and X are right angled. pplying Pythagorean Theorem to both M 2 + MQ 2 = Q X 2 = X 2 M = = 2 M 2 = 1 2 = 9 M = 4 (M > 0) = 3 ( > 0) Therefore, M = M + = 7, and so the distance between the chords is 7 units. X P M 8 Q Y. The two parallel chords and CD are a distance of 14 apart. If has length 12 and the radius of the circle is 10, calculate the length of CD. Let M be the perpendicular bisector of, and hence the perpendicular bisector of CD. The length of M is the distance between the chords, so M = 14. If = 12, then M = M =. bserve and C are radii. Thus = C = 10. pply the Pythagorean Theorem to M. C M 8 D M 2 + M 2 = 2 M = 10 2 M 2 = 4 M = 8 (M > 0) 2

3 The CETRE for EDUCTI in MTHEMTICS and CMPUTIG Since 14 = M = M + = 8 +, then =. pply the Pythagorean Theorem to C 2 + C 2 = C C 2 = 10 2 C 2 = 4 C = 8 (C > 0) So C = 8. ut is the midpoint of CD, so C = D and therefore CD = C +D = 1.. Two circles with centre and have radii and 8, respectively. The circles intersect at the points X and Y. If XY = 8, determine the length of, the distance between the centres. In the diagram, radii Y and Y are drawn as dotted lines. Let Z mark the intersection of and XY. X y x 8 Z x y 8 Since X and Y are radii of the same circle, X = Y and hence XY is isosceles. So XY = Y X = x. Similarly, since X = Y, XY is isosceles and XY = Y X = y. bserve X = x + y = Y. Then by SS, X = Y. It follows that, X = Y. y S, XZ = Y Z (X = Y, XZ = Y Z, X = Y ) and hence ZX = ZY. ut XY is a straight line, so ZX = ZY = 90, and similarly ZX = ZY = 90. Thus XZ and XZ are right angled. Furthermore, XZ = ZY = 4 since XY = 8. pplying the Pythagorean Theorem to the two triangles gives XZ 2 + Z 2 = X 2 XZ 2 + Z 2 = X Z 2 = Z 2 = 8 2 Z 2 = 9 Z 2 = 48 Z = 3 (Z > 0) Z = 48 ( > 0) Therefore the distance between the centres of the circles is = Z + Z = (ote: 48 can be simplified to so = ). Y 3

4 The CETRE for EDUCTI in MTHEMTICS and CMPUTIG 7. In the diagram, P = and Q = 20, where P and Q are the centres of the circles. Determine the length of if P Q = 21. bserve that this problem is similar to Problem. Thus, using the same reasoning, T = T = y, T P = T P = T Q = T Q = 90, and T P, T Q are right angled. Let x = P T. Since P Q = 21 and P T = x, then QT = 21 x. y the Pythagorean Theorem: P T 2 + T 2 = P 2 QT 2 + T 2 = Q 2 x 2 + y 2 = 2 (1) (21 x) 2 + y 2 = 20 2 From (1), substitute 2 for x 2 + y 2 in (2) to get Substitute x = into (1) to solve for y Therefore = T + T = 2y = x + 2 = x = 231 x 2 + y 2 = 2 P x y T y 21 - x x + x 2 + y 2 = 20 2 (2) 42x = x = 210 x = y 2 = 2 2 y 2 = 144 y = 12 (y > 0) 8. In the diagram, C is inscribed in the semicircle with centre D. If = D, determine the measure of CD Q ny angle inscribed in a semi-circle is right angled, so C = 90. Since = D, D is isosceles and D = D = x. Then D = 180 2x, and DC = 90 D = 2x x 2x - 90 x x y D D and DC are radii, so D = DC and thus DC is isosceles. Therefore CD = CD (1) and y = 2x 90 (2) follows. Setting (1) and (2) equal to each other gives 2x 90 = 90 x. Then 3x = 180 and x = 0 follows. Substituting x = 0 into (1), y = 2(0) 90 = 30. Therefore, CD = C

5 The CETRE for EDUCTI in MTHEMTICS and CMPUTIG lternate Since D and D are radii, D = D. ut D = D. Therefore, D = D = D and D is equilateral. It follows that each angle in D is 0. Since D and DC are radii, DC is isosceles and CD = DC = y. bserve that D is exterior to DC. Then 0 = D = DC+ DC = y+y. So 2y = 0 and hence y = In the diagram, XY Z is right-angled at Z. W is the midpoint of XY, and the circle with diameter ZW intersects W X at V. If XY = 0 and W V = 7, determine the length of XZ. Y 2 In the diagram, since W is the midpoint of XY, W Y = W X = 2. Furthermore, if W V = 7, then V X = 18. ecause V lies on the circle and ZW is a diameter, W V Z = ZV X = 90 (angle inscribed in a semi-circle). W 7 V 2 Consider W UX. Since U is on the circumference, W UZ = W UX = 90. Then W UX Y ZX by. So W X Y X = 2 0 = 1 2 = XZ. Hence XZ = 2UZ so UX = UZ. UZ Z U x 18 X ut then W UX = W UZ by SS since they share common side W U. So ZW = XW = 2. ow, W V Z, ZV X are right-angled. y the Pythagorean Theorem, and W V 2 + ZV 2 = ZW ZV 2 = 2 2 ZV 2 = ZV 2 = 7 ZX 2 = V X 2 + ZV 2 ZX 2 = ZX 2 = 900 ZX = 30 (ZX > 0)

6 The CETRE for EDUCTI in MTHEMTICS and CMPUTIG lternate In the previous solution, it was shown that V ZX was right angled. ut V ZX also shares common angle Y XZ with ZY X. Hence V ZX ZY X. Let x = ZX as shown. Using the properties of similar triangles, Hence x = ZX = 30. V X ZX = ZX Y X V X x = x Y X V X Y X = x = x = x 2 30 = x (x > 0)

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