Plane geometry Circles: Problems with some Solutions


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1 The University of Western ustralia SHL F MTHMTIS & STTISTIS UW MY FR YUNG MTHMTIINS Plane geometry ircles: Problems with some Solutions 1. Prove that for any triangle, the perpendicular bisectors of the three sides, and are concurrent, at say. educe that = =, and hence is the centre of a circle K of radius R = that passes through each of the vertices,, of. (K,, R are respectively the circumcircle, circumcentre and circumradius of, and K is circumscribed about.) Solution. raw and let the midpoints of sides, and be, and F, respectively. raw the perpendicular bisectors from sides and to meet at a point, and join to F (we must show F is also a perpendicular bisector). Then = 90 = = is common =, = Similarly, = So now we have =. = F = F F is common F = F, by the SS Rule by the SSS Rule F = F = 90, since F is a straight angle (180 ) Thus, F is the perpendicular bisector of, and hence the perpendicular bisectors of concur at. lso, we showed that = =, so that, and lie on a circle centred at with radius R =. 2. Prove that the angle bisectors of a triangle are concurrent, and that their common point I is equidistant from the sides of. (Let r be the common distance of I from the three sides of. Then the circle K(I, r) with centre I and radius r is inscribed in, and K(I, r), I, r are respectively the incircle, incentre and inradius of.) Solution. raw and draw the angle bisectors from and to meet at I. rop perpendiculars from I to points, and F, of the sides, and, respectively, and join I to (we must show I is also an angle bisector). Then F 1
2 IF = I IF = 90 = I I is common IF = I, IF = I Similarly, I = I I = I. So now we have IF = 90 = I I is common I = IF IF = I, IF = I by the S Rule F I by the RHS Rule Thus, I is the angle bisector of, and hence the angle bisectors of concur at I. lso, we showed I = I = IF, so that, and F lie on a circle centred at I, and since I, I and IF are perpendiculars to the sides, and, respectively, the circle K(I, r), where r = I, is tangent to the sides of at, and F, respectively. 3. If is a chord of a circle K(, r) where r = 5 cm, M is the foot of the perpendicular from to and M = 3 cm, find the length of chord. Solution. 90 = M = M, since M =, radii of circle K M is common M = M, by the RHS Rule M M = M = 2 M M 2 = 2 M 2, by Pythagoras Theorem = = 25 9 = 16 = 4 2 M = 4 = 2 M = 8 cm 4. Three points,, lie on the circumference of a circle K with centre. The points and are on the same side of chord. Given = 15, find the size of. Solution. t the final step we use Theorem 19 of the Plane Geometry notes, the part we need of which states: The angle subtended at the circumference is half the angle subtended at the 2
3 = is isosceles 15 = = = 180 = = 150 = 1 2, by Theorem 19, common arc: = = Two nonintersecting chords and are drawn in a circle. If = 20 and = 70, find the size of. 6. The radius of a circle K is = 5 cm, is a diameter of K and is a point on K, and = 6 cm, find the length of chord. 7. Points,,, are points in order on the circumference of a circle. The chord is produced and a point on the line is marked so that is between and. If = 48, find the size of. 8. In a quadrilateral, = 30, = 80 = 50, find the sizes of and. Solution. Firstly, = = = 30. Now, 3 noncollinear points lie on a unique circle. Thus there is a unique circle through the points, and. Now Theorem 19 states that angles subtended at the circumference standing on the same arc are equal. In fact, the converse of Theorem 19 is also true, namely that: If X, Y and Z lie on a circle K and for some other point U, YXZ = Y UZ then U also lies on K, so that XY ZU is a cyclic quadrilateral. We may apply the theorem with X =, Y =, Z =, U =, noting that Thus is a cyclic quadrilateral. Now, = 30 =. 50 = =, by Theorem 19, common arc: = 180, = = 100. by Theorem 27 (opposite angles of a cyclic quadrilateral are supplementary) 3
4 9. Two chords and of a circle with centre, intersect at. Show that = ( + ). 1 2 Solution. We use Theorem 19 (twice) and Theorem 7 of the Plane Geometry notes. (Part of) Theorem 19. The angle subtended at the circumference is half the angle subtended at the Theorem 7. n exterior angle of a triangle equals the sum of the two nonadjacent interior angles. = 1 2, by Theorem 19, arc: = 1 2, by Theorem 19, arc: = +, = + = 1 2 ( + ) by Theorem 7, is exterior to Note. This problem has some similarity with Problem 15, but note that the chords intersect internally in this problem and externally in Problem In quadrilateral, = 140, = 70 and =, prove that bisects. Solution. = is isosceles 70 = = = 180, = = 40 Now observe that opposite angles of are supplementary: + = = 180 is cyclic, by Theorem = =, by Theorem 19, common arc: = 1 2 bisects. 11. The three vertices of an acuteangled triangle lie on a circle K. The three altitudes of are produced so that they meet the circumference at points X, Y, Z. Prove that X, Y, Z are the bisectors of the angles of XY Z. 4
5 12. Show that the ends of two diameters of a circle are the vertices of a rectangle. Solution. We use Theorem 20 of the Plane Geometry notes (except we have added an arc symbol () here to distinguish the arc, from the chord), which states: If is a semicircular arc of a circle and is any point on the circumference of the circle then is a right angle. Note that Theorem 20 follows immediately from Theorem 19. In the diagram the angle at the circumference is half the angle at the centre which is a straight angle (180 ). Therefore, = 90. nother way of saying this is that is an angle in a semicircle and such angles are right angles. = 90, angle in semicircle = 90, angle in semicircle = 90, angle in semicircle = 90, angle in semicircle is a rectangle. 13. Two chords of a circle K subtend equal angles at the circumference of K. Prove that the lengths of the chords are equal. 14. Three points,, on a circle K are vertices of an equilateral triangle. If, distinct from,, also lies on K, find the size of. 15. hords and of a circle K are produced to meet at X. If is the centre of K, prove that X = 1 ( ). 2 Solution. s in Problem 9, we use Theorem 19 (twice) and Theorem 7 of the Plane Geometry notes. (Part of) Theorem 19. The angle subtended at the circumference is half the angle subtended at the Theorem 7. n exterior angle of a triangle equals the sum of the two nonadjacent interior angles. = 1 2, by Theorem 19, arc: = 1 2, by Theorem 19, arc: = X + X, = + X X = = 1 2 ( ) by Theorem 7, is exterior to X X 16. Find the distance between the centres of two circles that cut each other, if their radii are r and R and the length of their common chord is a. 5
6 17. Points,,,, are points in order on a circle, such that the chord is equal and parallel to chord. Prove that is a rectangle. 18. Two perpendicular chords and of circle K intersect at M. Given that M = 2 cm, M = 12 cm, and = 10 cm, find the radius of K. 19. straight line meets two circles at points,,,. The circles intersect at and F. Given F = α, find. 20. The diagonal of a cyclic quadrilateral bisects and. Prove that is a diameter of the circumcircle of. 21. In quadrilateral, = = 90 and =, find. 22. Points,,,, are points in order on a circle K. is produced to. The bisector of meets the circumference of K again at F. Prove that F bisects. 23. Show that in a cyclic quadrilateral with circumcircle K the bisectors of any two opposite angles meet K at ends of a diameter of K. Solution. Let the cyclic quadrilateral be and its circumcircle be K. onsider the opposite angles at and. Let the bisector of intersect K at, and let the bisector of intersect K at F. Let x = and let y = F. In the diagram we have pretended the centre of K is such that F 180 ; we will show that in fact F = 180. We use both parts of Theorem 19 (which we have rewritten to indicate the parts) and Theorem 27 of the Plane Geometry notes. Theorem 19. (a) ngles at circumference standing on the same arc are equal. (b) The angle subtended at the circumference is half the angle subtended at the (Part of) Theorem 27. pposite angles of a cyclic quadrilateral sum to 180. x = =, y = F = F, F = + F by Theorem 19(a), arc: by Theorem 19(a), arc: F = x + y F = 1 2 F, by Theorem 19(b), arc: F F = 2x + 2y = 2x, = 2y, is angle bisector F is angle bisector 180 = +, by Theorem 27 = 2x + 2y F = 180 Thus F is a straight angle and hence F is a diameter of circle K. 24. In, points X and Y on and, respectively are the feet of altitudes of. If = 50, find XY. F 6
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nswers: (0 HKM Heat Events) reated by: Mr. Francis Hung Last updated: pril 0 09 099 00  Individual 9 0 0900  Group 0 0 9 0 0 Individual Events I How many pairs of distinct integers between and 0 inclusively
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