The High School Section

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Della Phillips
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1 Viète s Relations The Problems. 1. The equation 10/07/017 The High School Section Session 1 Solutions x 5 11x 4 + 4x 3 + 55x 4x + 175 = 0 has five distinct real roots x 1, x, x 3, x 4, x 5.

1 1 Viète s Relations The Problems. 1. The equation 10/07/017 The High School Section Session 1 Solutions x 5 11x 4 + 4x x 4x = 0 has five distinct real roots x 1, x, x 3, x 4, x 5. Find: x 1 + x + x 3 + x 4 + x 5. x 1 + x + x 3 + x 4 + x 5 = (x 1 + x + x 3 + x 4 + x 5 ) (x 1 x + + x 4 x 5 ) = 11 4 = 73.. The roots of the equation x 3 9x 11x + c = 0 are in arithmetic progression. Determine c and solve the equation. Let x 1 = a, x = a + r, x 3 = a + r be the roots. By Viète s relations 9 = x 1 + x + x 3 = 3a + 3r, 11 = x 1 x + x 1 x 3 + x x 3 = 3a + 6ar + r. From the first equation, r = (9 6a)/6; substituting into the second equation, and simplifying, a 3a 0 = 0. This last equation has the solutions, 5. Selecting a = we get r = 7/ and the roots work out to, 3/, and 5. Selecting a = 5 we get r = 7/, and the roots come out the same, but in decreasing order. By Viète s relation, c = x 1x x 3 = ( )( 3 )(5) = 15, thus c = It was mentioned that all symmetric functions can be written in terms of the elementary symmetric functions. There is a converse to this. All symmetric functions can be written in terms of sums of (equal) powers of the variables. For example, if n =, s 1 (x 1, x ) = x 1 + x

2 1 VIÈTE S RELATIONS is already so expressed; the power is 1. s (x 1, x ) = x 1 x = 1 ( (x1 + x ) (x 1 + x ) ). Do the same for n = 3. Of course s 1 is already a sum of powers, so concentrate on s, s 3. The objective is to express s (x 1, x, x 3 ) = x 1 x + x 1 x 3 + x x 3, s 3 (x 1, x, x 3 ) = x 1 x x 3 in terms of x 1 + x + x 3, x 1 + x + x 3, x x 3 + x 3 3. The sums of powers may appear in any way; they may appear squared or cubed, or multiplied with each other. Etc. Because (x 1 + x + x 3 ) = x 1 + x + x 3 + x 1 x + x 1 x 3 + x x 3 = x 1 + x + x 3 + s (x 1, x, x 3 ), we have s (x 1, x, x 3 ) = x 1 x + x 1 x 3 + x x 3 = 1 ( (x1 + x + x 3 ) (x 1 + x + x 3) ). The case of s 3 is trickier. We have (x 1 + x + x 3 ) 3 = ((x 1 + x ) + x 3 ) 3 = (x 1 + x ) 3 + 3(x 1 + x ) x 3 + 3(x 1 + x )x 3 + x 3 3 = x x 1x + 3x 1 x + x 3 + 3x 1x 3 + 6x 1 x x 3 +3x x 3 + 3x 1 x 3 + 3x x 3 + x 3 3 = x x 3 + x x 1 (x + x 3) + 3x (x 1 + x 3) +3x 3 (x 1 + x ) + 6x 1 x x 3 = x x 3 + x x 1 (x 1 + x + x 3) 3x 3 1 3x (x 1 + x + x 3) 3x 3 + 3x 3 (x 1 + x + x 3) 3x x 1 x x 3 = (x x 3 + x 3 3) + 3(x 1 + x + x 3 )(x 1 + x + x 3) + 6x 1 x x 3. Thus s 3 (x 1, x, x 3 ) = x 1 x x 3 = 1 6 ( (x1 + x + x 3 ) 3 + (x x 3 + x 3 3) 3(x 1 + x + x 3 )(x 1 + x + x 3) ). 4. Find all triples (x, y, z) such that x y z and x + y + z = 4 x + y + z = 46 x 3 + y 3 + z 3 = 190 Hint: Suppose P (t) = t 3 + at + bt + c is a polynomial with zeros x, y, z. Use the results from Problem 3 Suppose P (t) = (t x 1 )(t x )(t x 3 ) = t 3 + at + bt + c is the (monic) polynomial of roots x 1, x, x 3. By Problem 3 we have a = s 1 (x 1, x, x 3 ) = 4 b = s (x 1, x, x 3 ) = 1 ( 4 46 ) = 15 c = s 3 (x 1, x, x 3 ) = 1 ( ) = Thus P (t) = t 3 4t 15t + 18, and it easy to see that one of the zeros of P (t) is t = 1. Dividing by t 1 we get the quadratic polynomial t 3t 18 with zeros 3, 6. The answer is x = 3, y = 1, z = 6.

3 CHASING ANGLES 3 Chasing Angles 1. Suppose AB is a chord of a circle, not the diameter, and suppose C is on the circle, on the same side of the line determined by A, B as the center O of the circle. Show that ACB = 1 AOB. Pictures for the problem. There are actually two possibilities: The center is inside the triangle ACB, or outside of it. Let s ignore the limiting case of the center on the segment AC or BC. Case 1 Case I ll give you some help for case 1. Case is similar, perhaps even easier. Here is the picture for Case 1, with a few more segments marked. Perhaps you can answer the questions below. Hints and Questions Hint. Angles OCA and CAO look equal. Prove they are equal. Same for OCB and CBO. Question. What can you say about AOC + OCA = AOC + OCA + CAO? Question. What can you say about BOC + OCB = BOC + OCB + CBO? Question. Noticing that ACB = OCA + OCB, can you express ACB in terms of AOC + BOC? Question. How is AOC + BOC related to AOB?

4 CHASING ANGLES 4 Case 1. OCA = CAO because they are the base angles of the isosceles triangle ACO, OA = OC since they are both radii of the circle. Similarly, triangle BCO is isosceles, hence OCB = CBO. AOC + OCA = AOC + OCA + CAO = 180 by the theorem about the sum of angle of a triangle. For the same reason, From this we get BOC + OCB = BOC + OCB + CBO = 180. ACB = OCA + OCB = (180 AOC) + (180 BOC) = 360 ( AOC + BOC). But 360 ( AOC + BOC) = AOB, so we showed that ACB = AOB. Case. Here is the figure for Case, with a few more lines. We have to show that ACB = 1 AOB. Now OA = OB = OC, they are all radii of the circle. We also see that in this case, AOC = AOB + BOC. Because triangle AOC is isosceles, OCA = CAO so that (by the sum of the angle of a triangle adding up to 180 ), 180 = AOC + OCA + CAO = AOC + OCA. Triangle BOC is also isosceles, so OBC = OCB = OCA + ACB and by the sum of angles of a triangle applied to triangle BOC, 180 = BOC + OBC + OCB = BOC + ( OCA + ACB). It is time to put all this together. We have and we are done. AOB = AOC BOC = (180 OCA) [180 ( OCA + ACB)] = ACB. Suppose the point C on the circle is on the other side of the line through AB as the center O of the circle. Show that ACB = AOB. Consider the picture below. We have to show θ = AOB = 180 (ψ ϕ ).

5 CHASING ANGLES 5 We see several isosceles triangles in the picture below. Using the theorem about sum of theangles of a triangle we get Adding we get the desired result. (θ β) + ϕ = 180 β + ψ = Show that all angles subtended by the diameter are right angles. That is, if AB is a diameter of the circle, C a point on the circle, C A, B, then ACB = 90. Conversely, show that if A, B, C are distinct (no two equal) points on the circle, and ACB is a right angle, then AB is a diameter of the circle. This is partially a limiting case of the previous results. 4. A cyclic quadrilateral is a quadrilateral that can be inscribed in a circle. Assume ABCD is a cyclic quadrilateral. Show that opposite angles add up to 180. That is, the angle at A plus the angle at C add up to 180, and the same is true for the angles at B, D. Draw the diagonal BD. If this diagonal is a diameter of the circle, then the angles at A and at C are both right angles and add up to 180. If not, the center O is on one side of AC and then one of the angles at A or at C equals 1 AOB, the other one equals AOB, so that they add up to 180. The converse of this result is also true: 5. Prove: If a convex quadrilateral ABCD satisfies that a pair of opposite angles (say at A and at C) add up to 180, then it is cyclic. Hint: Consider the circle through A, B, C. If D is not on the circle, where can it be? If D is not on the circle it is either outside the circle or inside the circle. In either case draw the line through B and D; it intersects the circle (of course) at B and at a point E. We then have one of the two following pictures:

CHASING ANGLES 6 Now we must have ADC = AEC, since both added to ABC give 180 ; the ADC by hypothesis, AEC because ABCE is a cyclic quadrilateral. But this is impossible.

6 CHASING ANGLES 6 Now we must have ADC = AEC, since both added to ABC give 180 ; the ADC by hypothesis, AEC because ABCE is a cyclic quadrilateral. But this is impossible. In the case of the first figure, CAE < CAD and ACE < ACD; the angles of triangles ACE and ACD must add up to 180, forcing angle ADC to be less than angle AEC. A similar argument shows that one also gets a contradiction assuming D is inside the circle; the hypothesis implies ADC = AEC, but the picture shows that angle ADC is larger than angle AEC.

21. Prove that If one side of the cyclic quadrilateral is produced then the exterior angle is equal to the interior opposite angle.

21. Prove that If one side of the cyclic quadrilateral is produced then the exterior angle is equal to the interior opposite angle. 22. Prove that If two sides of a cyclic quadrilateral are parallel, then