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1998-1999 Olympiad Correspondence Problems

Set 1

1.
ABC is an isosceles triangle with A = 100 and AB = AC. The bisector of angle B meets AC in D. Show that BD + AD = BC.
View solution
2.
Let I be the incentre of triangle ABC. Let the lines AI, BI and CI produced intersect the circumcircle of triangle ABC at D, E and F respectively. Prove that EF is perpendicular to AD.
View solution
3.
Let PQR be an arbitrary triangle. Points A, B and C external to the triangle are determine for which
AQR = ARQ = 15
QPC = RPB = 30
PQC = PRB = 45 .
Prove that

       (a) AC = AB;

       (b) BAC = 90.

View solution

4.
Let a and b be two positive real numbers. Suppose that ABC is a triangle and D a point in side AC for which
BCA = 90
|AD | = a              |DC | = b .
Let |BC | = x and ABD = q. Determine the values of x and q for the configuration in which q assumes its maximum value.

View solution

5.
Let \frakC be a circle with centre O and radius k. For each point P O, we define a mapping P P where P is that point on OP produced for which
|OP ||OP| = k2  .
In particular, each point on \frakC remains fixed, and the mapping at other points has period 2. This mapping is called inversion in the circle \frakC with centre O, and takes the union of the sets of circles and lines in the plane to itself. (You might want to see why this is so. Analytic geometry is one route.)

(a) Suppose that A and B are two points in the plane for which |AB | = d, |OA | = r and |OB | = s, and let their respective images under the inversion be A and B. Prove that
|AB| = k2 d
rs
  .

(b) Using (a), or otherwise, show that there exists a sequence { Xn } of distinct points in the plane with no three collinear for which all distances between pairs of them are rational.
View solution
6.
Solve each of the following two systems of equations:
(a)             x + xy + y = 2 + 32 ,      x2 + y2 = 6;
(b)
x2 + y2 + 2xy
x + y
= 1 ,
  ____
x + y
 
= x2 - y  .

View solution


Solutions

Problem 1

1.
First solution. See Figure 1.1. Let |AB | = u. Then
|BD |+ |AD |
= u sin100
sin60
+ u sin20
sin60
= u
sin60
(sin100 + sin20)
= 2usin60cos40
sin60
= 2ucos40 = |BC | .

1.
Second solution. See Figure 1.2. Let E and F be located in BC so that DEC = DFB = 80. Then DABD DEBD while triangles DEF, DCF and DBF are isosceles. Hence
BC = BF + FC = BD + DF = BD + DE = BD + AD .

1.
Third solution. See Figure 1.3. Select G on BC so that BG = BD and H on AB so that HD || BC. Then, by filling in the angles, we see that triangles BDG, HBD, AHD and DGC are isosceles. Also triangles AHD and GCD are similar and DC = BH = HD. Hence DAHD DGCD and AD = GC. Hence BC = BG + GC = BD + AD.
1.
Fourth solution. See Figure 1.3. Define G as in the third solution. Triangles ABC and GCD are similar. Hence, using the angle bisection theorem, we have that GC:CD = AB:BC = AD:DC, so that GC = AD. Hence BC = BG + GC = BD + AD.
1.
Fifth solution. See Figure 1.3. Define G and H as in the previous solution. ABGD is concyclic, so AD = DG since they are chords subtending equal angles of 20 at the circumference. It follows from the similarity of triangles AHD and GDC that they are congruent, so AD = DG = GC. Since also BD = BG, the result follows.


Problem 2.

2.
First solution. See Figure 2.1. Let AD and EF meet at H. Then
DHF
= HEI + HIE
= FEB + AIE
= FCB + (ABI + BAI)
= 1
2
ACB + 1
2
ABC + 1
2
BAC = 90
as desired.

2.
Second solution. Identify points in the complex plane, with the circumcircle of DABC being the unit circle with centre 0. Let A ~ 1, B ~ cos2b+ isin2b, C ~ cos2g+ i sin2g. Observe that AD bisects the arc BC, BE the arc CA and CF the arc AB, so that
D ~ cos(b+ g) + i sin(b+ g) ,
E ~ cos(g+ p) + i sin(g+ p) = - [cosg+ i sing] ,
F ~ cosb+ i sinb .
The vector EF is given by
(cosb+ cosg) + i(sinb+ sing) = cos b- g
2


cos b+ g
2
+ i sin b+ g
2


and the vector AD by
(cos(b+ g) - 1)
+ i sin(b+ g)
= 2 sin b+ g
2


-sin b+ g
2
+ i cos b+ g
2


= 2 sin b+ g
2


cos b+ g
2
+ i sin b+ g
2


i
from which it can be seen that they are perpendicular.

2.
Third solution. [D. Brox] See Figure 2.3. Since EFBC is concyclic, IFX = CFE = CBE = EBA = IBX. Hence FBIX is concyclic and so FXB = FIB. Similarly EYC = EIC.
Hence AXY = FXB = FIB = EIC = EYC = AYX, so that AX = XY. Since DAXY is isosceles and AD bisects XAY, then AD ^FE.

2.
Fourth solution. [L. Lessard] See Figure 2.4. Let O be the circumcentre of the triangle. Since F bisects arc AB, then OF right bisects AB. Also OE right bisects AC. Since DOEF is isosceles, OEF = OFE and so EUC = FVB. Hence AUV = AVU, so that DAUV is isosceles. Thus, the bisector of angle A right bisects UV and the result follows.

2.
Fifth solution. [H. Dong] See Figure 2.3. Using the fact that AFBCE is concyclic, we have that AEF = ACF = BCF = BEF. Also AFE = CFE. Hence DAFE DIFE (ASA) so that EA = EI. Thus DEAI is isosceles with apex angle AEI whose bisector EF must right bisect the base AI.
2.
Sixth solution. See Figures 2.6 and 2.3. We first note a preliminary result: If P, R, Q, S are four points on a circle and PQ and RS intersect inside the circle at T, then STQ = SPQ + PSR, which is equal to half the sum of the angles subtended at the centre by arcs PR and SQ. Now, ABE = 1/2ABC, so that arc AE subtends an angle equal to ABC at the centre. Similarly BF subtends an angle equal to ACB at the centre and BD subtends an angle equal to BAC at the centre. Hence FD subtends an angle equal to ACB + BAC at the centre. By the preliminary result, AHE is equal to half the sum of the angles subtended at the centre by arcs AE and FD, namely half of 180. The result follows.

Problem 3.

3.
First solution. See Figure 3.1. (a) Let point D be selected on the same side of QR as A so that triangle QDR is equilateral. Then DA ^QR so that QDA = RDA = 30 and DQA = DRA = 45. Hence
DPCQ ~ DDAQ      and      DPBR ~ DDAR .

A rotation about Q followed by a dilatation takes C to A and P to D so that
CQ:PQ = AQ:DQ .
Since CQA = PQD, DCAQ ~ DPDQ so that AC:PD = AQ:DQ. Similarly, DBAR ~ DPDR and AB:PD = AR:DR. Since AQ = AR and DQ = DR, it follows that AC = AB.

(b) By the similar triangles identified in (a), CAQ = PDQ and BAR = PDR. Hence
CAB
= QAR - (CAQ + BAR)
= QAR - (PDQ + PDR)
= QAR - QDR = 150 - 60 = 90 .

3.
Second solution. See Figure 3.2. Let S be the image of R under a counterclockwise rotation about A through 90. Since AS = AR, ASR = ARS = 45 so QRS = 30. Since QA = AR = AS, and since QAS = 150 - 90 = 60, AQS = ASQ = 60 and so SQR = 45. Hence triangles CQP, BRP, SQR are similar, and CQ:PQ = QS:QR. Also CQS = 45 PQS = PQR ( according as S lies inside or outside of DPQR). Hence DCQS ~ DPQR. Therefore CSQ = PRQ CSA = CSQ + 60 = PRQ + 60 = ARB. Also CS:RP = QC:QP = RB:RP CS = RB. Since in addition SA = RA, DCSA DBRA (SAS), so that AC = AB and SAC = RAB. Finally,
BAC = BAS + SAC = BAS + RAB = RAS = 90 .

3.
Third solution. [D. Brox] Note that sin75 = cos15 = (6 + 2)/4 and sin15 = (6 - 2)/4. Let
a =

6 - 2
2




2
2


= 3 - 1
2
 .
We solve the problem using vectors in the complex plane. Let lower case letters correspond to the points in the plane given in upper case, so that a corresponds to A, et cetera.

Since |r - a | = |q - a | = |r - q |/2 cos15 = 1/2(6 -2) |r - q |, we have that
q - a = 6 - 2
2


6 + 2
4
- 6 - 2
4
i

(q - r) = a

3 + 1
2
- 3 - 1
2
i

((q - p) - (r - p))
and
r - a = 6 - 2
2


6 + 2
4
+ 6 - 2
4
i

(r - q) = a

3 + 1
2
+ 3 - 1
2
i

((r - p) - (q - p)) .
Applying the sine law yields
|b - r | = |r - p |sin30/sin105 = 1
2
(6 - 2) |p - r |    and    |c - q | = 1
2
(6 - 2)|q - p |
so that
b - r = 6 - 2
2


2
2
- 2
2
i

(p - r) = a(i - 1)(r - p)
and
c - q = 6 - 2
2


2
2
+ 2
2
i

(p - q) = -a(i + 1)(q - p) .
Hence
c - a
= (q - a) + (c - q) = a



3 - 1
2
- 3 + 1
2
i

(q - p)-

3 + 1
2
- 3 - 1
2
i

(r - p)

= ia

-

3 + 1
2
+ 3 - 1
2
i

(q - p) +

3 - 1
2
+ 3 + 1
2
i

(r - p)

= i[(r - a) + (b - r)] = i(b - a)
from which it follows that AC and AB are perpendicular segments of equal length.

3.
Fourth solution. [D. Pritchard]
Lemma: For any triangle with angles a, b, g,
sin2 a+ sin2 g- 2sinasingcos(b+ 60) = sin2 b+ sin2 g- 2 sinbsingcos(a+ 60) .

Proof: Taking the difference between the two sides yields
sin2 a-
sin2 b- sing[2 sinacos(b+ 60) - 2 sinbcos(a+ 60)]
= 1
2
[cos2b- cos2a]- sin(a+ b)[ sin(a+ b+ 60)+ sin(a- b- 60)
                        - sin(b+ a+ 60) - sin(b- a- 60)]
= sin(a+ b) sin(a- b)- sin(a+ b)[sin(a- b)cos60 - cos(a-b) sin60
                        - sin(b- a)cos60 + cos(b- a) sin60]
= sin(a+ b) [sin(a- b) -2sin(a- b) cos60 ] = 0   .

(a) Wolog, we can let the lengths of PQ, PR and QR be sinR, sinQ and sinP, respectively, since by the Law of Sines, these lengths are proportional to these quantities. Then
|AR | = sinP sin15
sin150
= 3 - 1
2
sinP
|RB | = sinQ sin30
sin105
= 3- 1
2
sinQ
|AB |2 =

3 - 1
2


2

 
[sin2 P + sin2 Q - 2 sinP sinQ cos(R + 60)]
Similarly
|AC |2 =

3 - 1
2


2

 
[sin2 P + sin2 R - 2 sinP sinR cos(Q + 60)]
Hence AB = AC.

(b) |BP | = sinQ sin45/sin105 = (3 - 1)sinQ, |CP | = (3 - 1)sinR and
|BC |2 = (3 - 1)2[(sin2 R + sin2 Q - 2sinR sinQ cos(P + 60)]
from which |BC |2 = 2|AB |2, and the result follows.

3.
Fifth solution. [D. Nicholson] See Figure 3.5. Let |AQ | = |AR | = u, |BR | = v, |CQ | = w, a = QPR, b = PQR and g = PRQ. Let D, E, F be the respective feet of the perpendiculars from A, B, C to QR, PR and PQ.
|PQ | = |QC |cos45+ |CF |cot30 = 1
2
(1 + 3)w
|PR | = 1
2
(1 + 3)v
and
|QR | = 2ucos15 = 1
2
(1 + 3)u .
Hence
cosa = v2 + w2 - u2
2vw
cosb = u2 + w2 - v2
2uw
cosg = u2 + v2 - w2
2uv
 .
Then
|AC |2
= u2 + w2 - 2uw cos(b+ 60)
= 1
2
(u2 + v2 + w2) +3 uw sinb
|AB |2
= u2 + v2 - 2uv cos(g+ 60)
= 1
2
(u2 + v2 + w2) +3 uv sing  .
Now |PC | = |FC |csc 30 = 2|FC | = 2w and |PB | = 2v. Hence
|BC |2
= 2v2 + 2w2 - 4vw cos(a+ 60)
= u2 + v2 + w2 + 23 vw sina .
Since
sina: sinb: sing = |QR |: |PR |: |PQ | = u : v : w  ,
uw sinb = uv sing = vw sina .
Thus |AC | = |AB | and |BC |2 = |AC |2 + |AB |2, yielding the desired result.

3.
Sixth solution. [T. Costin] See Figure 3.1. First, we establish a preliminary result. Let ABC be an arbitrary triangle. Using the standard notation and applying the sine law, we find that
a2
+ b2 - 2ab cos(C + 60) = b2 sin2 A
sin2 B
+ b2 - 2b2 sinA
sinB


cosC
2
- 3sinC
2


= b2
sin2 B
[ sin2 A + sin2 B -sinA sinB cosC + 3sinA sinB sinC ]
= b2
sin2 B
[sin2 B + sin2 A -sinA cosC (sinA cosC + cosA sinC) +3sinA sinB sinC ]
= b2
sin2 B
[sin2 B + sin2 A sin2 C- sinA cosA sinC cosC + 3sinA sinB sinC ] .
Since this is symmetric in A and C, it is equal to b2 + c2 - 2bc cos(A + 60), and, by further symmetry, to c2 + a2 - 2ac cos(B + 60).

(a) Let D be constructed as in the figure, and note that QC:QA = r:p by similar triangles. By the Law of Cosines,
AC2
= QC2 + AQ2 - 2QC·QA cos(Q +60)
= QA2
p2
[ r2 + p2 - 2pr cos(Q + 60)]
and
AB2 = RA2
p2
[q2 + p2 - 2qp cos(R +60) ]  .
An application of the preliminary result to DPQR yields AB = AC.

(b) PC:AD = r:p and PB:AD = q:p. Hence, by the Law of Cosines applied to DPCB,
BC2 = AD2
p2
[ r2 + q2 - 2qr cos(P +60)] .
Since p = QR = 2QAcos15 and
AD
= QDcos30 - QAsin15 = p[cos30 - 1
2
tan15]
= p
2
[ 3 - (2 - 3)] = (3- 1)p ,
so BC2 = (4 - 23)[r2 + q2 - 2qrcos(P + 60)]. But
AB2 + AC2 = 2QA2
p2
[r2 + p2 - 2prcos(Q + 60)] = 1
2
sec2 15[r2 + q2 - 2qr cos(P + 60)] = BC2
as desired.

3.
Seventh solution. [Y. Shen] (a) From the Sine Law,
AR = AQ = QR (2 sin15)
BR = PR/2 sin75
and
CQ = PQ/2 sin75 .
Hence
AB2
= AR2 + BR2 - 2AR·BR cos(60 + PRQ)
= 4 sin2 15 QR2 + PR2
2(1 - cos150)
-2 RQ ·PR tan15

1
2
· PR2 + QR2 - PQ2
2 PR ·QR
- 3
2
·sinPRQ

= (2 - 3) QR2 + (2 - 3) PR2 - 1
2
(2 - 3)(PR2 + QR2 - PQ2) +(4 3 - 6)[PQR]
= 1
2
(2 - 3)(PQ2 + QR2 + PR2) +(43 - 6)[PQR] .
By symmetry, AC2 = AB2, so AC = AB.

(b)
BP = 2
2
· RP
sin75
 ,       CP = 2
2
· PQ
sin75
 .
Then
BC2
= BP2 + CP2 - 2BP ·CP cos(60 + QPR)
= 2 PR2 (2 - 3) + 2PQ2(2 -3) - (2 - 3)(PR2 + PQ2 - QR2)+ (83 - 12)[PQR]
= AB2 + AC2
so that BAC = 90.


Problem 4.

4.
First solution. See Figure 4.1. q assumes its maximum value when the circumcircle of DADB is tangent to BC. For, if X is any other point on CB produced, X lies outside this circle and AD subtends at X a smaller angle than it subtends at the circumference of the circle.
Let O be the centre of the circumcircle and r its radius. Then O lies on the right bisector of AD and two radii are OD and OB. Since BC is tangent, OB ^BC. Hence x2 + (a/2)2 = r2 = (b + (a/2))2 so that x = [(b(a+b))]. Since q = arctan[(a+b)/x]- arctanb/x,
tanq = a/x
1 + [b(a+b)/x2]
= a
2x
= a
2   _____
b(a+b)
 
 .

Comment. The angle can also be identified by noting that
ABD = 1
2
AOD = EOD = arcsin (1/2)a
r
= arcsin a
a + 2b
 .

4.
Second solution. See Figure 4.2. (using calculus) As in (a), for general position of B we can calculate
tanq = a/x
1 + [b(a+b)/x2]
= ax
x2 + b(a+b)
 .
Since 0 < q < p/2, q is maximized when tanq is maximized or when cotq = 1/tanq is minimized. Now
d
dx
cotq
= d
dx


x
a
+ b(b+a)
ax


= 1
a
- b(b+a)
ax2
= x2 - b(a+b)
ax2
 .
Thus, cotq is decreasing when x2 < b(a+b) and increasing when x2 > b(b+a) and so achieves its minimum value of (2[(b(a+b))])/a when x = [(b(a+b))].

4.
Third solution. [D. Brox] As in the previous solution we find that
cotq = 1
a


x + b(a+b)
x


 .
Using the arithmetic-geometric means inequality, we find that cotq 2/a[(b(a+b))] with equality if and only if x2 = b(a+b). Hence the maximum value of q = arccot (2/a[(b(a+b))]) is assumed when x = [(b(a+b))].

4.
Fourth solution. [K. Choi] Comparing two expressions for the area of triangle ABD, we have that
ax = sinq   ______
x2 + b2
 
  __________
x2 + (a + b)2
 
or
csc2 q
= 1
a2


x2 + (b2 + (a+b)2) + b2(a+b)2
x2


= 1
a2




x - b(a+b)
x


2

 
+(a + 2b)2

  .
Since 0 < q < 90, q is maximum when csc q is minimum, i.e., when x = [(b(a+b))]. For this value of x,
csc q = a+2b
a
   whence   q = arcsin a
a+2b
 .

4.
Fifth solution. By the Law of Cosines,
a2 = [(a+b)2 + x2] + [b2 + x2] -2   ___________________________
b2(a+b)2 + [a2 + 2b(a+b)]x2 + x4
 
cosq
whence
cos2 q = [b(a+b) + x2]2
[b(a+b) + x2]2 + a2 x2
and
sin2 q = a2 x2
[b(a+b) + x2]2 + a2 x2
 .
Thus
csc2 q = 1 +

b(a+b) + x2
ax


2

 
 .
Since 0 < q < 90, to maximize q, we must minimize csc2 q, and hence minimize
1
a


b(a+b)
x
+ x

 .
This can be done as in the third solution. When x = [(b(a+b))], we find that
sin2 q = a2
(2b+a)2
and so the maximum angle is arcsin(a/(2b+a)).

4.
Sixth solution. [Y. Shen] As in the second solution,
tanq = ax
x2 + b(b+a)
whence (tanq)x2 - ax + b(b+a)tanq = 0. This quadratic is satisfiable by real values of x if and only if a2 4b(b + a)tan2 q or
tanq a
2   _____
b(b+a)
 
 .
When tanq = a/(2[(b(b+a))]), then x = [(b(b+a))], and we obtain the same result as before.


Problem 5.

5.
First solution. See Figure 5.1. Since O is common and OA:OB = OB:OA, triangles OAB and OBA are similar. Hence
r:s:d = k2
s
: k2
r
:|AB|
from which it follows that |AB| = (k2d)/(rs).

(b) We first show that the inversion with respect to a circle \frakC of a line not passing through its centre O is a circle passing through O. Let F be the foot of the perpendicular from the point O to the line, and let P be any other point on the line. Let F and P be their respective images with respect to the inversion. As in (a), we have that DOPF ~ DOFP whence OPF = OFP = 90. Hence P lies on the circle with diameter OF.
Note that applying the inversion twice yields the identity, so that the image of F is F. Let P be any point distinct from O on the circle of diameter OF. Its image P must satisfy OFP = OPF = 90 and so it lies on the line perpendicular to OF. Hence the image of the line is the entire circle apart from O.
To construct our example, let \frakL be a line at distance 1 from O, with F the foot of the perpendicular from O to the line. Let q be any rational number with 0 < q < 1 and let Pq be selected on \frakL with PqOF = 2arctanq. Then tanPq OF = 2q(1 - q2)-1 so that |FPq | = 2q(1 - q2)-1 and |OPq | = (1 + q2)(1 - q2)-1. Hence all pairs of points Pq are rational distance apart and each Pq is a rational distance from O. Invert this line with respect to any circle with centre O and rational radius to obtain a circle through O. All images of points Pq lie on this circle, so no three are collinear. We can arrange these points in a sequence which satisfies the requirements.
5.
Second solution. See Figure 5.1. (a) Suppose that |AB| = d and let AOB = a. By the Law of Cosines, d2 = r2 + s2 - 2rscosa. Since |OA| = k2/r and |OB| = k2/s,
d2 = k4
r2
+ k4
s2
- 2k4
rs
cosa = k4
r2 s2
[r2 + s2 - 2rscosa] = k4d2
r2 s2
as desired. Observe that if k, r, s are rational, then d is rational if and only if d is rational.

(b) In the cartesian plane, let \frakC be the circle of radius 1 and centre (0, 0). Consider the line \frakL with equation x = 1. A point (x, y) is on the image of this line under inversion with respect to \frakC if and only if x > 0 and the point (1, y/x) on the ray through (0, 0) and (x, y) satisfies (x2 + y2)(1 + y2/x2) = 1. This is equivalent to x2 + y2 = x or (x - 1/2)2 + y2 = 1/4. Thus, the image of the line \frakL is the circle of radius 1/2 and centre (1/2, 0), and no three points on this circle are collinear.
To solve the problem, we select a sequence { Un } of points on \frakL for which |OUn | is rational and Un has rational coordinates, and let { Xn } be the images of these with respect to inversion in \frakC. But such a selection is possible since there are infinitely many rational pythagorean triples whose smallest number is 1. For example, we can take
Un ~

1, 2n2 + 2n
2n + 1


so that |OUn | = (2n2 + 2n + 1)/(2n + 1).


5.
Third solution. [Y. Shen] An alternative approach to the solution of the second part comes by the use of Ptolemy's Theorem. See Figure 5.3. Let a circle of diameter l be given and two additional points on the circle be given whose chords make angles a and b with the diameter. If the points are on the same side of the diameter and are distant l apart, we have, by the Law of Cosines,
l2
= cos2 a+ cos2 b-2cosacosb(cosacosb- sinasinb)
= cos2 a(1 - cos2 b) +cos2 b(1 - cos2 a) + 2cosacosbsinasinb)
= (cosasinb+ sinacosb)2
so that l = sinacosb+ cosasinb. On the other hand, if the points are on the same side of the diameter, then l2 = cos2 a+ cos2 b- 2cosacosb(cos(a- b)) so that l = cosasinb-cosbsina. We can locate infinitely many points on the circumference of the circle for which the sine and cosine of the angle its chord makes with a given diameter are rational, and the distance between any pair of these points will be rational.


Problem 6.

6.
(a) First solution. Let u = x + y and v = xy. Then u + v = 2 + 32 and u2 - 2v = 6. Thus, u2 + 2u = 10 + 62 so that (u + 1)2 = 11 + 62 = (3 + 2)2. Therefore
(u, v) = (2 + 2, 22)    or    (u, v) = (-4 -2, 6 + 42) .
In the first case, x and y are roots of the quadratic equation
0 = t2 - (2 + 2)t + 22 = (t - 2)(t - 2)
so that (x, y) = (2, 2), (2, 2). In the second case, x and y are roots of the quadratic equation
0 = t2 + (4 + 2)t + (6 + 42) ,
and these are nonreal.

6.
(a) Second solution. [D. Nicholson] Observe that (x + y + 1)2 = (x2 + y2) + 2(x + y + xy) + 1 = 11 + 62 = (3 + 2)2. Thus x + y = 2 + 2 or -4 - 22, and we can proceed as in the first solution.
6.
(b) First solution. The first equation can be transformed to
(x + y)2 - 2xy + 2xy
x + y
= 1
or to
0
= (x + y)3 - (x + y) - 2xy[(x + y) - 1]
= [(x + y) - 1][(x + y)2 + (x + y) - 2xy]
= [(x + y) - 1][(x2 + y2) + (x + y)]  .

Suppose that x + y = 1. Then the second equation becomes x2 - (1 - x) = 1 or 0 = x2 + x - 2 = (x + 2)(x - 1). Hence (x, y) = (-2, 3), (1, 0) and both these solutions check out.
Suppose that (x2 + y2) + (x + y) = 0. Then x and y cannot be real. Otherwise, x + y = -(x2 + y2) 0. By the first equation, x + y 0 and by the second x + y > 0 and we obtain a contradiction. From the second equation
x + y = x4 - 2x2 y + y2
0
= x4 - 2yx2 - x + (y2 - y)
= (x2 - x - y)(x2 + x - y + 1)
so that y = x2 - x or y = x2 + x + 1.

Let x2 + y2 + x + y = 0 and y = x2 - x. Then x2 = x + y = -(x2 + y2) y2 = -2x2 y = i2x x2 = (1 i2)x. Since x + y must be nonzero, x = 0 is inadmissible. Hence (x, y) = (1 + i2, -2 + i2), (1 - i2,-2 - i2), and both of these check out.
Let x2 + y2 + x + y = 0 and y = x2 + x + 1. Then y2 + 2y - 1 = 0 or (y + 1)2 = 2. Hence y = -1 2, so that x2 + x + (2 2) = 0. Hence
x =
-1 i


7 42

2
and we obtain a pair of complex solutions for (x, y). Note that, when y = x2 + x + 1, we have x + y = (x + 1)2 and so must take [(x + y)] = -(x + 1) in order to corroborate the equation [(x + y)] = x2 - y.

6.
(b) Second solution. Let x + y = u2 so that u = x2 - y = x2 -(u2 - x), Therefore 0 = x2 - u2 + (x - u) = (x - u)(x + u + 1), whence y = x2 - x or y = x2 + x + 1.
Plugging y = x2 - x into the first equation yields
0 = x4 - 2x3 + 2x2 + 2x - 3 = (x - 1)(x + 1)(x2 - 2x + 3)
so that
(x, y) = (1, 0), (-1, 2), (1 + i2,-2 + i2), (1 - i2,-2 - i2) .
All of these work except (x, y) = (-1, 2) which is extraneous.

Plugging y = x2 + x + 1 = (x+1)2 - x into the first equation and using x + y = (x + 1)2 yields
0
= x2 (x + 1)2 + [(x+1)2 - x]2(x+1)2+ 2x[(x + 1)2 - x] - (x + 1)2
= (x+1)6 - 2x(x + 1)4+ (2x2 + 2x - 1)(x + 1)2 - 2x2
= [(x+1)2 - 1][(x + 1)4 - (2x - 1)(x + 1)2 + 2x2]
= x(x + 2)(x4 + 2x3 + 5x2 + 4x + 2) .
The first two factors yield the possibilities (x, y) = (0, 1), (-2, 3); the solution (x, y) = (0, 1) is extraneous, but (x, y) = (-2, 3) checks out. Since
x4 + 2x3 + 5x2 + 4x + 2 = (x2 + x + 2)2 - 2 = (y + 1)2 - 2 ,
additional pairs of complex solutions can be obtained by solving
x2 + x + 2 = 2      y = -1 2 .

Comment. Since
x4 + 2x3 + 5x2 + 4x + 2 = (x2 + x + 1)2 + x2+ (x + 1)2 = (x2 + x)2 + (2x + 1)2 + 1 ,
we see that the quartic factor will yield no more real solutions. By noting that
x4 + 2x3 + 5x2 + 4x + 2 = (x2 + x)2 + 4(x2 + x) + 2 ,
we can find these solutions by obtaining and then solving x2 + x = -2 2.

We can get a picture of the situation for real solutions. See Figure 6(b). The equation [(x+y)] = x2 - y requires x + y 0 and y x2 to be viable for a real solution. The locus of this pair of inequalities is hatched. The first of the two given equations also requires x + y 0. Hence there are no real solutions with x2 + y2 + x + y = 0. However, when x + y = 1 there is one real solution (x, y) = (1, 0) for which y = x2 - x and one real solutions (x, y) = (-2, 3) for which y = x2 + x + 1. The putative solutions (x, y) = (-1, 2), (0, 1) lie outside the hatched region and so are extraneous.


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