advanced algebra and geometry workbook...

Advanced Algebra and GeometryWorkbook (Solutions)

Paul Yiu

Department of MathematicsFlorida Atlantic University

Student:

Fall 2016Last update: December 8, 2016

Advanced Algebra and Geometry 2016Workbook Checklist

Name:

A1 � A2 � A3 A4 � A5 � A6 � A7B1 B2 � B3 � B4 � B5 B6 � B7 �C1 � C2 � C3 � C4 C5 C6 C7 �D1 D2 D3 D4 D5 � D6 � D7 �E1 � E2 � E3 E4 E5 E6 E7E8 E9 � E10 � E11 � E12 �F1 � F2 � F3 � F4 F5 � F6 � F7 �F8 F9 F10 �G1 G2 � G3 � G4 � G5 � G6 G7 �G8 � G9 �H1 � H2 � H3 � H4 � H5 � H6 � H7 �

Total:

Problem A1 (Solution).(Two circles and a square)

Find the ratio of the radii of the two circles.

ba

b

2√ab

Solution. Let the radii be a and b.Each side of the square has length 2a.

(2√ab− a)2 + (2a− b)2 = b2

=⇒ 5a2 − 4a√ab = 0 =⇒ a : b = 16 : 25.

Remark. This configuration is contained in a right triangle with legs

Problem A2 (Solution).Given a point C on the diameter AB of a semicircle, a line CP perpen-

dicular to AB intersects the semicircle again at P . A circle is tangent to thesemicircle, CP , and CB at X . Prove that AX = AP .

R a r

r

XA C

P

BO

Solution. Let R be the radius of the semicircle, and OC = a.If the circle has radius r, then

r2 + (a+ r)2 = (R− r)2,

r2 + 2(R + a)r = R2 − a2 = CP 2,

(R + a+ r)2 = (R + a)2 + CP 2 = AP 2.

Therefore, AX = R + a+ r = AP .

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Problem A3 (Geogebra exercise).Four congruent incircles in an equilateral triangle: cosXBC = 3

4.

Make use of this to construct the following diagram.

A

B C

Z

X

Y

Problem A4 (Solution).Find the length of the side of the square in terms of r.

r − 2ar

r − a

2a

(r − a)2 + (r − 2a)2 = r2 =⇒ r = 5a.

Problem A5 (Solution).

BA

O

Suppose the chord AB is at distance h from the center O, andthe radius of the circle and the side of the square are both a.Then, the large circle has radius 2a+ h, and

(2a)2 + (a+ h)2 = (2a+ h)2 =⇒ a = 2h.

The distance from the center to the chord AB is 15

of the radius of the largecircle.

Problem A6 (Solution).Given a circle (O), a chord AB and two equilateral triangles as shown in

the diagram below,show that the distance of AB from the center is 1

7of the radius.

M X

Y

BA

O

Suppose the chord AB is at a distance h from the center O,the circle has radius 3a, and the equilateral triangle has side 2b.Then, the large circle has radius 6a+ h and MX =

√3a.

Considering the points B and Y on the large circle, we have

(√3a+ 2b)2 + h2 = (6a+ h)2,

(√3a+ b)2 + (

√3b+ h)2 = (6a+ h)2.

From these,

2√3ab− 2

√3bh = 0 =⇒ a = h and 2b = 3

√3h.

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Problem A7.Given three squares arranged in a rectangle of dimensions u× v, find the

area of the smallest square in terms of u and v.

u

v

Answer. 13u2−24uv+13v2

25.

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Problem B1. (Geogebra exercise)A chain of three circles tangent to chord and arc:

(R− r)2 = (R− h+ r)2 + 2hr =⇒ r =h(2R− h)

4R=

d2

4R.

Make use of this to construct the following diagram.

r

h

d d

Problem B2 (Geogebra exercise) (Solution).Three circles with centers on chord and tangent to arc:

(R− r)2 = (h+ r)2 + (R− h)2 =⇒ r =(R− h)h

R + h.

Make use of this to construct the following diagram.

O

C

MA B

D

P

Q

N

K

Solution. Let M be the midpoint of AB,and OC, OD the radii perpendicular and parallel to AB.Construct CD to intersect AB at P ,and let N be the midpoint of DP .Complete the square CMPQ.Construct QN to intersect AB at K.The circles M(P ) and K(P ) are tangent to each other,and each to the arc AB.

Problem B2 (Geogebra exercise) (Solution).Three circles with centers on chord and tangent to arc:

(R−r)2 = (h+r)2+(R−h)2 =⇒ r =(R− h)h

R + h=⇒ r+h =

2Rh

R + h.

Make use of this to construct the following diagram.

O

C

MA

BP

Q

X Y

ZK

Solution. Let R be the radius of the arc,and M , C the midpoints of the chord and the arc AB.Complete the squares CMPQand MXY Z with X on the line OC and MX = XY = R.Join QY to intersect AB at K.Then MK = 2Rh

R+h, and

the circles M(P ), K(P ) touch each other, and also the arc AB.

Problem B3 (Solution).

r

r

A

B CT

Z

X

Y

θ

Solution. (a) AZ + ZC = AC =⇒ r cot 30◦ + r cot θ = 2 =⇒ r =2

cot 30◦+cot θ.

(b) ∠BCY = 12(60◦ − 2θ) = 30◦ − θ.

(c) In triangle Y CT ,

r = tan(30◦ − θ) =tan 30◦ − tan θ

1 + tan 30◦ tan θ=

cot θ − cot 30◦

cot 30◦ cot θ + 1.

(d) Note that cot 30◦ =√3. Putting x = cot θ in the expressions for r, we

have 2√3+x

= x−√3√

3x+1=⇒ x2 − 3 = 2

√3x+ 2 =⇒ x2 − 2

√3x− 5 = 0,

and x =√3 + 2

√2, rejecting the negative root.

(e) From (a), r = 2√3+x

= 1√3+

√2=

√3−√

2.

Problem B4 (Solution).An equilateral triangle of side 2a is partitioned symmetrically into a

quadrilateral, an isosceles triangle, and two other congruent triangles. Itis known that if the incircles of the quadrilateral and the isosceles are con-gruent, then their common radius is (

√3−√

2)a.Make use of this information to construct the diagram.

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Problem B5.ABC is a right triangle with points X on AB, Y , Z on BC forming an

equilateral triangle. XPQR is a square with vertices P and Q on AB andBC respectively. Suppose each side of the equilateral triangle has length 1.

C Y ZA

B

Q

R

P

X

I J

(a) Show that A = 30◦.(b) Calculate the lengths of AR and RY .(c) Calculate the radius of the circle (I) inscribed in the quadrilateralQRY C.(d) Calculate the radius of the circle (J) inscribed in the triangle AXZ.(e) Show that the points CX , I , X are collinear.

Problem B6 (Geogebra exercise) (Solution).Construct the following diagram in which the circles and lines intersect

tangentially.

Solution. Let the radii be a > b > c.a = 2b+ c =⇒ a+ c = 2(b+ c).Three centers form a right triangle with shorter sides in the ratio 1 : 2.The points of tangency of the incircle of the right triangle with the sidesdetermine the radii of the three circles.

Problem B7 (Geogebra exercise) (Solution).Construct the following diagram in which the circles and lines intersect

tangentially.

Solution. The radii are in the ratio 1 : 2 : 6.Three centers form an equilateral triangle.

Problem C1 (Solution).Solution. (a) Let R be the radius of the circle (O), and r that of the re-quired circle.

r2 + (R− h+ r)2 = (R− r)2

=⇒ r2 + 2(2R− h)r = 2Rh− h2

=⇒ (r + 2R− h)2 = 2Rh− h2 + (2R− h)2 = 2R(2R− h).

Therefore,

r =√

2R(2R− h)− (2R− h)

=√CD · CM − CM

= CA− CM = CP − CM = MP.

O

B A

C

C

P

M Q

K

T

(b) Complete the arc AB into a circle, center O.Let CD be the diameter perpendicular to AB, with C on the opposite sideof the given arc.Construct the arc, center C, through A and B, to intersect CD at P .Let M be the midpoint of AB. Complete MP into a square MPKQ.The circle, center K, radius KP , is tangent to AB at Q, CD at P , and thegiven arc AB.

Problem C2 (Solution).In the following diagram the arcs BC and AC have centers A and B re-

spectively, and PQ is parallel to AB. If AB = R and the distance betweenPQ and AB is h, calculate the radius of the small circle.

QP

C

A B

Solution. Let AB = R, and r be the radius of the circle.

(R− r)2 =

(R

2

)2

+ (r + h)2.

From this,

r =3R2 − 4h2

8(R + h).

Problem C3 (Geogebra exercise) (Solution).Construct the following diagram in which the arcs BC and AC have

centers A and B respectively, and PQ is parallel to AB.

A B

C

C

M

P QYX

E

F

K

Solution. Let D and M be the midpoints of AB and PQ.

Construct pointsX and Y on the line PQ such thatDX = DY = DC(=

√3R2

).

Note that MX = MY =√

3R2

4− h2.

Let E be a point on the line CD such that DE = AB and C, E are onopposite sides of AB.Construct the circle through E, X , Y to intersect the line CD at F .By the intersecting chords theorem,

MF ·ME = MX ·MY =⇒ MF =3R2 − 4h2

4(R + h)= 2r.

Therefore, MF is a diameter of the required circle.

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Problem C4.In the following diagram, the three small circles are congruent, and one

them touches the chord AB at its midpoint. Suppose circle (O) has radiusR and MN = h. Prove that NP =

√2h(R− h).

A B

O

P

N

M

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Problem C5 (Geogebra exercise).In the following diagram, the three small circles are congruent, and one

them touches the chord AB at its midpoint N . Construct the circle (K) byfirst locating the point of tangency P .

A B

O

K

P

N

M

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Problem C6 (Geogebra exercise).Construct the following diagram with three circles each tangent internally

to the circumcircle at a vertex, and externally to the incircle of a giventriangle.

OI

A

B C

Problem C7 (Solution).Two chords AX and BY intersect at a point P on CM , where C and

M are the midpoints of the arc AB and the chord AB. The two circles aretangent to the chords AX , BY , and also to the arc AB.If CP = u and PM = v, calculate the common radii of the circles.

x

u

v r

R

P

O

A B

XY

M

C

x2 = (R− r)2 − (R− u)2,

r2

x2=

v2

v2 +R2 − (R− u− v)2.

((u+ v)r − uv)((2R− u− v)r + (2R− u)v) = 0,

r =uv

u+ vor − (2R− u)v

2R− u− v.

Since r is positive, it must be uvu+v

.

Problem D5 (Solution).(a) Show that the equation of the circle with (x1, y1) and (x2, y2) as end-

points of a diameter is

(x− x1)(x− x2) + (y − y1)(y − y2) = 0.

(b) Find the coordinates of the endpoints of the diameter perpendicularto the one in (a).Solution. (a) Let A(x1, y1) and B(x2, y2) be the given points.A point P is on the circle with diameter AB if and only ifAP ⊥ BP , i.e., the product of the slopes of AP and BP is equal to −1:

y − y1x− x1

· y − y2x− x2

= −1.

Clearing denominators, we have

(x− x1)(x− x2) + (y − y1)(y − y2) = 0.

(b) The center M of the circle is the midpoint of AB:

M =

(x1 + x2

2,y1 + y2

2

).

The endpoints of the diameter perpendicular to AB are obtained by rotatingA and B about M through 90◦. These are

A′ =(x1 + x2

2,y1 + y2

2

)+

(−y1 − y2

2,x1 − x2

2

)

=1

2(x1 + x2 − y1 + y2, x1 − x2 + y1 + y2) ,

B′ =(x1 + x2

2,y1 + y2

2

)+

(−y2 − y1

2,x2 − x1

2

)

=1

2(x1 + x2 + y1 − y2, −x1 + x2 + y1 + y2) .

Problem D6 (Solution).(a) Find the equations of the four tangents to the circles

x2 + y2 − 2x− 4y = 20,

x2 + y2 − 10x+ 4y = 20

at their points of intersection.(b) Show that these tangents are tangent to a circle with center

(83, 1

3

).

AD

B

D

P

Solution. (a) The radical axis of the two circles is the line y = x.This intersects the circles at A(5, 5) and B(−2,−2).The tangents at A are 5x + 5y − (x + 5) − 2(y + 5) = 20, 4x + 3y = 35,and 5x+ 5y − 5(x+ 5) + 2(y + 5) = 20, y = 5.The tangents at B are −2x−2y− (x−2)−2(y−2) = 20, 3x+4y = −14,and −2x− 2y − 5(x− 2) + 2(y − 2) = 20, x = −2.

(b) These four lines bound a quadrilateral with vertices

A(5, 5), C(−2, 5), B(−2,−2), D(26,−23).

This is a symmetric quadrilateral with AC = BC = 7 and AD = BD =35. There is a circle tangent to the four lines.It center is the point P on the diagonal CD, dividing CD in the ratio CP :PD = 7 : 35 = 1 : 5.This is the point 5(−2,5)+(26,−23)

6=

(83, 1

3

).

The radius of the circle is 143

. The equation of the circle is9x2 + 9y2 − 48x− 6y = 131.

Problem D7 (Solution).Find the equation of the circle which passes through the common points

of the circles

x2 + y2 + 4x− 6y − 3 = 0,

x2 + y2 − 2x+ 8y − 8 = 0

and cuts the first circle orthogonally.Solution. A circle through the common points of the given circles hasequation

x2 + y2 + t(4x− 6y − 3) + (1− t)(−2x+ 8y − 8) = 0

for some t.

x2+ y2− 2(1− 3t)x− 2(−4+7t)y+(5t− 8) = 0; (x− 1+3t)2+(y+4 − 7t)2 = 25 − 67t + 58t2. This has center (1 − 3t,−4 + 7t) and radius√25− 67t+ 58t2.This is orthogonal to the first circle (with center (−2, 3) and radius 4)

if and only if (3− 3t)2 + (−7 + 7t)2 = 42 + 25− 67t+ 58t2.Equivalently, 58(1− t)2 = 41− 67t+ 58t2, 17− 49t = 0, t = 17

49.

This gives the circle 49x2 + 49y2 + 4x+ 154y − 307 = 0.

Problem E1 (Solution).Find the values of a, b, c so that the polynomial

x5 + ax4 + bx3 + cx2 + 3x+ 2

is divisible by x3 + 1.Solution. Divide.

x5+ax4+bx3+cx2+3x+2 = (x3+1)(x2+ax+b)+(c−1)x2+(3−a)x+(2−b).

For exact divisibility, the remainder must be zero. Therefore, a = 3,b = 2, c = 1.

Problem E2 (Solution).Solve the equation

(x+ 1)(x+ 2)(x+ 3)(x+ 4) = 24.

Solution.

(x+ 1)(x+ 2)(x+ 3)(x+ 4) = 24

(x+ 1)(x+ 4)(x+ 2)(x+ 3) = 24

(x2 + 5x+ 4)(x2 + 5x+ 6) = 24

(x2 + 5x+ 4)2 + 2(x2 + 5x+ 4)− 24 = 0

(x2 + 5x+ 4− 4)(x2 + 5x+ 4 + 6) = 0

x(x+ 5)(x2 + 5x+ 10) = 0.

Therefore, x = 0, 5, or 12(−5±√−15).

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Problem E3.(a) Let a, b, c be distinct real numbers. Show that every quadratic poly-

nomial f(x) can be written in the form

f(x) = A(x− b)(x− c) +B(x− c)(x− a) + C(x− a)(x− b)

for appropriate choice of constants A, B, C.(b) F (x) is a polynomial which leaves remainders 1, 2, and 3 respectively

when divided by x − 1, x − 2, and x − 3. Find the remainder when F (x)is divided by (x − 1)(x − 2)(x − 3). [Note: this remainder is a quadraticpolynomial in x.]

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Problem E4.Factorize the following symmetric and alternating functions(1)

∑a(b− c)3

(2)∑

a4(b2 − c2)

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Problem E5.Factorize the following symmetric and alternating functions(1) (a+ b+ c)3 − (b+ c− a)3 − (c+ a− b)3 − (a+ b− c)3.(2) a(b− c)2 + b(c− a)2 + c(a− b)2 + 8abc.

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Problem E6.Factorize 2x2 − 5xy − 3y2 − x− 25y − 28.

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Problem E7.Show that a2 + b2 + c2 − bc− ca− ab cannot be negative. Hence, show

that the roots of equation

3x2 − 2(a+ b+ c)x+ (bc+ ca+ ab) = 0

must be real.

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Problem E8.If ω is a complex cube root of unity, and

x = a+ b, y = aω + bω2, z = aω2 + bω,

show that (i) xyz = a3 + b3, and (ii) x2 + y2 + z2 = 6ab.

Problem E9 (Solution).(a) If two adjacent vertices of a square have complex affixes z1 and z2,

what are the complex affixes of the other two vertices?

z1

z2

z1

z2

(b) If two opposite vertices of a square have complex affixes z1 and z2,what are the complex affixes of the other two vertices?Solution.

(a) (1) z4 = z1 + (z2 − z1)i = (1− i)z1 + iz2.(2) z1+z3 = z2+z4 =⇒ z3 = −z1+z2+z4 = −z1+z2+(1−i)z1+iz2 =−iz1 + (1 + i)z2.

(b) Let z0 = 12(z1 + z2) be the midpoint of z1z2.

w1 is obtained by rotating z1 counterclockwise about z0 through 90◦ coun-terclockwise.Since a counterclockwise rotation through 90◦ is multiplication by i,

w1 = z0 + (z1 − z0)i =z1 + z2

2+

(z1 − z2)i

2=

(1 + i)z1 + (1− i)z22

.

Similarly,

w2 = z0 + (z2 − z0)i =z1 + z2

2+

(z2 − z1)i

2=

(1− i)z1 + (1 + i)z22

.

Problem E10 (Solution).Without using calculus, find the maximum and minimum values of

y =7x2 + 4x− 1

x2 + x+ 1,

where a, b, c are constants.Solution. Note that x2 + x+ 1 > 0 for all x. Therefore,

(7− y)x2 + (4− y)x− (1 + y) = 0.

For this to have real solutions in x,

(4− y)2 + 4(7− y)(1 + y) ≥ 0.

This gives (y + 2)(y − 22

3

) ≤ 0. Therefore, −2 ≤ y ≤ 223

. The minimumand maximum of y are −2 and 22

3respectively

Problem E11 (Solution).Find a polynomial of lowest degree, and with integer coefficients, which

admits 1 +√2 +

√3 as a root.

Solution. The polynomial must also admit 1 −√2 +

√3, 1 − √

2 +√3,

and 1−√2−√

3 as its roots. The one with lowest degree is the quartic

(x− 1−√2−

√3)(x− 1−

√2 +

√3)(x− 1 +

√2−

√3)(x− 1 +

√2 +

√3)

= x4 − 4x3 − 4x2 + 16x− 8.

Problem E12 (Solution).Solve the equation

(x2 − 9x+ 18)(x2 − 9x+ 20) = 24.

Solution. Let y = x2−9x+18. We have y(y+2) = 24 =⇒ (y−4)(y+6) = 0, y = 4 or −6.

If y = 4, x2− 9x+18 = 4, x2− 9x+14 = 0, (x− 2)(x+7) = 0, x = 2or 7.

If y = −6, x2 − 9x+ 18 = −6, x2 − 9x+ 24 = 0, x = 12(9±√−15).

Problem F1 (Solution).Construct a cubic polynomial with integer coefficients, one of whose

roots is −3, and a second one is −12(1 +

√−23).Solution. (x+ 3)(x2 + x+ 6) = x3 + 4x2 + 9x+ 18.

Problem F2 (Solution).Show that roots of the cubic equation

x3 + 3ax2 + 3bx+ c = 0

are in arithmetic progression if and only if

2a3 − 3ab+ c = 0.

Solution. Suppose the roots are α, β, γ with 2β = α + γ. Thenα + β + γ = −3a =⇒ 3β = −3a, and β = −a.It follows that (−a)3+3a·(−a)2+3b(−a)+ c = 0, and 2a3− 3ab+ c = 0.

Problem F3 (Solution).One root of the cubic equation

x3 + 4x2 − 12x− 27 = 0

is the geometric mean of the other two roots. Solve the equation.Solution. Suppose the roots are α, β, γ with γ2 = αβ. Then(i) α + β + γ = −4,(ii) αβ + αγ + βγ = −12 =⇒ γ(α + β + γ) = 12 =⇒ 4γ = 12, andγ = 3.

The cubic factors as (x − 3)(x2 + 7x + 9) = 0. The other two roots are12(−7±√

13).

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Problem F4.Show that if

x+ y + z = 3, x2 + y2 + z2 = 5, x3 + y3 + z3 = 7,

then x4 + y4 + z4 = 9.

Problem F5 (Solution).Solve the cubic equations(1) x3 − 6x2 + 11x− 6 = 0.(2) (x− 1)3 + (x− 2)3 = 0.

Solution. (1) Clearly, x = 1 is a solution. Dividing the cubic by x− 1, wehave x2 − 5x+ 6 = (x− 2)(x− 3).The other two roots are x = 2 and x− 3.

(2) Solution. Put y = x−12−x

. The equation becomes y3 = 1.The solutions are y = 1, ω, ω2.Since y = x−1

2−x=⇒ x = 2y+1

y+1, the roots of the given cubic equation are 3

2,

2ω+11+ω

= −2ω+1ω2 = −ω(2ω + 1) = −(2ω2 + ω) = −(−1 + ω2) = 1 − ω2,

and2ω2+11+ω2 = 1− ω.

Problem F6 (Solution).Solve the cubic equation y3 + 9y + 6 = 0.

Answer. 3√3− 3

√4, − 3

√3ω − 3

√4ω2, 3

√3ω2 − 3

√4ω.

Problem F7 (Solution).Solve the cubic equation y3 − 3y + 1 = 0 by reducing it to an angle

trisection problem.Answer. 2 sin 10◦, 2 sin 50◦, and −2 sin 70◦.

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Problem F8.If α, β, γ are the roots of the equation x3 − x2 − 2 = 0, construct the

equation whose roots are

α

β + γ − α,

β

γ + α− β,

γ

α + β − γ.

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Problem F9.ABCD is a square of unit side. P is a point on BC so that the incircle

of triangle ABP and the circle tangent to the lines AP , PC and CD haveequal radii. Show that the length of BP satisfies the equation

2x3 − 2x2 + 2x− 1 = 0.

A B

P

CD

0,0270

Problem F10 (Solution).Suppose each side of the equilateral triangle ABC has length 4, and that

the three incircles are congruent. Prove that their common radius is 312 −3

16 .

y

x4− 2x

4

r r r

A

B CP Q

Solution. Suppose each side of the equilateral triangle has length 4. LetCQ = x and AQ = y. From the congruent incircles,

x

4 + x+ y=

4− 2x

4− 2x+ 2y=⇒ y =

4− 2x

x− 1.

In triangle ACY , y2 = x2 − 4x+ 16. Eliminating y, we obtain,

x3 − 6x2 + 21x− 20 = 0.

With x = t+ 2, the cubic equation is transformed into

t3 + 9t+ 6 = 0.

This has a unique real root t = −323 + 3

13 . From this,

x = 2− 323 + 3

13 , y = 3

23 + 3

13 .

The area of the equilateral triangle being Δ = 4√3, the common inradius r

is given by Δ = 12(12 + 4y)r = 2(3 + y)r, r = Δ

2(3+y). Therefore,

r =2√3

3 + 323 + 3

13

=2√3 · 3− 1

3

323 + 3

13 + 1

=2 · 3 1

6 (313 − 1)

(313 − 1)(3

23 + 3

13 + 1)

=2 · 3 1

6 (313 − 1)

(313 )3 − 1

= 312 − 3

16 .

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Problem G1.Find the roots of the quartic polynomials given that they are factorizable.(1) x4 − 4x3 + 3x2 + 2x− 1 = 0.(2) x4 − 6x3 + 14x2 − 20x+ 8 = 0.

Problem G2 (Solution).Solve the quartic equation

(x− 1)4 + (x− 2)4 = 1.

Solution. Clearly x = 1 and x = 2 are solutions of the equation. Dividing(x − 1)4 + (x − 2)4 − 1 by (x − 1)(x − 2) we have 2(x2 − 3x + 4). Theother two roots are x = 1

2(3±√−7).

Problem G3 (Solution). Solve the quartic equation

x4 − 4x3 + 7x2 − 6x− 28 = 0,

given that one of its roots is 1 +√5.

Solution. A second root is 1−√5. The quartic has a quadratic factor

(x− 1−√5)(x− 1 +

√5) = (x− 1)2 − 5 = x2 − 2x− 4.

Division gives the other factor as x2 − 2x+ 7.The other two roots are 1±√−6.

Problem G4 (Solution).Construct a quartic polynomial with integer coefficients, one of whose

roots is√5−√

2.Solution. The other three roots must be

√5+

√2, −√

5−√2, and −√

5+√2.The quartic polynomial is

(x−√5 +

√2)(x−

√5−

√2)(x+

√5 +

√2)(x+

√5−

√2)

= x4 − 14x2 + 9.

Problem G5 (Solution).Solve the quartic equation x4 − 4x3 + 3x2 + 2x+ 2 = 0.

Solution. x = 12(2±√−1±√

7).

Problem G6 (Solution).Solve the quartic equation 2x4 − 4x3 + 4x2 − 2x+ 2 = 0.

Answer.

Problem G7 (Solution).Solve the quartic equation x4 − 3x2 + 4x− 3 = 0.

Answer. x = −ω, −ω2, 12(−1±√

13).

Name:

Problem G8.Solve the quartic equation x4 − x2 + 2x+ 2 = 0.

Solution. x = −1, −1, 1±√−1.

Problem G9 (Solution).Solve the quartic equation x4 + 3x− 2 = 0.

Answer. x = 12(−1±√

5), 12(1±√−7).

Problem H1 (Solution). (a) Suppose you want to solve the quarticequation

x4 + ax2 + bx+ c = 0

by intersecting the parabola y = x2 with a circle, so that the x-coordinatesof the intersections are the roots of the equation. What are the center andthe radius of the circle you should draw?Answer. Center

(− b2, 1

2− a

2

), and square radius (a−1)2+b2−4c

4.

(b) Let d > 0. Suppose you want to solve the quartic equation

x4 + ax3 + bx2 + cx+ d = 0

by intersecting the hyperbola xy =√dwith a circle so that the x-coordinates

of the intersections are the roots of the equation. What are the center andthe radius of the circle you should draw?

Answer. Center(−a

2, − c

2√d

), and square radius

a2−4b+ c2

d

4.

Problem H2 (Solution). Consider the problem of construction of tri-angle ABC from the vertex A, the orthocenter H , and the trace Tb of thebisector of angle B.Put the origin at A, and let Tb = (a, 0), H = (h, k).

(a) Show that the x-coordinate of B is h.The line ATb contains the vertex C, and is the x-axis. Therefore, B lies

on the vertical line through H(h, k), and has x-coordinate h.

(b) Suppose B = (h, q). Show that C is the point(

h2+kqh

, 0)

.

AB is the line qx− hy = 0. It has slope qh. The perpendicular to from H

to AB is the line y − k = −hq(x− h),

h(x− h) + q(y − k) = 0.

This intersects AB at C, which has coordinates given by(

h2+kqh

, 0)

.

(c) Use the angle bisector theorem to show that

kq3 + h(h− 2a)q2 + (h2 − a2)kq + h2(h− a)2 = 0.

Since ATb = a, TbC = h2+kqh

− a = h2−ah+kqh

, by the angle bisectortheorem, AB2 : CB2 = AT 2

b : TbC2.

h2 + q2 :k2q2

h2+ q2 = a2 :

(h2 − ah+ kq)2

h2,

(h2 + q2)(h2 − ah+ kq)2 = a2(k2q2 + h2q2),

(h2 + q2)(kq3 + h(h− 2a)q2 + (h2 − a2)kq + h2(h− a)2) = 0.

Since h2 + q2 �= 0, we must have

kq3 + h(h− 2a)q2 + (h2 − a2)kq + h2(h− a)2 = 0.

Problem H3 (Solution). [Continuing Problem H2]Consider the problem of construction of triangle ABC from the vertex

A, the orthocenter H , and the trace Tb of the bisector of angle B.Put the origin at A, and let Tb = (a, 0), H = (h, k).

It is known (from Problem H2) that if B = (h, q), then

kq3 + h(h− 2a)q2 + (h2 − a2)kq + h2(h− a)2 = 0.

(1) Let (a, h, k) = (1, 2, 2).(a) Show that the equation has no rational roots.

The equation becomes q3 + 3q + 2 = 0. Since ±1, ±2 do notsatisfy the equation, it has no rational roots.

(b) Show that the roots are the intersections of the the rectangularhyperbola xy = 1 and the parabola y2 + 2x+ 3 = 0.Eliminating x, we have y2 + 2

y+ 3 = 0 =⇒ y3 + 3y + 2 = 0,

whose roots are the q’s in (a).

(2) Let (a, h, k) = (1, 2, 1). Show that there is one real solution, whichis constructible. Compute the coordinates of the vertices.

The equation becomes (q + 1)(q2 − q + 4) = 0. It has a rationalroot q = −1.B = (2,−1), C =

(32, 0

).

(3) Let (a, h, k) = (3, 2, 1). Show that there are three real solutions.One of these has vertices with rational coordinates. Compute thecoordinates of the vertices.

The equation becomes (q + 1)(q2 − 9q + 4) = 0. It has rootsq = −1, 1

2(9±√

65).

q B C

(i) −1 (2,−1)(32, 0

)(ii) 1

2(9 +

√65) (2,−1)

(14(17 +

√65), 0

)(iii) 1

2(9−√

65) (2,−1)(14(17−√

65), 0)

Problem H4 (Solution). (Geogebra exercise) Illustrate the followingstatement:Two parabolas have a common focus F ; their directrices intersects at apoint A. The perpendicular bisector of AF is the common tangent of thetwo parabolas.Solution. Let F = (a, 0) and A = (−a, 0) so that the perpendicular bisec-tor of AF is the y-axis.

Consider a line of slope m through A. It has equation y−m(x+ a) = 0.The parabola with F as focus and this line as directrix is given by

(x− a)2 + y2 =(y −m(x+ a))2

1 +m2.

Simplifying, we have

(1 +m2)((x− a)2 + y2)− (mx− y +ma)2 = 0.

With x = 0, this reduces to

0 = (1 +m2)(a2 + y2)− (y −ma)2 = a2 + 2amy +m2y2 = (a+my)2.

The parabola is tangent to the y-axis at the point(0, − a

m

).

Summary: Given two distinct points F and A, the parabola with focus Fand directrix a line through A is tangent to the perpendicular bisector ofAF . This perpendicular bisector is the common tangent of all parabolaswith common focus F and directrices lines through A.

Problem H5 (Solution). (Geogebra exercise) Illustrate the followingstatement:Tangents drawn to two confocal parabolas from a point on the commontangent intersect at the same angle as the axes of the parabolas.Solution. Let F = (a, 0) and A = (−a, 0). The parabola with focus Fand directrix the line through A with slope m has equation

(1 +m2)((x− a)2 + y2)− (mx− y +ma)2 = 0. (1)

Equivalently,

x2 + 2mxy +m2y2 − 2a(1 + 2m2)x+ 2may + a2 = 0

Let P (0, t) be a point on the y-axis. We find the equation of the tangentsfrom P to the parabola. One of them is the y-axis x = 0. Let y = kx+ t bethe other, where k is the slope of the tangent. Substitution into (1) gives

(1 + km)2x2 + 2(−a+ akm− 2am2 +mt+ km2t)x+ (a+mt)2 = 0.

This has equal roots if and only if

(−a+ akm− 2am2 +mt+ km2t)2 − (1 + km)2(a+mt)2 = 0.

Equivalently,

ma(1 +m2)((am− t)− (a+mt)k) = 0.

Therefore, k = ma−ta+mt

.The axis of the parabola, being perpendicular to the the directrix, has

slope − 1m

.The angle between the tangent and the axis is therefore

arctanma−ta+mt

+ 1m

1− ma−ta+mt

· 1m

= arctana

t.

This angle is independent of the choice of the directrix.

Problem H6 (Solution). Prove that the locus of points from which thetwo tangents to E are perpendicular is the circle

x2 + y2 = a2 + b2.

Solution. Let (u, v) be a point whose tangents to the ellipse x2

a2+ y2

b2= 1

are perpendicular. If the points of tangency are (p1, q1) and (p2, q2), thecondition for perpendicularity is

(p1 − p2)2 + (q1 − q2)

2 = (u− p1)2 + (v − q1)

2 + (u− p2)2 + (v − q2)

2,

u2 + v2 − (p1 + P2)u− (q1 + q2)v + (p1p2 + q1q2) = 0.

Now, these points of tangency are the intersections of the ellipse and thepolar of (u, v), which is the line

ux

a2+

vy

b2= 0.

Eliminating y from this and the ellipse equation, we have

(a2v2 + b2u2)x2 − 2a2b2ux+ a4(b2 − v2) = 0.

From this,

p1 + p2 =2a2b2u

a2v2 + b2u2, p1p2 =

a4(b2 − v2)

a2v2 + b2u2.

Similarly, eliminating x from the polar and ellipse equations, we have

(a2v2 + b2u2)y2 − 2a2b2vy + b4(a2 − u2) = 0,

and

q1 + q2 =2a2b2v

a2v2 + b2u2, q1q2 =

b4(a2 − u2)

a2v2 + b2u2.

Therefore, (u, v) satisfies

u2 + v2 − 2a2b2u2

a2v2 + b2u2− 2a2b2v2

a2v2 + b2u2

+a4(b2 − v2)

a2v2 + b2u2+

b4(a2 − u2)

a2v2 + b2u2= 0,

(a2v2 + b2u2)(u2 + v2)− 2a2b2(u2 + v2) + a4(b2 − v2) + b4(a2 − u2) = 0.

This factors as

(x2 + y2 − a2 − b2)(a2b2 − b2u2 − a2v2) = 0.

The second factor represents the given ellipse. Thus the locus of the pointsfrom which the tangents to the ellipse are perpendicular is the circle

x2 + y2 = a2 + b2.

Problem H7 (Solution). In a right triangle of sides a and b, an ellipseof semi-axes h, k (parallel to the sides) is inscribed. Show that

2(a− h)(b− k) = ab.

C B

A

Z

X

Y

Solution. The line AB has equation xa+ y

b= 1, or y = b(a−x)

a. The x-

coordinates of the intersections of this line with the ellipse

(x− h)2

h2+

(y − k)2

k21

are the roots of(x− h)2

h2+

(b(a− x)− ak)2

a2k2= 1,

or

(b2h2 + a2k2)x2 − 2ah(b2h− bhk + ak2)x+ a2h2(b− k)2 = 0.

For tangency, these quadratic equation has two equal roots, i.e.

a2h2(b2h− bhk + ak2)2 = (b2h2 + a2k2)a2h2(b− k)2

(b2h− bhk + ak2)2 − (b− k)2(b2h2 + a2k2) = 0

abk2(−ab+ 2bh+ 2ak − 2hk) = 0

ab− 2bh− 2ak + 2hk = 0

2(a− h)(b− k) = ab.