Problem Set 4
 19.

Let f(1) = 1 and f(2) = 3. Suppose that, for
n ³ 3, f(n) = max{ f(r) + f(nr) : 1 £ r £ n1 }.
Determine necessary and sufficient conditions on the pair (a, b)
that f(a + b) = f(a) + f(b).
View solution
 20.

Suppose that a and c are fixed real numbers with
a £ 1 £ c. Determine the largest value of b which is
compatible with a + bc £ b + ac £ c + ab.
View solution
 21.

Solve the following system for real values of x and
y:
2^{x2 + y} + 2^{x + y2} = 8 

View solution
 22.

Let ABC be a triangle with BC = 2·AC  2·AB and D be a point on the side BC. Prove that
ÐABD = 2ÐADB if and only if BD = 3CD.
View solution
 23.

Two unequal spheres in contact have a common tangent cone.
The three surfaces divide space into various parts, only one of which
is bounded by all three surfaces; it is ``ringshaped''. Being
given the radii r and R of the spheres with r < R, find the
volume of the ``ringshaped'' region in terms of r and R.
View solution
 24.

Let ABC be a triangle. Select points D, E, F
outside of DABC such that DDBC, DEAC,
DFAB are all isosceles with the equal sides meeting
at these outside points and with ÐD = ÐE = ÐF. Prove that the lines AD, BE and CF all
intersect in a common point.
View solution
Solutions to Problem Set 4
19.
 19.

First solution. From the first few values of
f(n), we conjecture that f(2k) = 3k and f(2k + 1) = 3k + 1 for each positive integer k. We establish this by
induction. It is easily checked for k = 1. Suppose that it holds
up to k = m.


Suppose that 2m + 2 is the sum of two positive even numbers
2x and 2y. Then f(2x) + f(2y) = 3(x + y) = 3(m + 1). If
2m + 2 is the sum of two positive odd numbers 2u + 1 and 2v + 1,
then
f(2u + 1) + f(2v + 1) = (3u + 1) + (3v + 1) = 3(u + v) + 2 < 3(u + v + 1) = 3(m + 1) . 

Hence f(2(m + 1)) = 3(m + 1).


Suppose 2m + 3 is the sum of 2z and 2w + 1. Then
z + w = m + 1 and
f(2z) + f(2w + 1) = 3z + 3w + 1 = 3(z + w) + 1 = 3(m + 1) + 1 . 

Hence f(2(m+1) + 1) = 3(m+1) + 1.
The conjecture is established by induction.


By checking cases on the parity of a and b, one verifies that
f(a + b) = f(a) + f(b) if and only if at least one of a and
b is even. (If a and b are both odd, the left side is
divisible by 3 while the right side is not.)
 19.

Second solution. [K. Yeats] By inspection, we
conjecture that f(n+1) = f(n) + 2 when n is odd, and
f(n+1) = f(n) + 1 when n is even. This is true for
n = 1, 2. Suppose it holds up to n = 2k. If 2k + 1 = i + j with i even and j odd, then f(i1) + f(j+1) = f(i)  2 + f(j) + 2 = f(i) + f(j) and f(i+1) + f(j1) = f(i) + 1 + f(j)  1 = f(i) + f(j) (where defined),
so in particular f(2k+1) = f(2k) + f(1) = f(2k) + 1.
Note that this also tells us that f(2k+1) = f(i) + f(j)
whenever i + j = 2k+1. Now consider 2k+2 = i + j.
If i and j are both even, then
f(i+1) + f(j1) = f(i) + 1  f(j)  2 = f(i) + f(j)  1 

while if i and j are both odd, then
f(i+1) + f(j1) = f(i) + 2  f(j)  1 = f(i) + f(j) + 1 . 

Thus, f(2k+2) = f(i) + f(j) if and only if i and j are
both even. In particular, f(2k+2) = f(2k) + f(2) = f(2k+1)  1 + 3 = f(2k) + 2. We thus find that
f(a+b) = f(a) + f(b) if and only if at least one of a and
b is even.
20.
 20.

First solution. Observe that
(b + ac)  (a + bc) = (c  1)(a  b) and (c + ab)  (b + ac) = (1  a)(c  b), so that the inequalities are equivalent
to
(c  1)(a  b) ³ 0 and (1  a)(c  b) ³ 0 . 

If a = c = 1, then the inequalities hold for all values of b and
so there is no largest value of b. If a = 1 < c, then the
inequalities hold if and only if b
£ a = 1, so that largest
value of b is 1. If a < 1 = c, then the inequalities hold
if and only if b
£ c = 1, so the largest value of b is 1.
Finally, if a < 1 < c, then the inequalities hold if and only
if b
£ a, so the largest value of b is a.
21.
 21.

Preliminary comments. With the surds in the
second equation, we must restrict ourselves to nonnegative
values of x. Because of the complexity of
the expressions, it is probably impossible to eliminate one of the
variables and solve for the other. Let us make a few preliminary
observations:


(i) (x, y) = (1, 1) is an obvious solution;


(ii) Both equations are symmetric in x and y;


(iii) Taking f(x, y) = 2^{x2 + y} + 2^{x + y2} and
g(x, y) = Öx + Öy, we have that
f(0, y) = 2^{y} + 2^{y2} and g(0, y) = Öy; thus,
f(0, y) = 8 Þ 1 < y < 2 and g(0, y) = 2Û y = 4. The graphs of f(x, y) = 8 and
g(x, y) = 2 may look like the figures in the diagram.


This suggests that f(x, y) = 8 Þ x + y £ 2
and g(x, y) = 2 Þ x + y ³ 2 with equality for both
Û (x, y) = (1, 1). Hence we look for a relationship
among f(x, y), g(x, y) and x + y.


First solution.
(Öx + Öy)^{2} = x + 2 
 __ Öxy

+ y £ x + (x + y) + y = 2(x + y) 

by the ArithmeticGeometric Means Inequality. Hence
Öx + Öy £ 
 ______ Ö2(x + y)

. 

Also, by the same AGM inequality,
2^{x2 + y} + 2^{x + y2} ³ 2  Ö

2^{x2 + y + x + y2}

. 

Now, using the inequality again, we find that
x^{2} + y + x + y^{2} = (x^{2} + y^{2}) + (x + y) ³ 
1 2

(x + y)^{2} + (x + y) 

so that
2^{x2 + y} + 2^{x + y2} ³ 2^{1 + 1/4(x + y)2 +1/2(x + y)} = 2^{1/4[(x + y + 1)2 + 3]} . 

Suppose the (x, y) satisfies the system. Then

 ______ Ö2(x + y)

³ 2 Þ (x + y) ³ 2 

and

1 4

[(x + y + 1)^{2} + 3] £ 3 Þ (x + y + 1)^{2} £ 9 Þ x + y + 1 £ 3 Þ x + y £ 2 . 

Hence x + y = 2 and all inequalities are equalities. Therefore
x = y = 1.
 21.

Second solution. [A. Rodriguez] Wolog, we may assume
that x ³ 1. Let Öx + Öy = 2; then y = (2  Öx)^{2}. Define


= x + y^{2} + y + x^{2} = (2  Öx)^{4} + x^{2} + x + (2  Öx)^{2} 
 
= 2x^{2}  8x^{3/2} + 26x  36x^{1/2} + 20 . 

 

Then


= 4x  12x^{1/2} + 26  18x^{1/2} = 2x^{1/2} (2x^{3/2}  6x + 13x^{1/2}  9) 
 
= 2x^{1/2}(x^{1/2}  1)(2x  4x^{1/2} + 9) = 2x^{1/2}(x^{1/2}  1)[2(x^{1/2}  1)^{2} + 7] > 0 

 

for x > 1. Hence g(x) is strictly increasing for x > 1, so that
g(x)
³ g(1) = 4 for x
³ 1 with equality if and only if
x = 1. Thus, if the first equation holds, then
8 = 2^{x2 + y} + 2^{x + y2} ³ 2  Ö

2^{g(x)}

Þ 16 ³ 2^{g(x)} Þ g(x) £ 4 . 

Hence g(x) = 4, so that x = 1 and y = 1. Thus, (x, y) = (1, 1)
is the only solution.
 21.

Third solution. [S. Yazdani] Set Öx = 1 + u
and Öy = 1  u. Then x^{2} + y = (1 + u)^{4} + (1  u)^{2}
and x + y^{2} = (1  u)^{4} + (1 + u)^{2}, so
8 = 2^{x2 + y} + 2^{x + y2} = 2^{u4 + 7u2 + 2} 
æ ç
è

2^{4u3 + 2u} + 
1 2^{4u3 + 2u}


ö ÷
ø

³ 2^{2} (2) = 8 

with equality if and only if u = 0. Since the extremes of this
inequality are equal, we must have u = 0, so x = y = 1.
 21.

Fourth solution. [C. Hsia] With Öx = 1 + u
and Öy = 1  u, we can write the first equation as
2^{4u3 + 2u} + 
1 2^{4u3 + 2u}

= 2^{1  7u2  u4} . 

Let z = 2
^{4u3 + 2u}. We note that the quadratic
z
^{2}  2
^{1  7u2  u4} z + 1 = 0 is solvable, and so has
nonnegative discriminant. Hence
2^{2  14u2  2u4} ³ 4 = 2^{2} Þ 14u^{2}  2u^{4} ³ 0Þ u = 0 . 

Hence x = y = 1.
 21.

Fifth solution. [M. Boase] 2(x + y) ³ (x + y)+ 2Ö[xy] = (Öx + Öy)^{2} = 4 so that
x + y ³ 2. Let f(t) = t(t + 1). For positive values of
t, f(t) is an increasing strictly convex function of t. Hence
f(x) + f(y) ³ 2f( 
1 2

(x + y)) ³ 2f(1) = 4 

so that x
^{2} + x + y
^{2} + y
³ 4. Equality occurs if and only
if x = y = 1. Applying the ArithmeticGeometric Means Inequality,
we find that
4 = 
1 2

(2^{x2 + y} + 2^{x + y2}) ³ 2^{1/2(x2 + y2 + x + y)} 

so that
x
^{2} + x + y
^{2} + y
£ 4. Hence x
^{2} + x + y
^{2} + y = 4 and
so x = y = 1.


Comment. Note that 2(x^{2} + y^{2}) £ (x + y)^{2}
with equality if and only if x = y. Hence
x^{2} + y^{2} + x + y ³ 
1 2

(x + y)^{2} + (x + y) ³ 4 

with equality if and only if x = y = 1. This avoids the use of
the convexity of the function f.
 21.

Sixth solution. [J. Chui] Wolog, let x ³ y
so that Öx ³ 1 ³ Öy. Suppose that
Öx = 1 + u and Öy = 1  u. Then x + y = 2 +2u^{2} ³ 2 and xy = (1  u^{2})^{2} £ 1. Thus


= 2^{x2 + y} + 2^{x + y2} ³ 2  Ö

2^{x2 + y + x + y2}


 
= 2  Ö

2^{(x+y)(x+y+1)  2xy}

³ 2  Ö

2^{2·3  2·1}

= 2^{3} = 8 

 

with equality if and only if x = y.
22.
 22.

First solution. Case (i): Suppose that ÐB
is acute. Let AH ^BC and E lie on CH such that
AE = AB.


AC^{2}  CH^{2} = AB^{2}  BH^{2} implies that
AC^{2}  AB^{2} = CH^{2}  BH^{2} = (CH  BH)(CH + BH) = (CH  HE)BC = CE ·BC = CE[2(AC  AB)] . 

Hence AC + AB = 2CE. Also AC  AB =
^{1}/
_{2}BC. Therefore
2AB +
^{1}/
_{2}BC = 2CE.


Suppose that ÐABD = 2ÐADB. Then
ÐAEB = 2ÐADB Þ DADE is isosceles.
Hence
AB = AE = DE Þ 2DE + 
1 2

BC = 2CEÞ BC = 4(CE  DE) = 4CD Þ BD = 3CD . 

Conversely, suppose that BD = 3CD. Then
BC = 4CD Þ 
1 4

BC = CE  DE . 

From the above,
AB = CE  
1 4

BC = DE Þ AE = DE 



Case (ii): Suppose ÐB = 90^{°}. Then
AC^{2}  AB^{2} = BC^{2} = 2(AC  AB)·BC ÞAC + AB = 2BC 

Þ 
1 2

BC + AB + AB = 2BCÞ AB = 
3 4

BC . 

ÐABD = 2ÐADB Þ ÐADB = 45^{°} = ÐBAD Þ AB = BD 

Þ BD = 
3 4

BC Þ BD = 3CD . 

BD = 3CD Þ BD = 
3 4

BC = ABÞ ÐADB = ÐBAD = 45^{°} = 
1 2

ÐABD . 



Case (iii): Suppose ÐB exceeds 90^{°}. Let
AH ^BC and E be on CH produced such that AE = AB.
Then
AC^{2}  CH^{2} = AB^{2}  BH^{2} Þ(AC  AB)(AC + AB) = CH^{2}  BH^{2} = (CH  BH)(CH + BH) = CB·CE 

Also
AC  AB = 
1 2

BC Þ 2AB + 
1 2

BC = 2CE Þ AB + 
1 4

BC = CE . 



Let ÐABD = 2ÐADB. Then
180^{°}  ÐABE = 2ÐADB ÞÐAEB + 2ÐADE = ÐABE + 2ÐADB = 180^{°} . 

Also
ÐAEB + ÐEAD + ÐADE = 180^{°} ÞÐEAD = ÐADE Þ AE = ED . 

Hence
AB = ED Þ 2ED + 
1 2

BC = 2CE ÞBC = 4(CE  DE) = 4CD Þ BD = 3CD . 

Conversely, suppose that BD = 3CD. Then BC = 4CD and
ED = CE  CD = CE 
^{1}/
_{4}BC = AB so that
ED = AE and
ÐEAD =
ÐADE. Therefore
ÐABD = 180^{°}  ÐAED = ÐEAD + ÐADE = 2ÐADE = 2ÐADB . 

 22.

Second solution. [R. Hoshino] Let ÐABD = 2q. By the Law of Cosines, with the usual conventions for
a, b, c,


= cos2q = 
c^{2} + 4(bc)^{2}  b^{2} 4c(bc)


 
= 
bc c

 
b+c 4c

= 
3b5c 4c

(since b ¹ c) 

 



Þ 3(bc) = 6c  8csin^{2} q 
 
Þ 
3(bc) 2

sinq = c(3sinq4sin^{3} q) = c sin3q 
 
Þ 
sinq c

= 
2 sin3 q 3(bc)

. (*) 

 



Suppose now that D is selected so that
ÐADB = q. Then, by the Law of Sines,

sinq c

= 
sin(180^{°}  3q) x

= 
sin3q x



where x =
BD
. Comparison with (*) yields
x =
^{1}/
_{2}(3(bc)) so 4BD = 3BC
Þ BD = 3CD as
desired.


On the other hand,
suppose D is selected so that BD = 3CD. Then
BD = ^{3}/_{2}(bc). Let ÐADB = f. Then

sinf c

= 
sin(180^{°}  f 2q)

= 
sin(f+ 2q)

. 

Hence


sin(f+ 2q) sinf

= 
sin3q sinq



Þ sinqsin(f+ 2q) = sin3qsinf 
 
Þ 
1 2

[cos(q+ f)  cos(3q+ f)] = 
1 2

[cos(3q f)  cos(3q+ f)] 
 
 
Þ q+ f = ±(3q f) or q+ f+ 3q f = 360^{°} . 

 

The only viable possibility is
q+
f = 3
q
fÞ q =
f as desired.
 22.

Third solution. [J. Chui] First, recall
Stewart's Theorem. Let XYZ be a triangle with sides
x, y, z respectively opposite XYZ. Let W be a point
on YZ so that XW  = u, YW  = v and
ZW  = w. Then x(u^{2} + vw) = vy^{2} + wz^{2}. This
is an immediate consequence of the Law of Cosines. Let
q = ÐYWX. Then z^{2} = u^{2} + v^{2}  2uv cosq
and y^{2} = u^{2} + w^{2} + 2uw cosq. Multiply these equations
by u and v respectively, add and use x = v + w to obtain
the result.


Now to the problem. Suppose BD = 3CD. Let AC  = 2b, AB  = 2c, so that BC  = 4(b  c),
BD  = 3(b  c) and CD  = b  c.
If AD  = d, then an application of Stewart's Theorem
yields d^{2} = 2c(3b  c). Applying the Law of Cosines to
DABC and DABD respectively yields
cosÐABC = 
3b  5c 4c

and cosÐADB = 
3b  c

. 

Then
cos2
ÐADB = (3b  5c)/4c. Hence, either 2
ÐADB =
ÐABC or
ÐABC + 2
ÐADB = 360
^{°}. In the
latter case,
ÐABC +
ÐADB = 360
^{°} 
ÐADB > 180
^{°}, which is false. Hence
ÐABC = 2
ÐADB.


On the other hand, let 2ÐADB = ÐABC. If D¢
is a point on BC with BD¢ = 3CD¢, the 2 ÐAD¢B = ÐABC = 2 ÐADB, so that D = D¢. The result follows.
 22.

Fourth solution. Let AB  = a,
AC  = a + 2, BD  = 3, CD  = 1,
ÐABD = 2q, ÐADB = f. Then
(a + 2)^{2} = a^{2} + 16  8acos2q, whence a = 3(1 + 2cos2q)^{1}
(so 0 < q < 60^{°}). By the Law of Sines,

sin(2q+ f) 3

= 
(1 + 2cos2q)sinf 3



so that


= sinf+ 2sinfcos2q sin(2q+ f) 
 
= sinf+ sinfcos2q sin2qcosf 
 
= sinf+ sin(f 2q) = 2sin(f q) cosq . 

 

Since 0
£ f
q < 180
^{°}, we find that
f =
q as desired. The converse can be obtained as in the third
solution.
 22.

Fifth solution. [A. Birka] First, note that, when
BD = 3CD, we must have ÐADB < 90^{°}, since AC > AB
and D is on the same side of the altitude from A as C. Also,
when ÐABD = 2ÐADB, ÐADB < 90^{°}. Thus,
we can assume that ÐADB is acute throughout.


We can select positive numbers u, v and w so that
BC  = v + w, AC  = u + w and
AB  = u + v. By hypothesis,
v + w = 2(w  v), so that w = 3v.


Suppose that BD = 3CD. Then BC = 4CD, whence
CD  = v. Hence BD  = 3v. By the Law
of Cosines,
(u + 3v)^{2} = (u + v)^{2} + (4v)^{2}  8v(u + v)cosB 

so that
cosB = 
8v^{2}  4uv 8v(u + v)

= 
2v  u 2(u + v)

. 

Hence
AD ^{2} = (u+v)^{2} + (3v)^{2}  6v(u+v)cosB = u^{2} + 5uv + 4v^{2} = (u + 4v)(u + v) . 



Since sin^{2} ÐABD = 1  cos^{2} B = [3u(u+4v)]/[4(u+v)^{2}], and, by the Law of Sines,

sin^{2} ÐADB sin^{2} ÐABD

= 
u+v u+4v

, 

we have that
sin^{2} ÐADB = 
3u 4(u + v)

and cos^{2} ÐADB = 
u + 4v 4(u + v)

. 

Thus sin
^{2} ÐABD = 4sin
^{2} ÐADB cos
^{2} ÐADB
so that either
ÐABD = 2
ÐADB or
ÐABD + 2
ÐADB = 180
^{°}. The latter case would yield
ÐADB =
ÐBAD, so that AB = BD. This would make
DABC a
345 right triangle and
DABD an isosceles right triangle,
whence 90
^{°} =
ÐABD = 2
ÐADB. The converse can be
shown as in the previous solutions. The result follows.
23.
 23.

First solution. The configuration described is
obtained by rotating this configuration about the line passing
through the centres P and Q of the spheres.


From a consideration of similar triangles and pythagoras
theorem, we find that
OP  = r ( [(R+r)/(Rr)] ) 
OU  = [(4Rr^{2})/(R^{2}  r^{2})] 
UP  = r ( [(Rr)/(R+r)] ) 
AU  = [(2r)/(R+r)] Ö[(Rr)] 
OQ  = R ( [(R+r)/(Rr)] ) 
OV  = [(4R^{2}r)/(R^{2}  r^{2})] 
VQ  = R ( [(Rr)/(R+r)] ) 
BV  = [(2R)/(R+r)] Ö[(Rr)] 
The volume of the cone obtained by rotating OBV is

1 3

pBV ^{2} OV  = 
16 pR^{5} r^{2} 3(R+r)^{3} (Rr)



and the volume of the cone obtained by rotating
OAU is

16 pR^{2} r^{5} 3(R+r)^{3} (Rr)



so that the volume of the frustum obtained by rotating
AUVB is

16 pR^{2} r^{2} (R^{3}  r^{3}) 3 (R + r)^{3} (R  r)

= 
16 pR^{2} r^{2} 3 (R + r)^{3}

(R^{2} + Rr + r^{2}) . 



The volume of a slice of a sphere of radius a and
height h from the equatorial plane is
p 
ó õ

h
0

(a^{2}  t^{2})dt = p[a^{2} h  h^{3}/3] . 

The portion of the larger sphere included within the frustum
has volume


 p 
é ê
ë

R^{3} 
æ ç
è


Rr R+r


ö ÷
ø

 
R^{3} 3


æ ç
è


Rr R+r


ö ÷
ø

3


ù ú
û


 
= 
pR^{3} 3


é ê
ë

2  3 
æ ç
è


Rr R+r


ö ÷
ø

+ 
æ ç
è


Rr R+r


ö ÷
ø

3


ù ú
û


 
= 
pR^{3} 3 (R+r)^{3}

[ 4r^{3} + 12Rr^{2} ] = 
4pR^{2} r^{2} 3 (R+r)^{3}

[Rr + 3R^{2}] 

 

and the portion of the smaller sphere included within the frustum has
volume

2pr^{3} 3

+ p 
é ê
ë

r^{3} 
æ ç
è


Rr R+r


ö ÷
ø

 
r^{3} 3


æ ç
è


Rr R+r


ö ÷
ø

3


ù ú
û

= 
4pR^{2} r^{2} 3 (R + r)^{3}

[Rr + 3r^{2}] . 

Hence, the portions of the sphere lying within the frustum have
total volume

4pR^{2} r^{2} 3 (R + r)^{3}

[3R^{2} + 2Rr + 3r^{2}] . 

Subtracting this from the volume of the frustum yields the
volume of the ringshaped region

4pR^{2} r^{2} 3 (R + r)^{3}

[(4R^{2} + 4Rr + 4r^{2})  (3R^{2} + 2Rr + 3r^{2})] = 
4pR^{2} r^{2} 3 (R + r)^{3}

[R^{2} + 2Rr + r^{2}] = 
4 pR^{2} r^{2} 3(R + r)

. 



Comment. The volume of a slice of a sphere of radius
a and height h from the equatorial plane can be obtained from the
volume of a right circular cone and a cylinder using the method of
Cavalieri. The area of a crosssection of the slice at height t
from the equator is p(a^{2}  t^{2}) = pa^{2}  pt^{2}. The
term pa^{2} represents the crosssection of a cylinder of
radius a and height h while pt^{2} represents the area of
the cross section of a cone of base radius h at distance t from
the vertex. Thus the area of the each crosssection of the cylinder is
the sum of the areas of the corresponding crosssections of the
spherical slice and cone. Cavalieri's principle says that the
volumes of the solids bear the same relation. Thus the volume of
the spherical slice is
24.
 24.

First solution.
Suppose AD splits BC into
segments of lengths a_{1} and a_{2} as in the diagram and angle
BDC into subangles a_{1} and a_{2}, and let
b_{1}, b_{2}, b_{1}, b_{2}, c_{1}, c_{2}, g_{1}, g_{2} be
similarly defined as in the diagram.


Applying the Law of Sines to DBPD and DCPD,
we find that

a_{1} sina_{1}

= 
a_{2} sina_{2}



and similarly that

b_{1} sinb_{1}

= 
b_{2} sinb_{2}

and 
c_{1} sing_{1}

= 
c_{2} sing_{2}

. 

Let
a =
ÐBAE. Then
a =
ÐFAC since
ÐFAB =
ÐEAC. Similarly, let
b =
ÐFBC =
ÐABD and
g =
ÐBCE =
ÐACD.


Let AB  = c, BC  = a, AC  = b, AD  = u, BE  = v, CF  = w.
By the Law of Sines, we find that

v sina

= 
c sinb_{1}

and 
v sing

= 
a sinb_{2}



so that

c sina sinb_{1}

= 
a sing sinb_{2}

Þ 
sinb_{1} sinb_{2}

= 
c a

· 
sina sing

. 

Similarly

sina_{1} sina_{2}

= 
b c

· 
sing sinb

and 
sing_{1} sing_{2}

= 
a b

· 
sinb sina

. 



Putting this altogether yields

a_{1} a_{2}

· 
b_{1} b_{2}

· 
c_{1} c_{2}

= 
sina_{1} sina_{2}

· 
sinb_{1} sinb_{2}

· 
sing_{1} sing_{2}

= 
b c

· 
c a

· 
a b

· 
sing sinb

· 
sina sing

· 
sinb sina

= 1 . 

By the converse of Ceva's Theorem, the cevians
AP,
BQ and
CR are
concurrent and the result follows.