elementary triangle goemetry-solutions
TRANSCRIPT
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Elementary Triangle GeometryMark DabbsSolutions to Exercises
Version 1.0 April 2004(www.mfdabbs.com)
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Problem 1. Prove:4 A B C r r r r R+ + =
Solution:
( ) ( )
( )( )
( )
( )
( ) ( ) ( )(
( ) ( )( )
)
( )
( )( )( )
( )( ) ( )
3 2
3 2
2
,
,
2,
2,
2 2,
,
,
4 , as required (on us
lhss a s b s c s
s b s a s s c
s a s b s s c
s a b s s c c s a s b
s s a s b s c
s s a b c abc
s s a s b s c
s s s abc
s s a s b s c
abc
abc
R
= + +
+ = +
+ =
+ + +=
+=
=
=
ing (5.2)).
using (6.2) and (6.10)
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Problem 2. Prove:1 1 1 1
A B C r r r r + + =
Solution:
( )3,
3 2,
,
1, as required (on using (6.2)).
s a s b s clhs
s a b c
s s
s
r
= + +
+ +=
=
=
, using (6.10)
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Problem 3. Prove:1 1 1
A B C h h h
1
r+ + =
Solution:
2 2 2
1 1 1,
2 sin sin 2 sin sin 2 sin sin
1 1 1,
sin sin sin sin sin sinsin sin sin
sin sin sin,
sin sin sin sin sin sin
2sin 2sin 2sin
2 sin sin 2 sin sin 2 sin
lhsR B C R C A R A B
a b c B C C A A
A B C
A B C
a B C b C A c A B
a A b B c C
a B C b C A c A
= + +
= + +
= + +
= + +
B
( )
2 2
,sin
1 1 1,
2 2 2
1,
2
,
1
B
a b c
b c
a b c
s
r
= + +
+ + =
=
using (9.2)
using (4.7)
using the forms of (5.3)
, as required (on using (6.2))
Notice the much simpler method based upon the observation that
1 1 12 2 2A Bah bh ch = C ,
which leads immediately to ( above.)
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Problem 4. Prove:2
A B Cr r r rs=2
A B Cr r r r =
Solution to r r r :2 A B C rs=
( )( ) ( )
( ) ( ) ( )
( )
3
3
3
2
,
1,
,
,
lhss a s b s c
s a s b s c
s
s s a s b s c
s
s
rs s
rs
=
=
=
=
=
=
using (6.10)
using (5.6)
, using (6.1)
2, as required.
Solution to r r :2 A B C r r =
Using the above result we have on multiplication by that2
A B C r r r rs= r
( )
2 2
2
2
,
,
A B C r r r r r s
rs
=
=
, as required.
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Problem 5. Prove:( ) ( ) ( )
( ) ( ) ( )
1 1 12 2 2
1 1 12 2 2
4 cos cos cos
4 cos cos cos
s R A B C
Rr A B
=
= C
Solution to ( ) ( ) ( )1 1 12 2 24 cos cos cos s R A B C = :
Multiply together all cyclic forms of (6.7) to give
( ) ( ) ( )( )( ) ( ) ( )
21 1 12 2 23
1 1 12 2 2
sin sin sin
cos cos cos
abc A B C
A B C =r
( ) ( ) ( ) ( ) ( ) ( )( )
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( )
23 1 1 1 1 1 12 2 2 2 2 2
23 1 1 1
2 2 2
23 1 1 1
2 2 2
1 1 12 2 2
1 1 12 2 2
1 1 12 2 2
cos cos cos sin sin sin ,
cos cos cos ,4
cos cos cos ,4 4
1cos cos cos ,
4
1cos cos cos ,
4
4 cos cos cos
r A B C abc A B C
rr A B C abc
R
r abcr A B C
R R
r A B C R
r A B C rsR
R A B C s
=
=
=
=
=
=
using (6.8)
using (5.2)
using (6.1)
, as required.
Solution to ( ) ( ) ( )1 1 12 2 24 cos cos cos Rr A B C = :
Using the above result we have on multiplication by thatr
( ) ( ) ( )
( ) ( ) ( )
1 1 12 2 2
1 1 12 2 2
4 cos cos cos ,
4 cos cos cos ,
rs Rr A B C
Rr A B C
=
=
as required (on using (6.1)).
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Problem 6. Prove:cos cos cos 4 sin sin sina A b B c C R A B+ + = C
Solution:
( )
( )
( ) ( )( )
( ) ( )( )
( )( )
2 sin cos 2 sin cos 2 sin cos ,
2 sin cos sin cos sin cos ,
sin 2 sin 2 sin 2 ,
2 2 2 22sin cos sin2 ,
2 2
2sin cos sin 2 ,
2sin 180 cos sin 2 ,
2sin cos sin 2 ,
lhs R A A R B B R C C
R A A B B C C
R A B C
A B A BR C
R A B A B C
R C A B C
R C A B C
= + +
= + +
= + +
+ = +
= + +
= +
= +
( )
( )
( )( )
( )[ ]( )
2sin cos cos 2sin sin sin sin 2 ,
2sin cos cos 2sin sin sin 2sin cos ,
2sin cos cos 2sin sin sin 2sin cos 180 ,
2sin cos cos 2sin sin sin 2sin cos ,
(2sin cos cos 2sin sin sin
2si
R C A B C A B C
R C A B C A B C C
R C A B C A B C A B
R C A B C A B C A B
R C A B C A B
= + +
= + +
= + +
= + + = +
n cos cos 2sin sin sin ),
4 sin sin sin
C A B C A B
R A B C
+
=
+
using (4.7)
, as required.
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Problem 7. Prove:2
A B B C C Ar r r r r r s+ + =
Solution:
( ) ( ) ( )
( )( ) ( )
( )
( )( )( )
2
2
2
2
2
, using (6.10)
,
3,
3 2, using (5.6)
, as required.
lhss a s b s b s c s c s a
s a s b s c
s a s b s c
s a b cs
s s a s b s c
s ss
s
= + +
+ + =
+ +=
=
=
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Problem 8. Prove:( )2
1 1 1 1
4 A B C A B C
abcs
r r r r r r R r + + =
+ + +
Solution:
From Problem 2. and Problem 1. rearranged we have:
( )
( )
( )
( )
( )2
1 1,
4
4,
4
4,
4
, using (5.2)4
1, using (6.2)
4
, as required.4
lhsr R r
R r r
r R r
R
r R r
abc
r R r
abc
R rs
abcs
R r
= +
+ =
+
=+
=+
= +
= +
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Problem 9. Prove:( ) ( )2 1 12 24 cos cotB Cr r R A a A+ =
Solution:
( ) ( ) ( ) ( ) ( ) ( )
( ) ( ) ( ) ( ) ( )
( ) ( )
( ) ( )
( )
1 1 1 1 1 12 2 2 2 2 2
1 1 1 1 12 2 2 2 2
1 1 12 2 2
1 12 2
2 12
4 cos sin cos 4 cos cos sin , using (8.6)
4 cos sin cos cos sin ,
4 cos sin ,
4 cos sin 90 ,
4 cos , as required.
lhs R A B C R A B C
R A B C B C
R A B C
R A A
R A
= +
= +
= +
=
=
Further, we have
( )( )
( )
( )
2 12
1
2
4 cos 4 , using (7.5)
, using (5.2)
,
cot , as required (using (6.5)).
s s a R A R
bc
s s aabc
rs bc
s aa
r
a A
=
=
=
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Problem 10. Prove:( )( )( ) ( )4A B B C C A A B B C C Ar r r r r r R r r r r r r + + + = + +
Solution:
( ) ( ) ( )
( ) ( ) ( )( )
( ) ( ) ( )( )
( )
( )
2 2 21 1 12 2 2
23 1 1 12 2 2
21 1 12 2 2
2
4 cos 4 cos 4 cos , using above
64 cos cos cos ,
4 4 cos cos cos ,
4 , using above
4 , as required (using above).A B B C C A
lhs R A R B R C
R A B C
R R A B C
R s
R r r r r r r
=
=
=
=
= + +
Problem 10
Problem 5Problem 7
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Problem 11. Prove:( )( ) ( ) 24 A B C r r r r r r R r =
Solution:
( ) ( ) ( )3
3
2 2
2
2
2
2
2
, using (6.10) and (6.2)
,
, using (5.6)
,
4, using (5.2)
4 ,
4 , as required (on using (6.2)).
lhss a s s b s s c s
a b c
s s a s s b s s c
abc
s
abc
s
R
s
Rs
Rr
=
=
=
=
=
=
=
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Problem 12. Prove:3
2 2 2 2 2
1 1 1 1 1 1 1 1 1 1 1 1 64 4
A B B C C A A B C
R R
r r r r r r r r r r r r a b c r s
+ + + = =
Solution to3
2 2 2 2 2
1 1 1 1 1 1 64 4
A B B C C A
R R
r r r r r r a b c r s
+ + + =
:
( ) ( )( )
( )
( )
( )
( )
( )
( )
( )
2
2
2
2
2
22
22 2
2
,
,
4, using
4, using
4, using
4 4 ,
4,
B C C AA B
A B B C C A
A B B C C A
A B C
A B B C C A
A B C
A B C
r r r r r rlhs
r r r r r r
r r r r r r
r r r
R r r r r r r
r r r
Rs
r r r
Rs
rs
R Rr s rs
R
+ ++=
+ + +=
+ +=
=
=
=
=
Problem 10
Problem 7
Problem 4
2
3
2 2 2
using (6.1)
4, using (5.2)
4
64, as required.
R
abc
R
R
a b c
=
=
Solution to3
2 2 2 2 2
1 1 1 1 1 1 64 4
A B C
R R
r r r r r r a b c r s
=
:
We rearrange the results ofProblem 2 to give the forms below.1 1 1 1 1 1 1 1 1 1 1 1
, ,
A B C B C A C A B
r r r r r r r r r r r r = + = + = + .
The result follows by direct substitution.
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Problem 13. Prove:
( )( ) ( )( ) ( ) ( ) 21 1 1
s a s b s b s c s c s a r + +
1=
Solution:
( )
( )( ) ( )
( )
( )( )( )
( )
( ) ( )( )
( ) ( ) ( )
( ) ( ) ( )
( )[ ]
[ ]
2
2
22
2
2
,
,
3 -, using (5.6)
3 2,
,
1, as required (on using (6.2))
s s a s s b s s clhs
s s a s b s c s s a s b s c s s a s b s c
s s a s s b s s c
s s a s b s c
s s a b c
s s s
s s
r
= + +
+ + =
+ +=
=
=
=
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Problem 14. Prove:( )
22 2 2 2 2 24 1b c b c a 6= + +
Solution:
From the Cosine Rule (4.5) and Area Rule (5.1) we have both
2 2 22 cos (i)
and 2 sin 4 (ii)
bc A b c a
bc A
= +
=
Squaring and adding, , gives( ) ( )2
i ii+2
( ) ( )( )
22 2 2 2 2 2 2 2
22 2 2 2 2 24 cos sin 16 .
Hence, 4 16 , as required.
b c A A b c a
b c b c a
+ = + +
= + +
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Problem 15. Prove:( ) ( )2 24 sin 2 sin 2b A a = + B
Solution:
[ ]
( )
2 2
212
212
2 sin cos 2 sin cos ,
sin sin sin cos4
sin sin
sin sin sin cos4 ,
sin sin
sin cos sin cos4 4 , using (5.3)
sin sin
4 sin cos sin cos ,sin
4 4sin 180 sin ,
sin sin
4 ,
rhs b A A a B B
A C B Ab
B C
B C A Ba
A C
B A A B
C C
B A A BC
C CC C
= +
=
+
= +
= +
=
=
as required.
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Problem 16. Prove:( ) ( ) ( )2 1 1 12 2 2tan tan tan s A B = C
Solution:
From (6.8) and Problem 5 we have the results( ) ( ) ( )
( ) ( ) ( )
1 1 12 2 2
1 1 12 2 2
4 sin sin sin ,
4 cos cos cos .
r R A B C
s R A B C
=
=
On division we have
( ) ( ) ( )
( ) ( ) ( )
( ) ( ) ( )
1 1 12 2 2
1 1 12 2 22
2 1 1 12 2 2
tan tan tan
tan tan tan , using (6.2)
tan tan tan , as required.
r A B C
s
A B C s
s A B C
=
=
=
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Problem 17. Prove:22 sin sin sin R A B C =
Solution:
( ) ( )
12
12
212
2
sin , using (5.1)
2 sin 2 sin sin , using (4.7)
4 sin sin sin ,
2 sin sin sin , as required.
lhs ab C
R A R B C
R A B C
R A B C
=
=
=
=
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Problem 18. Prove:( )sin sin sin Rr A B C = + +
Solution:
( )
2 2 2, using (5.1)
2 2 2,
2,
22 ,
4,
1, using (5.2)
, as required (on using (6.1)).
rhs Rrbc ca ab
a b cRr
abc
Rra b c
abc
Rrs
abc
Rrs
abc
rs
rs
= + +
+ + =
= + +
=
=
=
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Problem 19. Prove:
( )
2 2 sin sin
2 sin
a b A B
A B
=
Solution:
From (5.3) we consider the two forms
2 21 12 2
sin sin sin sin, .
sin sin
B C C a b
A B = =
A
Rearranging gives:
2 21 12 2
sin sinsin , sin .
sin sin
A Ba C b
B A
= = C
Subtracting we have
( )
2 21 12 2
2 2 2 2
2 2
2 2
sin sinsin sin ,
sin sin
sin sinsin ,
sin sin 2
sin sin sin
2 sin sin
A Ba C b C
B A
A B a bC
A B
a b A B C
A B
=
=
=
where( )( )
( ) ( )
( ) ( )( )
2 2sin sin sin sin sin sin ,
2cos sin 2sin cos ,2 2 2 2
2sin cos 2sin cos ,2 2 2 2
sin sin ,
sin 180 sin ,
sin sin .
A B A B A B
A B A B A B A B
A B A B A B A B
A B A B
C A B
C A B
= +
+ + =
+ +
= +
Hence,2 2
2 2
sin sin sin
2 sin sin
a b A B C
A B
=
can be written:
( )
2 2 sin sin, as required.
2 sin
a b A B
A B
=
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Problem 20. Prove:( ) ( ) ( )( )3 2 2 232 sin 2 sin 2 sin 2a b c A B C = + +
Solution:
From Problem 6 we know that ( ) ( ) ( )n 2 sin 2 sin 2 4 sin sin sin A B C A Bsi C+ + ,therefore we have:
( )
( )
2 2 2
2 2 2
2
2 2
2
3
4 sin sin sin ,
4 , using above2
2 16 , using (5.2) squared
32 , as required.
rhs a b c A B C
a b cR
R
R
=
=
=
=
Problem 17
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Problem 21. Prove:( )
2 2 2
4cot cot cot
a b c
A B C
+ + =
+ +
Solution:
Rearranging the form of (5.9) we have that
2 2 2
2 2 2
2 2 2
4,
tan
4,
tan
4.
tan
b c aA
c a bB
a b cC
+ =
+ =
+ =
Addition of these three forms gives
( )
2 2 2
2 2 2
2 2 2
1 1 14 ,
tan tan tan
4 cot cot cot ,
4 , as required.cot cot cot
a b c A B C
a b c A B C
a b c
A B C
+ + = + +
+ + = + +
+ + =
+ +
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Problem 22. Prove:( ) ( )
( ) A B C A B C AB CB CA B B C C A
r r r r r r rra r
s rr r r r r r
+ += = = +
+ +r
r
Solution to ( ) A B C
A B B C C A
r r r
r r r r r r +=
+ +a :
( )
2
2
2
,
,
,
, using
, using .
A B A C
A B B C C A
A B A C B C B C
A B B C C A
A B B C C A B C
A B B C C A
B C
B C B C
A
r r r r rhs
r r r r r r
r r r r r r r r
r r r r r r
r r r r r r r r
r r r r r r
s r r
s
s r r r r rss s
s s r
+=
+ +
+ +
+ +
+ +
= + +
=
= = =
Problem 7
Problem 4
If we now equate the results of (6.1) and (6.9) it is seen that
( ) .AA
rsrs r s a s ar
= =
Hence
( )( ) , as required. A B C
A B B C C A
r r r s s a a
r r r r r r
+= =
+ +
Solution to( ) A B C r r r
s
+=a :
( ) ( )We have: , as required (on using )
A B C A B C
A B B C C A
r r r r r r a sr r r r r r
+ +
= + + Problem 7
Solution to ( ) AB CB C
rrr
r r= +a r :
( ) ( )
( ) ( )
We have: , using
, as required.
A B C A B C
A B C
B C AB C
B CB C
A
r r r r r r a
s r r r
r
r r rra r r
r rr r
rr
+ +=
+ = +
Problem 4
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Problem 23. Prove:( ) ( )2 A Ba r r r r = + C
Solution:
( ) ( )( )
( )
( )( )( )
( )
( )
( )
( )( )
( )
2
2
2
2
2
, using (6.2) and (6.10)
,
2,
2, using (5.6)
2 ,
2 ,
2 ,
2 2 ,
, as re
rhss a s s b s c
a s c s b
s s a s b s c
a s b c
s s a s b s c
a s b c
a s b c
a s a b c a
a s a b c a
a s s a
a
=
+ =
=
=
=
+
= + + +
= +
= quired.
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Problem 24. Prove:( )
( ) ( )12sin
A
A B A C
rA
r r r r =
+ +
Solution:
( ) ( )( )
( )
( )
( )
( )
( ) ( )
( )
2
2 12
12
1 12 2
12
,
, using4
, using4
4 cos , using4
cos ,
tan cos , using (6.12)
sin , as required.
A B C
A B B C C A
A B C
A B B C C A
A B C
A
A
r r rrhsr r r r r r
r r r
R r r r r r r
r r r
R s
r R As R
rA
s
A A
A
+=+ + +
+=
+ +
+=
=
=
=
=
Problem 10
Problem 7
Problem 9
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Problem 25. Prove:( )( )
2sin A A B B C C A
A B A C
r r r r r r r A
r r r r
+ +=
+ +
Solution:
( )( )( )( )
( )
( )
( )
( )
( )2 12
2,
2, using
4
2,
4
2
, using4
2 4 cos, using
4
2
A B C A B B C C A
A B A C B C
A B C A B B C C A
A B B C C A
A B C
A B B C C A
A B C
A
A
r r r r r r r r r rhs
r r r r r r
r r r r r r r r r
R r r r r r r
r r r
R r r r r r r
r r r
Rs
r R A
Rs
r
s
+ + +=+ + +
+ + +=
+ +
+=
+ +
+=
=
=
Problem 10
Problem 7Problem 9
( )
( ) ( )
( ) ( )
2 12
21 12 2
1 12 2
cos ,
tan cos , using (6.12)
2sin cos ,
sin , as required.
A
A A
A A
A
=
=
=
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Problem 26. Prove:( ) ( ) ( )3 2 2 21 1 12 2 2cot cot cot A B C r r r r A B C =
Solution:
( ) ( ) ( )
( ) ( ) ( )
( )( )( )[ ]
( )
2 2 2 2 2 21 1 12 2 23
2 2 2
3
2
2 3
4
2 3
4
2
2 3
1 cot cot cot ,
1, using (6.5)
,
, using (5.6)
, using (6.1)
, as required (on using )A B C
rhs r A r B r C r
s a s b s cr
s s a s b s c
s r
s r
srrs
s r
r r r
=
=
=
=
=
= Problem 4
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Problem 27. Prove:2 2
2 2 2 2 2
1 1 1 1
A B C
a b c
r r r r
2+ ++ + + =
Solution:
( ) ( ) ( )
( )
2 2 22
2 2 2
2 2
2 2 2 2 2 2
2 2
2 2 2 2
2 2
2 2 2 2 2
2 2
2 2 2 2
2 2 2
2
1 1 1 1 , using (6.10)
1,
1 2 2 2,
3 21
,
1 3 4,
1,
1 1
lhsr
s a s b s c
s a s b s c
r
s as a s bs b s cs c
r
s s a b c a b c
r
s s a b c
r
s a b c
r
r
= + + +
+ + = +
+ + + + += +
+ + + + += +
+ + += +
+ += +
= 2 2 2
2 2
2 2 2
2
, using (6.2)
, as required.
a b c
r
a b c
+ ++
+ +=
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Problem 28. Prove:r r r r 2 2 2 2 2 2 216 A B C R a b c+ + + =
2
Solution:
Note that the identity ( ) (2 2 2 2 2 ) y z x y z xy yz zx+ + + + + + + is of great use in
the following.
( )2 2 2 2 2
2 2 2 2 2 2
From : 4 ,
Squaring gives:
2 16 8 ,
2 16 8 , using
A B C
A B C A B B C C A
A B C
r r r R r
r r r r r r r r r R Rr r
r r r s R Rr r
+ + = +
+ + + + + = + +
+ + + = + +
Problem 1
Problem 7
( )
( )( )( )
( )
2 2 2 2 2 2 2
2 2 2
22 2 2
2
4 2
2
4
2
3 2
2
16 8 2 2 ,
16 2 4 ,
But from (6.2) and (8.21)
4 ,
1,
1, using (5.6)
1
A B C r r r r R Rr r s
R s r Rr
abc s r Rr s
s s
s abcss
s s s a s b s c abcss
s a b c s ab bc cs
+ + + = + +
=
=
=
=
+ + + +( )
( ) ( )
( ) ( )
( ) ( )
( )
( )
212
212
2 2 212
2 2 2 2 2 2 2 212
2 2 2 2
,
,
,
2 ,
.
Hence, 16 2 ,
16 , as required.
A B C
a
s a b c ab bc ca
a b c ab bc ca
a b c ab bc ca
a b c
r r r r R a b c
R a b c
= + + + +
= + + + +
= + + + +
+ +
+ + + = + +
=
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Problem 29. Prove: A B C ab r r rr = +
Solution:
( ) ( )( )[ ]( ) ( )( )
( ) ( )( )
( )
2
2
, using (6.10)
,
, using (5.6)
2 ,
, as required.
rhss a s b s s c
s s a s b s c
s s a s b s c
s s a s b s c
s s a b c ab
ab
= +
+ =
= +
= + + +
=
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Problem 30. Prove:1 1 1 1
2ab bc ca Rr + + =
Solution:
,
,
2, using (5.2)
4
, using (6.1)2
1 , as required.2
a b clhs abc abc abc
a b c
abc
s
R
s
Rrs
Rr
= + +
+ +=
=
=
=
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Problem 31. Prove:1 1
2CA B rr r
bc ca ab r R+ + =
Solution:
( )( ) ( ) ( ) ( )( )
( )( )( )
( )( ) ( ) ( ) ( )( )
( ) ( ) ( ) ( ){ }
2
2
,
1, using (6.10)
,
, using (5.6)
3 ,
A B C ar br cr lhsabc
a b c
abc s a s b s c
a s b s c b s c s a c s a s b
abc s a s b s c
a s b s c b s c s a c s a s bs
abc
ss a b c s a b c b c a c a b abc
abc
s
+ +=
= + +
+ + =
+ + =
= + + + + + + + +
= ( )32 2 3 , s s ab bc ca abcabc
+ + +
We now note the identity:
( ) ( ) ( ) ( ) ( )3 2
s a s b s c s a b c s ab bc ca s abc + + + + + Hence,
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
( ) ( ) ( ) ( )
3 3
3
3
3
2 ,
,
2 2 2 2 ,
abc 2 2 2 3
s a s b s c s s ab bc ca s abc
s a s b s c s ab bc ca s abc
s a s b s c s ab bc ca s abc
s a s b s c s ab bc ca s abc
= + + +
= + + +
= + + +
= + + +
Therefore, substituting gives:
( ) ( ) ( )[ ]
( )( )( )
2
2 ,
2,
2 1 2, using (5.6) and (6.1)
1 2 1 1, as required (on using (5.2)).
4 2
slhs abc s a s b s c
abc
s s a s b s cs
abc
s
abc r abc
abc
r abc R r R
=
=
=
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Problem 32. Prove:0
A B C
b c c a a b
r r r
+ + =
Solution:
( ) ( ) ( ) ( ) ( )( )
( ) ( ) ( ) ( ) ( ) ( )
, using (6.10)
,
,
0,
0, as required.
b c c a a blhs
s a s b s c
s a b c s b c a s c a b
s b c a b c s c a b c a s a b c a b
sb sc ab ac sc sa bc ba sa sb ca cb
= + +
+ + =
+ + =
+ + + + +=
=
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Problem 33. Prove:( ) ( ) ( )A B C B C A C A Ba rr r r b rr r r c rr r r + = + = +
Solution:
Consider the expression A B C r r rr + .
( )
( )
( )
, using
.
By symmetry: ,
and .
A B C
A B C
B C A
C A B
r r rr ab
c r r rr abc
a r r rr bca
b r r rr cab
+ =
+ =
+ =
+ =
Problem 29.
By equating the result follows immediately.
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Problem 34. Prove:A B B C C A A B C r r r r r r rr rr rr ab bc ca+ + + + + = + +
Solution:
, usingBy symmetry: ,
and .
A B C
B C A
C A B
r r rr ab
r r rr bc
r r rr ca
+ =
+ =
+ =
Problem 29.
The result follows on addition of these three terms.
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Problem 35. Prove:( ) ( ) ( ) ( )
2 2 28 2 A B C I I I I I I R R+ + = r
Solution:
( ) ( ) ( )
( )
( )
( )
( )
4 4 4 , using (8.22)
4 3 ,
4 4 3 , using
4 4 2 ,
8 2 , as required.
A B C
A B C
lhs R r r R r r R r r
R r r r r
R R r r
R R r
R R r
= + +
= + +
= +
=
=
Problem 1.
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Problem 36. Prove:216 A B C I I I I I I R r =
Solution:
( )( )( )
( )
( )( ) ( )
3
3
3
42 2
2
, using (8.19)
,
,
, using (5.6)
,
abc a abc b abc clhss s a s s b s s c
abc a b c
s s a s b s c
abc abcs
s s s a s b s c
abc abc ss
abc abc abc
ssr
=
=
=
= =
( )2
2
2
using (6.1)
4 , using (5.2)
16 , as required.
abcr R r
R r
= =
=
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Problem 37. Prove:( ) (
24 B C B C ) I I R r = + r
Solution:
( )( )
( )
( )
212
2 12
4 cos , using (9.13)
4 4 cos
4 , as required (on using ).B C
lhs R A
R R A
R r r
=
=
= + Problem 9
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Problem 38. Prove:( ) ( ) ( ) ( ) ( ) ( )
2 2 2 2 2
A B C B C A C A
2
B I I I I I I I I I I I I + = + = +
Solution:
First consider the term ( ) ( )2 2
A B C I I I I +
We have
( ) ( )( ) ( )
( ) ( )
( ) ( )
( ) ( )
( )
2 2
2 2
1 12 2
2 2 2 21 12 2
2 21 12 2
2
1 12 2
2 2
12
2
2
, using (9.12) and (9.13)cos sin
sin cos,
cos sin
cos sin
2 ,sin sin
2 2 , using (4.7)
16 .
A B C
a a I I I I
A A
a A a A
A A
a
A A
a a
A A
R
R
+ = +
+=
=
=
=
The results follow by symmetry.
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Problem 39. Prove:sin sin sinA B C B C A C A B I I I I I I I I I I I I
A B
= =
C
Solution:First consider the term
sin A B C I I I I
A
We have
( ) ( )
( ) ( )
( )
1 12 2
2
1 12 2
2
212
2
2
2
2
cos sin, using (9.12) and (9.13)
sin sin
cos sin
sin
sin 2,
sin sin
2 ,sin
2 2 , using (4.7)
8 .
A B C
a a
A AI I I I
A A
a
A A
A
a
A a
A A
a
A
R
R
=
=
=
=
=
The results follow by symmetry.
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Problem 40. Prove:, anda b c are the radii of three circles which touch one another externally and the
tangents at their points of contact meet in a point; prove that the distance of this
point fromeither
of their points of contact is
abc
a b c+ +
Solution:
Radius = c
Radius = b
Radius = a
I
D
F
C
A
B
E
From basic circle theory if t
circles are touching then a
line through their centres
crosses their point common
tangency at right angles.
wo
Therefore,
, say,
, say,
, say.
AB a b a
BC b c b
CA c a c
= +
= +
= +
Using the same argument as
on Page 70, Incentre Proof
we see that it must be the casethat
, say, DI EI FI r = =
where .inradius of triangler ABC =
The semi-perimeter for triangle ABC is given as
( ) ( )1 12 2s a b c a b b c c a a b c = + + + + + + + = + + .
( )( )( )
( )( ) ( ) ( )
( )( ) ( ) ( )
( )
( )
Also, from (5.6) ,
,
,
.
But from (6.1), .
Hence, , as required.
s s a s b s c
a b c s a b s b c s c a
a b c c a b
abc a b c
abc a b cr
s a b c
abc
r a b c
=
= + +
= + +
= + +
+ +=
+ +
=+ +
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