Elementary Triangle Geometry Mark Dabbs Solutions to Exercises Version 1.0 April 2004 (www.mfdabbs.com)
Elementary Triangle Geometry Mark Dabbs
Solutions to Exercises
Version 1.0 April 2004 (www.mfdabbs.com)
2
3
Problem 1. Prove: 4A B Cr 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 cs a s b s s c
s a b s s c c s a s bs s a s b s c
s s a b c abcs s a s b s c
s s s abcs s a s b s c
abc
abc
R
∆ ∆ ∆ ∆= + + −
− − −
− + − − −= ∆ + − − −
− − − + − −= ∆ − − −
− + + += ∆ − − −
− += ∆ − − −
= ∆ ∆
=∆
≡ ing (5.2)).
using (6.2) and (6.10)
4
Problem 2. Prove:
1 1 1 1
A B Cr 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)
5
Problem 3. Prove:
1 1 1
A B Ch h h1r
+ + =
Solution:
2 2 2
1 1 1 ,2 sin sin 2 sin sin 2 sin sin
1 1 1 ,sin sin sin sin sin sin
sin sin sin
sin sin sin ,sin sin sin sin sin sin
2sin 2sin 2sin2 sin sin 2 sin sin 2 sin
lhsR B C R C A R A B
a b cB C C A AA B C
A B Ca B C b C A c A B
a A b B c Ca B C b C A c A
= + +
= + +
= + +
= ⋅ + ⋅ + ⋅
B
( )
2 2
,sin
1 1 1 ,2 2 2
1 , 2
,
1
B
a b cb 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 1
2 2 2A Bah bh ch∆ = ≡ ≡ C , which leads immediately to ( above. )∗
6
Problem 4. Prove:
2A B Cr r r rs=
2A B Cr r r r = ∆
Solution to r r r : 2
A B C rs=
( )( )( )
( )( )( )
( )
3
3
32
,
1 ,
,
,
lhss a s b s c
s a s b s c
ss 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 Cr r = ∆ Using the above result we have on multiplication by that 2
A B Cr r r rs= r
( )
2 2
2
2
,
,
A B Cr r r r r s
rs
=
=
≡ ∆ , as required.
7
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 1
2 2 24 cos cos coss 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 sincos cos cos
abc A B CA 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 abcR
r abcr A B CR R
r A B CR
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 1
2 2 24 cos cos cosRr A B C∆ = : Using the above result we have on multiplication by that r
( ) ( ) ( )
( ) ( ) ( )
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)).
8
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 sin 2 ,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.
9
Problem 7. Prove: 2
A B B C C Ar r r r r r s+ + = Solution:
( ) ( ) ( )( )( )( )
( )( )( )( )
2
2
22
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 cs a s b s c
s a b cs
s s a s b s c
s ss
s
∆ ∆ ∆ ∆ ∆ ∆= ⋅ + ⋅ + ⋅
− − − − − −
− + − + −= ∆ − − −
− + += ∆ − − −
− = ∆ ∆
=
10
Problem 8. Prove:
( )2
1 1 1 14A B C A B C
abcsr 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 rr R r
Rr R r
abc
r R r
abc
R rs
abcsR r
= −+
+ −=
+
=+
∆ =
+
= ⋅∆∆ +
=∆ +
11
Problem 9. Prove: ( ) ( )2 1 1
2 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
12
4 cos 4 , using (7.5)
, using (5.2)
,
cot , as required (using (6.5)).
s s aR A R
bcs s aabc
rs bcs aa
ra A
−= ⋅
−= ⋅
−= ⋅
≡
12
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 5
Problem 7
13
Problem 11. Prove: ( )( )( ) 24A B Cr r r r r r R r− − − = Solution:
( ) ( ) ( )3
32 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 cs s a s s b s s c
abcs
abcs
Rs
Rs
Rr
∆ ∆ ∆ ∆ ∆ ∆ = − − − − − −
= ∆ ⋅ ⋅ ⋅− − −
= ∆∆
∆=
∆=
∆ = ⋅
=
14
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 Rr r r r r r r r r r r r a b c r s
+ + + = − − − = ≡
Solution to 3
2 2 2 2 2
1 1 1 1 1 1 64 4
A B B C C A
R Rr 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 rr rlhsr r r r r r
r r r r r rr r r
R r r r r r rr r r
Rsr r r
Rsrs
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.
Rabc
R
Ra b c
=
=
Solution to 3
2 2 2 2 2
1 1 1 1 1 1 64 4
A B C
R Rr r r r r r a b c r s
− − − = ≡
:
We rearrange the results of Problem 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 Br r r r r r r r r r r r
− = + − = + − = + .
The result follows by direct substitution.
15
Problem 13. Prove:
( )( ) ( )( ) ( )( ) 2
1 1 1s 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 cs s a s b s c
s s a b c
s s s
s s
r
− − −= + +
− − − − − − − − −
− + − + −=
− − −
+ +=
∆
−=
∆
= ≡ ∆ ∆
=
16
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 abc A
= + − −= ∆ −
Squaring and adding, , gives ( ) ( )2i ii+ 2
( ) ( )( )
22 2 2 2 2 2 2 2
22 2 2 2 2 2
4 cos sin 16 .
Hence, 4 16 , as required.
b c A A b c a
b c b c a
+ = + − + ∆
= + − + ∆
17
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 cos4sin 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 AbB C
B C A BaA C
B A A BC C
B A A BC
C CC C
= +
= × ×
+ × ×
= ×∆× + ×∆×
∆= +
∆ ∆= − ≡
= ∆ as required.
18
Problem 16. Prove: ( ) ( ) ( )2 1 1 1
2 2 2tan tan tans 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 Cs
A B Cs
s A B C
=
∆=
∴ ∆ =
19
Problem 17. Prove: 22 sin sin sinR 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
=
= ×
= ×
=
20
Problem 18. Prove: ( )sin sin sinRr A B C∆ = + + Solution:
( )
2 2 2 , using (5.1)
2 2 2 ,
2 ,
2 2 ,
4 ,
1 , using (5.2)
, as required (on using (6.1)).
rhs Rrbc ca ab
a b cRrabc
Rr a b cabc
Rr sabc
R rsabc
rs
rs
∆ ∆ ∆ = + +
∆ + ∆ + ∆ =
= ⋅∆ + +
= ⋅∆ ⋅
≡ ⋅ ∆
= ⋅ ∆ ∆
= ≡ ∆
21
Problem 19. Prove:
( )
2 2 sin sin2 sin
a b A BA B
−∆ = ⋅
−
Solution: From (5.3) we consider the two forms
2 21 12 2
sin sin sin sin, .sin sinB C Ca b
A B∆ = ∆ =
A
Rearranging gives:
2 21 12 2
sin sinsin , sin .sin sin
A Ba C bB A
∆ ∆= = C
Subtracting we have
( )
2 21 12 2
2 2 2 2
2 2
2 2
sin sin sin sin ,sin sin
sin sinsin ,
sin sin 2
sin sin sin 2 sin sin
A B a C b CB A
A B a b CA B
a b A B CA 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 sin2 sin sin
a b A B CA B
−∆ = ×
− can be written:
( )
2 2 sin sin , as required.2 sin
a b A BA B
−∆ = ⋅
−
22
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 sinA B C A Bsi C+ + ≡ , therefore we have:
( )
( )
2 2 2
2 2 22
2 22
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
RR
=
∆= ×
∆= ∆ ×
= ∆
Problem 17
23
Problem 21. Prove:
( )
2 2 2
4cot cot cot
a b cA 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 cA B C
a b c A B C
a b cA B C
+ + = ∆ + +
+ + = ∆ + +
+ +∴ ∆ =
+ +
24
Problem 22. Prove:
( ) ( ) ( )A B C A B C AB C
B CA B B C C A
r r r r r r rra rs rr r r r r r
+ += = = +
+ +r
r
Solution to ( )A B C
A B B C C A
r r rr 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 rrhsr r r r r r
r r r r r r r rr r r r r r
r r r r r r r rr r r r r r
s r rs
s r r r r rss ss 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 rs s a a
r r r r r r+
= − − =+ +
Solution to ( )A B Cr r rs+
=a :
( ) ( )We have: , as required (on using )A B C A B C
A B B C C A
r r r r r ra
sr r r r r r+ +
= ≡+ +
Problem 7
Solution to ( ) AB C
B C
rrrr 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 ra
s r r rr
r r rra r r
r rr rrr
+ += ≡
+∴ = ≡ +
Problem 4
25
Problem 23. Prove: ( )( )2
A Ba r r r r= − + C Solution:
( ) ( )( )
( )( )( )( )
( )
( )
( )
( )( )
( )
2
2
22
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 bs s a s b s c
a s b cs 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.
26
Problem 24. Prove:
( )( )( )
12sin A
A B A C
rAr r r r
=+ +
Solution:
( )( )( )
( )
( )
( )
( )
( ) ( )
( )
2
2 12
12
1 12 2
12
,
, using 4
, using 4
4 cos, using
4
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 rrhs
r r r r r r
r r rR r r r r r r
r r r
R s
r R As R
r As
A A
A
+=
+ + +
+=
+ +
+=
=
=
=
=
Problem 10
Problem 7
Problem 9
27
Problem 25. Prove:
( )( )
2sin A A B B C C A
A B A C
r r r r r r rA
r r r r+ +
=+ +
Solution:
( )( )( )( )
( )( )
( )
( )
( )2 12
2,
2, using
4
2,
4
2, using
4
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 rrhs
r r r r r r
r r r r r r r r rR r r r r r r
r r rR r r r r r r
r r rRs
r R ARs
rs
+ + +=
+ + +
+ + +=
+ +
+=
+ +
+=
×=
=
Problem 10
Problem 7
Problem 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
=
=
=
28
Problem 26. Prove: ( ) ( ) ( )3 2 2 21 1 1
2 2 2cot cot cotA B Cr r r r A B C= Solution:
( ) ( ) ( )
( ) ( ) ( )
( )( )( )[ ]
( )
2 2 2 2 2 21 1 12 2 23
2 2 23
2
2 3
4
2 3
42
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 Cr
s a s b s cr
s s a s b s cs r
s r
sr rss r
r r r
= × × ×
= × − × − × −
− − −=
∆=
= ≡
= Problem 4
29
Problem 27. Prove:
2 2
2 2 2 2 2
1 1 1 1
A B C
a b cr 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 cr
s as a s bs b s cs cr
s s a b c a b cr
s s a b cr
s a b cr
r
= + + +∆ ∆ ∆
− − −
− + − + −= +
∆
− + + − + + − += +
∆
− + + + + += +
∆
− + + += +
∆
+ += − +
∆ ∆
= −2 2 2
2 2
2 2 2
2
, using (6.2)
, as required.
a b cr
a b c
+ ++
∆
+ +=
∆
30
Problem 28. Prove: r r r r 2 2 2 2 2 2 216A B C R a b c+ + + = − − − 2
Solution: Note that the identity ( ) (2 2 2 2 2 )x 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 22
42
3 22
16 8 2 2 ,
16 2 4 ,
But from (6.2) and (8.21)
4 ,
1 ,
1 , using (5.6)
1
A B Cr r r r R Rr r s
R s r Rr
abcs r Rr ss 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
= + + − + +
= + + − + +
= + + − + +
≡ + +
+ + + = − × + +
= − − −
31
Problem 29. Prove: A B Cab 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 cs s a s b s c
s s a s b s c
s s a b c ab
ab
∆ ∆ ∆ ∆= ⋅ + ⋅
− − −
∆ − + − −=
− − −
= − + − −
= − + + +
=
32
Problem 30. Prove:
1 1 1 12ab bc ca Rr
+ + =
Solution:
,
,
2 , using (5.2)4
, using (6.1)2
1 , as required.2
a b clhsabc abc abc
a b cabc
sR
sRrs
Rr
= + +
+ +=
=∆
=
=
33
Problem 31. Prove:
1 12
CA B rr rbc ca ab r R
+ + = −
Solution:
( )( ) ( )( ) ( )( )( )( )( )
( )( ) ( )( ) ( )( )
( ) ( ) ( ) ( ){ }
2
2
,
1 , using (6.10)
,
, using (5.6)
3 ,
A B Car br crlhsabc
a b cabc s a s b s c
a s b s c b s c s a c s a s babc s a s b s c
a s b s c b s c s a c s a s bsabc
s s a b c s a b c b c a c a b abcabc
s
+ +=
∆ ∆ ∆ = + + − − −
− − + − − + − −∆= − − −
− − + − − + − −∆ = ∆
= + + − + + + + + + ∆
= ( )32 2 3 ,s s ab bc ca abcabc
− + + + ∆
We now note the identity: ( )( )( ) ( ) ( )3 2s 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 abcs a s b s c s ab bc ca s abcs a s b s c s ab bc ca s abcs 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 cabc
s s a s b s csabc
sabc r abc
abcr abc R r R
≡ − − − −∆
− − −= −∆ ∆
∆ ∆= − ≡ −∆ ∆
= − ⋅ ≡ −
34
Problem 32. Prove:
0A B C
b c c a a br 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
− − −= + +
∆ ∆ ∆ − − −
− − + − − + − −=
∆
− − − + − − − + − − −=
∆
− − + + − − + + − − +=
∆
=∆
≡
35
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 Cr 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.
36
Problem 34. Prove: A B B C C A A B Cr r r r r r rr rr rr ab bc ca+ + + + + = + + Solution:
, using
By 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.
37
Problem 35. Prove: ( ) ( ) ( ) ( )2 2 2 8 2A B CI 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.
38
Problem 36. Prove: 216A B CI 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 cs s a s b s c
abc abcss s s a s b s c
abc abc ss
abc abc abcss
r
= × × − − −
= × − − −
≡ ×
− − −
×= × ∆
= ≡ =∆∆ ∆ ∆
( )2
2
2
using (6.1)
4 , using (5.2)
16 , as required.
abc r R r
R r
= = ∆
=
39
Problem 37. Prove: ( ) (2 4B 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
40
Problem 38. Prove: ( ) ( ) ( ) ( ) ( ) ( )2 2 2 2 2
A B C B C A C A2
BI I I I I I I I I I I I+ = + = + Solution: First consider the term ( ) ( )2 2
A B CI I I I+ We have
( ) ( ) ( ) ( )
( ) ( )( ) ( )
( ) ( )
( )
2 22 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 Ca aI I I I
A A
a A a AA A
aA A
a aA A
R
R
+ = +
+=
≡
≡ = ×
= ×
=
The results follow by symmetry.
41
Problem 39. Prove:
sin sin sinA B C B C A C A BI I I I I I I I I I I I
A B⋅ ⋅ ⋅
= =C
Solution:
First consider the term sinA B CI 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 sinsin
sin 2 ,sin sin
2 ,sin
2 2 , using (4.7)
8 .
A B C
a aA AI I I I
A A
aA A
A
aA a
A A
aA
R
R
⋅
⋅ =
≡
≡ =
= ×
= ×
= The results follow by symmetry.
42
Problem 40. Prove:
, and a 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 from either of their points of contact is
abca b c+ +
Solution:
Radius = c
Radius = b
Radius = a
I
D
F
C
A
B
E
From basic circle theory if tcircles are touching then a line through their centres crosses their point common tangency at right angles.
wo
Therefore,
, say,, say,, say.
AB a b aBC b c bCA c a c
′= + ≡′= + ≡′= + ≡
Using the same argument as on Page 70, “Incentre Proof” we see that it must be the case that , say,DI EI FI r= = ≡
where . inradius of triangle r 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
abcra b c
′ ′ ′∆ = − − −
= + + − − − − − −
= + +
= + +
+ +∆= ≡
+ +
=+ +
43