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