Elementary Triangle Goemetry-Solutions

8/3/2019 Elementary Triangle Goemetry-Solutions

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Elementary Triangle GeometryMark DabbsSolutions to Exercises

Version 1.0 April 2004(www.mfdabbs.com)


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3

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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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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( ) ( )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



= + + +

= + + +

= + + +

= + + +

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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Elementary Triangle Goemetry-Solutions

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Transcript of Elementary Triangle Goemetry-Solutions