SOLUTION OF TRIANGLE - Full Length Tests - IIT JEEcompetishun.com/.../uploads/2015/03/Solution-Of-Triangle.pdfSOLUTION OF TRIANGLE PAGE # 1 SOLUTION OF TRIANGLE GENERAL NOTATION :

PAGE # 1SOLUTION OF TRIANGLE

SOLUTION OF TRIANGLE GENERAL NOTATION :

1. In a triangle ABC angles at vertices are usually denoted by A, B, C

sides opposite to these vertices are denoted by a, b, c respectively.

2. Circumradius is denoted by R

3. Inradius is denoted by ‘r’.

4. Ex-radii are denoted by 1 2 3r , r , r opposite to vertices A, B, C respectively..

5. 2s = a + b + c2s – 2c = a + b – ca + b – c = 2(s – c)a + c – b = 2(s – b)


2 2a b c = a b c a b c

= 4s(s – c)

2 2a b c a b c a b c

= 4 s a s b

s a , s b , s c will be always +ve.

6. ’’ or S is used for area of triangle.

7.1

pa2

2pa

perpedicular = 2 Ar. ofopp. side

.

8. 2 2 2 2AB AC 2 AD BD

2

2 2 2 ac b 2 AD4

2 2 2

22 b c a

2AD2

length of median, AD = 2 2 22 b c a

2

.

2. In a triangle the angle opp. to greatest side will be greatest.

3. In a centroid and incenter will always lie within it whereas orthocentre and circumcentre may lie outsidethe .

4. The ratio of the area of the triangles made on same base (altitudes) will be equals to ratio of altitudes (base)

Ar ABD BDAr ADC DC


2

1

Ar ABC pAr DBC p

4. In aABC(i) sin A, sin B, sin C > 0(ii) (a) sin(A + B) = sin C(b) cos(A + B) = –cos C(c) tan(A + B) = –tan C

(iii) (a) A B Csin cos2 2 2

(b) A B Ccos sin2 2 2

(c) A B Ctan cot2 2 2

5. # Sine Rule :

In ABC

a b c 2RsinA sinB sinC

In ABD & ACD :PROOF : AD = AD

c sinB bsinC

In ABD ACD

b a

sinB sinA O is circumcentre

In BOM BM = a2

asinA2

a = 2R sin AA

we have a b c 2R

sinA sinB sinC .


PATTERN IDENTIFICATION :

(i) a sin AA (i.e. a and sin(A) are mutually convertible)

(ii) 2 2a b sin A B .sin A B

sinC.sin A B

PROOF : 2 2a b = 2 2 24R sin A sin B

= 24R sin A B .sin A B

= 24R sin A B .sinC .

(iii) a b sinA sinB

c sinC

a b 2R sinA sinB

= A B A B4Rsin cos

2 2

= C A B4Rsin sin2 2

(iv) acosB sin2A

(v) a tanA

cos A .

6. To find a side indentify the in which it lies preferably right angle or the triangle one side of which and two of theangles are known.

7. If in the ques. two of the angles say A & B and one of the corresponding sides say a is given this may impliessine rute.

NAPIER’S ANALOGY :

1.b c sinB sinCb c sinB sinC

=

B C B C2sin cos2 2

B C B C2sin cos2 2

= B C B Ctan cot

2 2

= B C Atan .tan

2 2

B C b c Atan cot2 b c 2

PAGE # 5SOLUTION OF TRIANGLEwhich gives the result :

B C b c Atan tan2 b c 2

Similarly A C a c Btan tan

2 a c 2 &

A Btan2

= a b C

tana b 2

..

NOTE : If trigonometric ratio of difference of two of the angles (say A – B) and the corresponding sides (a & b) are givenin the ques this may imply Napier ’s analogy.

PROJECTION FORMULA :

BC = BD + DC

a c cosB bcosC

a c cosB bcosC

Similarly b = acosC c cos A

& c = acosB bcosA .

COSINE RULE :

1. In ABD (refer above figure)2 2 2AB AD AD

22 2 2C a bcosC b sin C

= 2 2 2 2 2a b cos C 2abcosC b sin C 2 2 2C a b 2abcosC 2 2 2b a c 2ac cosB 2 2 2a b c 2bc cos A

2 2 2a b c

cos C2ab

Similarly 2 2 2b c a

cosA2bc

2 2 2a c bcosB

2ac

.

GENERAL NOTE :

(i) for three quantities a, b, c.

(i) a b 0 (ii) a b 2 a (iii) b c a 0


(iv) a b c a (v) ab a b a b .

NOTE:

1. If two of the sides (say a & b) and the third angle (C) is given in the ques. then this may imply application ofcosine rule.

2. bc cos A = 2 2 21 b c a2

3. cot A

= cot Asin A

= 2 2 2b c a2bc sinA

= 2 2 2b c a

4

= 2 2 21a c a

4

= 2a

4 =

= 2 2 2a b c

4

.

4. If ABCD is cyclic quadrilateral :2 2BD BD

2 2 2 2a d 2adcos A b c 2bccosA

2 2 2 22cos A ad bc a b b c

2 2 2 2a d b ccosA

2 ad bc

2 2 2 2a b c dcosB

2 ab cd

cos C = 2 2 2 2b c a d

2 bc ad

2 ab cd ac bdBD

ad bc

Similarly

2 ac bd ad bc

ACab cd


22 2BD .AC ac bd

BD.AC = ac + bdBD.AC = AB.CD + AD.BC(Ptolemy’s Theorem)

NOTE : If A = 3

then cos A cos3

= 2 2 2b c a

2bc

.

2 2 21 b c a2 2bc

2 2 2a b c bc .

Q.1 (i) If a = 5, b = 4 & cos (A – B) = 3132

find C.

Q.2 P.T. b c cos A a .

Q.3 P.T. a bcos c ccos B = 2 2b c .

Q.4 If a 1 3, b 2 & c 60 , Find other sides & angles.

Q.5 If C2

, a = 3, b = 4 and

D is on A B such that BCD6

Find CD.

6. If , , are legnth of altitudes of ABC.

T.P.T. 2

cot A1

.

7. P.T.

A Aasin B b c sin2 2

.

8. If b c a c a b11 12 13

T.P.T. cos A cosB cosC

7 12 25 .

9. P.T. 2 2

2b c .sin 2A 0

a

.

10. P.T. a

2asec A tanBtanCcosB cosC

.


1.

2 1 cos A BA Btan2 1 cos A B

1 cos A BA Btan

2 1 cos A B

= 1 31/ 321 31/32

= 1

3 7.

Using Napier’s

A B a b ctan cot2 a b 2

= 1 1 c

cot9 23 7

c 7

tan2 3

1 7 /9cosc1 7 / 9

1

cosc8

.

2 2 2c a b 2abcosc

= 125 16 2 5 48

= 36

c = 6.

Q.2 L.H.S. = bcos A ccos A + acos B ccos B + acos C bcos C

= c + b + a = a .

Q.3 abcosC ac cosB

= 2 2 2 2 2 2a b c a c b

2 2

= 2 2

2 22 b c

b c2

.

Q.4 cosine Rule :2 2 2c a b 2abcosc

= 2 11 3 4 2.2. 1 3

2


= 1 3 3 1 2 4

= 3 1 3 1 4 = 6

= c = 6 .

Q.5 2 BCD :

CD BCsinB

sin B6

CD = 3sinB

sin B6

=

435

1 3 3 42 5 2 5

.

Q.62a

2b

2c

2

2 2

a14

= 2 2 2a b c 1

4

= 1 cot A .

Q.7 consider

b c sinB sinCa sinA

=

B C B C2sin cos2 2

A A2sin cos2 2

=

B Ccos2Asin2

(As A B Ccos sin2 2

)


=

C Bcos2Asin2

=

A B Bcos2Asin2

A Ab c sin acos B2 2 2

= Aa sin B2

.

Q.8b c a c a b11 12 13

= 2 a b c

36

(Using C & D)

c b a5 6 7

c 5

b 6 a = 7.Apply cosine rule

2 2 2b c acosA

2bc

= 36 25 49

60

etc.

Q.92 2

2b c sin2A

a

= 2 2

2b c 2sinA cosA

a

= 2 2b c

cos AR.a

= 2 2 2 2 2b c b c a

2Rabc

= 2 4 2 2 21 b c a b c2Rabc

= 0.

10.a

cosBcosC

= 1

acos AcosA cosBcosC


= 1

2RsinA cos AcosA cosBcosC

= R4sinA sinBsinC

cosA cosBcosC (using sin 2A 4 sin A )

= 2asec A tanB tanC

HALF ANGLE FORMULAE :

2 2 2b c a

cos A2bc

2 2 2

2 A b c a2cos 12 2bc

2 2 2

2 A b c a2cos 12 2bc

=

2 2 2b c a 2bc2bc

= 2 2b c a

2bc

= 2s.2 s a

2bc

s s aAcos

2 bc

Similarly s s bBcos

2 ac ,

s s cCcos2 ab

Again 2 Acos A 1 2sin2

= 2 2 2b c a

2bc

2 2 2

2 A b c a2sin 12 2bc

= 22a b c

2bc

s b s cAsin

2 bc

Similarly s a s cBsin

2 ac

also,

s b s cA

tan2 s s a

,

s a s cBtan

2 s s b

.

AREA OF TRIANGLE :

1.1 ht base2

= 1 c sinB a2


= 1 ac sinB2

= 1 bc sinA2

[using a sin B = b sin A]

= 12RsinB 2R sinC sinA

2

= C

2

A

2R sinA .

Also

1bc sinA2

= 1 A Ab.c.2sin cos2 2 2

= s b s c s s a

b.c. .bc bc

= S s a s b s c

1absinC

2

= 22R sinA= S s a s b s c .

*

s b s cA

tan2 S s a

= s b s c

= s s a

.

*

S s bB

cot2 s a s c

= s a s c

= s s b

.

*2

sin Abc

.

* If ques. contains half s and sides, then use above formula for manipulation.

COT - m-n THEOREM :

If base BC is divided by pt. D in ratio m : n, then


(1) (m + n) cot = m cot – n cot (2) (m + n) cot = n cot B – m cot C.Proof : in ABD :

AD BD

sin sin

...(1)

In ACD :

AD DC

sinsin

...(2)

(1) (2)

sin BD sinsin DC sin

sin m sinsin n sin

nsin sin cos cos sin msin sin cos cos sin

Dividing by sin .sin .sin , we get

(m + n) cot = m cot – n cot

* If in question, one of the side is bisected/trisected etc. then this may imply application of cot m – n theorem.

Q.1 P.T.

Acot2

cot A

=

2

2

a

a

.

Q.2 If 22a b c , then find tan A.

Q.3 Let C3

, A 75 , If D is on ACAC

Ar. (BAD) = 3 Ar. (BCD). Find ABD.

Q.4 If median AD AB, in a , P.T. tan A + 2 tan B = 0.

Q.5 If medians AD & BE intersects at 2

, then P.T..

2 2 2a b 5c .

Q.6 Find ‘A’, if (a + b + c) (b + c – a) = 3bc.

Q.7 If A3

, T.P.T. 2 2 24AD b bc c , where AD is median.

Q.8 If A, B, C are in A.P., and if 2 23c 2b , then find angles of .


Q.9 In a , P.T. A B 2c1 tan tan2 2 a b c

.

Q.10 If median AD is 'r to AC, T.P.T.s

2 22 c acosA cosC

3ac

.

Ans.1Acot2

=

s s a = 1 s s a

= 213s s a

= 2a b c4

also 2a

cot A4

Acot2

cot A

=

22

a

a .

Alt : A

cot2 =

Acot

2 = s s a

= 3s s a s b s c

=

2s

.

2. a b c a b c

= 4 s b s c

s b s c

= 4 A 1tan2 4

s b s cAtan

2 s s a

= s b s c

2

A2tan2tan AA1 tan2

=

12411

16

=

815

.

3. cot m – n th :

3 1 75 1.cot 75 3 cot60


4.

11 1 cot B 1.cot A cot2 2 2 2

2tanB tan A

tanA 2 tanB 0 .

5. Dist. of A frm. centroid = AG = 2AD

3 =

2 2 22 b c a

3

2 2 2AE AG GE (where is mid pt. of AC)

2 2 22 2 b c ab

2 9

+

2 2 22 a c b

36

.

6. s s a 32s2 s a 3bc

bc 4

= 2 Acos2

= 34

A 3

cos2 2 A = A =

3

Alt :

2 2b c a 3bc

2 2 2b c a

1bc

2 2 2b c a 1

2bc 2

= cos A A3

.

7. A = 3

2 2 2b c bc a Also


AD = 2 2 22 b c a

2

2 2 2 24AD 2 b c a

2 2 2 22 b c b c bc

= 2 2b c bc .

8. A, B, C A.P..

B3

2 23c 2b

c 2b 3 =

sinCsinB

2 3 2 1sinC sin .

3 3 2 3 2

C4

A = 3 4 .

9.A B1 tan tan2 2

=

s b s c s a s c

1s s a s s b

= s c

1c

= c 2cs a b c

.

10. 2 2 22

2 2 22 b c aa b AD b

2 4

2 2 2 2 2a 4b 2b 2c a

2 2 2a c 3b

2 2 2b c acosA

2bc

2 2 2b a ccosC

2ab


= 2 2b 3b2ab

2 2 22 a c 4b

cos AcosC3.2bc 2ab

= 2 22 c a

3ac

.

Q.1 P.T.

2

2

B Ccos2

b c

+

2

2

B Csin2

b c

= 2

1a

.

Q.2 ABCD is a trapezium such that AB & DC are || & BC is 'r to then. If ADB = , BC = p & CD = q,

T.P.T. 2 2p q sin

ABbcos qsin

.

Q.3 If 2 ab c

tan sina b 2

,

T.P.T. c a b sec .

Q.4 P.T. 2 22 2 2c cc a b cos a b sin2 2

.

Q.5 P.T. a cosBcosC cosA = c cosAcosB cosC = b cos A .cos C cos B .

Q.6 2 2 2a cos2B b cos 2A 2abcos A B c .

Q.7 2 2Abccos s2

.

Q.8 2b c Acos 0a 2

Q.9A2 acos2

= a

× .

Q.10

Atan 12a b a c

.

Sol.1 L.H.S =

2 2

2 2

B C B Ccos sin2 2

b c b c


=

2

22

B Ccos2

4R sin B sin C

+

2

22

B Csin2

4R sin B sin C

= 22 2

1 1 1B C B C4R 4sin 4cos

2 2

=

2 2

22 2

B C B Csin cos1 2 2

B C B C4R 4sin cos2 2

= 2 2

1 14R sin B C

= 2

1a

.

Sol.2 In BAD

AB BD

sin sin

AB =

sin .BD

sin .cos cos .sin

=

2 2

2 2 2 2

sin . p qq psin .cos

p q p q

=

2 2p q .sin

qsin pcos

.

Sol.3 2 2sec 1 tan

=

2

2

C4absin21

a b

=

2

2

a b 2ab 1 cos C

a b

=

2 2

2

a b 2abcos C

a b

=

2

2C

a b .


Sol.4 2 22 2C Ca b .cos a b .sin2 2

= 2 2 2 2C Ca b cos sin2 2

– 2 2C C2ab cos sin

2 2

= 2 2a b 2abcos C = C2.

Sol.5 a cos B cos C cos A

= a cos B cos C cos B C

= a cos B .cos C cos B cos C sin B sin C

= a sin B sin C

= 2Rsin A sin B sin C

Using symt. all terms are equals to 2R sin A .

Sol.6 2 2a cos 2B b cos 2A 2abcos A B

= 2 2 2a cos B sin B + 2 2 2b cos A sin A + 2ab cos A cos B sin A sin B

= 2 2 2 2a cos B b cos A 2abcos A cos B – 2 2 2 2a sin B b sin A 2absin A sin B

= 2 2acos B bcos A asin B bsin A

= 2 2c 0 c .

Sol.7 2 Abc.cos2

= s s a

bc.bc = s s a = s(s) = s2.

Sol.8 2b c A.cosa 2

= s s ab c

.a bc

= sb c s a

abc

= ss b c a b c 0

abc .


DIFFERENT CENTRES OF A TRIANGLE :

A. Circum centre :

(1) abcR4

(2) OM = Rcos A = Dist. of BC frm. c.c. Dist of side frm c.c. = R cos A, R cos B, R cos C.

B. Incentre :

(1) BD cDC b

.

(2) (a) BD = c

ab c

||ly DC = ab

b c(3) In BAD :

IA ABID BD

= c

acb c

= b c

a

.

(4) AIB = A B2 2

= c

2 2 .

(5) (i) In BAD :

AD BDAsinB sin2

AD = sinB C

sin A / 2 b c

[using a sin B = b sin A]

= 2bc A.cosb c 2

= length of r bisector..

similarly, 2ac BBE .cosa c 2

.

(ii) Ar ABC Ar IBC Ar (IAC) + Ar(IAB)

1 1 1ar br ar2 2 2

= r.s

r = s


(iii) rs

= r . s as s a

= As a .tan2

= Bs b .tan2

= Cs c .tan2

.

(iv) A B C4Rsin .sin .sin2 2 2

s b s c s a s c

4R .bc ac

.

s a s bab

= 4R. s a s b s c

abc

= s s a4R.

abc s

= 21

. rs s

= Ar 4R sin2

.

(i) rs

= As a .tan2

= A4R sin2

(iv) In BPI

BI = Br cosec2

= A C4Rsin sin2 2

= Dis. of vertex B from incenter.

Dis. of vertices from incenter rcscA2

, rcsc

B2

and rcsc

C2

.

PAGE # 22SOLUTION OF TRIANGLEIn BPI

B rtan2 BP

BP =

r s bBtan2

= length of tangent from B to incircle. BP = BR = s – bCP = CQ = s – cAQ = AR = s – a

(vi) IAO = A B2 2

= A B C

B2 2 2 2

= B C

2

AI = B C4R sin sin2 2

AO = R.In AIO

2 2 2IO AI OA 2.OA.AI.cos IAO .

2 2 2 2 2B CIO R 16R sin .sin2 2

– B C B C2.R.4R sin sin .cos2 2 2

= 2 2 B C B C B CR 8R sin sin 2sin sin cos

2 2 2 2 2

= 2 2 AR 8R sin

2

IO = AR 1 8 sin2

= 2R 2Rr(i) R 2r

(ii) A 1sin2 8

Orhocentre :

(i) BD = c.cos B

(ii) DC = b.cos C


(iii)

c cos BBDDC bcos C

=

tan Ctan B

(iv) BAD

BH = c.cos B .cosec C

= c .cos Bsin c

= 2Rcos B

HD = ccos B .cot C

=

cos Cc.cos B .

sin C

= 2Rcos B .cos C

Hence Dis. of verteces from orthocenter will be 2R cos (A)

2Rcos B , 2Rcos C and the distance of the sides from orthocenter will be 2R cos(B) cos(C),

2Rcos C cos A and 2Rcos A cos B

HAO B C AH = 2R cos(A)AO = R

HAC C2

AOM B

AOM B2

HAO C B2 2 = B – C.

2 2 2OH AO AH 2OA.AHcos B C

= 2 2 2R 4R cos A 2.R.2Rcos A cos B C

= 2R 1 4cos A cos A cos B C

= 2R 1 4cos A cos B C cos B C

= 2 2R 8R cos A = OH = R 1 8 cos A .

(iii) EX-CIRCLE :

1. (i) 1r s a


(ii) 2r s b

(iii) 3r s c

2. 1A B Cr 4R sin .cos cos2 2 2

2A B Cr 4Rcos .sin .cos2 2 2

3. (a) 1Ar s tan2

(b) 2Br s tan2

(c) 3Cr s tan2

4. 21 2

A Br r s tan tan2 2

21 2r r s .

5.1

1 1r r

Proof:s a

= 1 s a

= 3s 2s

=

s 1r

.

6. 1 2 3r r rr

Proof: . .s a s b s c

= 2

2s ss a

.

Q.1 Prove that rcos A 1R


Sol. cos A = A1 4 sin2

=

A4R sin21

R

= 1 + rR

.

Q.2 Prove that if 1 2 3r , r , r in H.P. then ABC are in A.P

Sol.1 2 3

1 1 1, ,r r r A.P..

s a s b s c, , .....r

A.P.

s – a, s – b, s – c A.P.–a, –b, –c A.P..a, b, c A.P..

Q.3 Let dis. of orthocenter from vertices is p, q, r then prove that

Proof: a.q.r abc p 2Rcos A

q 2Rcos B

r 2Rcos C

L.H.S. = aqr= 2Rsin A .2Rcos B .2Rcos C= 38R sin A .cos B .cos C= 38R . sinA abcALT :

1Ar ABC + 2 + 3

= 1 qr.sin B C2

= a

qr4R

1 2 3

abc aqr bpr cpr4R 4R 4R 4R


= abc = aqr .

Q.1 If 1 2 3r r r r

P.T. is right angled.Q.2 If B / 2

P.T. r = a b c

2

.

Q.3 If altitudes from A, B, C are produced to meet circum centre, if length of produced parts is , , then prove that

centre, if length of produced parts is , , then prove that a2 tan A

.

Q.4 If , , are length of internal bisectors T.P.T. 1 A.cos

2 =

1a .

Q.51 1

ab 2Rr .

Q.61

b c 0r

.

Q.7 If , , are the distance of the vertex of a from the corresponding pts. of contect with the in-circle

T.P.T. 2r

.

A.1s a s b s c s

1 1 1 1s a s s b s c

s s as s a

= 2s a b cs b s c

2 2s as s b c s bc

s b c a bc

a b c b c a bc

2 2 2b c 2bc a 2bc

2 2 2b c a .

4. Br s b .tan2

= s b .tan4 = s – b


= 2 s b

2

= a c b

2

.

5.1

ab = c

abc = 1

. cabc

= 2s

abc =

4 s 1. .

abc 2

=

1 1 1.

R r 2 .

b c

s a

= b c s a

= 1 b c b c = 0.

6.1

b cr =

b c. s a

= s b c – 1 a b c

= 0.

7. s a

s b

s c s

=

=

s a

s

= 2

s. s a

s

= 2

2rs

2r

.

4.2bc A.cosb c 2

Acosb c 1 1 122bc 2 b c

= Acos2

=

1 1 12 b c


= 1 1 12 b c

= 1a .

3. As BPA = BCA

BPA c Also,

In BDP

BDcot C

=

cos Cc.cos B .

sin C

= 2Rcos B cos C

a a

2Rcos B cos C

=

sin A

cos B .cos C

=

sin B CC B .C C

= tan B tan C

= q = t B t C = 2 t A .

Q. If 1 2 3p , p , p are length of 'r from vertices to sides then P.T..

(i) 1

1 1p r

(ii) 1

cos A 1p R

(iii) 1bpc

+ 2cpa

+ 3cpb

= 2 2 2a b c

2R

.

Q.2 P.T. pro. IA = abcA B Ctan tan tan2 2 2

.

Q.3 If x, y, z are respectively distance of the vertex from its orthocentre then prove that

x 2 R r .

Q.4 If x, y, z respectively are the 'r from the circum center to the sides of the triangle ABC then P.T. a abcx 4xyz .

Q.5 (i) P.T. in a triangle 2 Atan 1

2 .


(ii)

2 Aacos324a

.

(iii) 2s

3 3

(iv)

2 Aasin124a

.

Ans.1 12

pa

1

1p =

a2 =

a

2 =

2s2

= 1r

.

(iii)

1

cos A

p

= cos A .a

2

= R sin 2A2 = R .4 sin A

2

= 22R sin A

R .

(iii) 1b 2bpc a

2

1b .2

bpabc

1pbc =

2abc

. 2p = 21 a2R .

A.2. T.P. IA = Aa. tan2

L.H.S. = B C4Rsin sin2 2

= 3 A B C A B C64R s s s .s s s

2 2 2 2 2 2

= 3 s A .s B .s C A B C64R .t . .

8 2 2 2 = Aabc tan

2

.

Q.3 x 2Rcos A


2Rsin Aa tan A

x 2Rcos A

ax = tan A = tan A =

ax

x = 2Rcos A

= A2R 1 4 sin2

= A2 R 4R sin2

= 2 R r .

Q.4 x = R cos A

2Rsin Aax Rcos A = 2 tan(A)

a2 tan A

x = 2 tan A

= a22x

=

1 a4 x

.

Q.5 Apply A.M. = G.M.

on 2 Atan

2 ,

2 Btan2

2 2A B A Btan tan tan .tan

2 2 2 2

similarly

adding 2 Atan 1

2 (As

A Btan tan 12 2

)

(ii) 2 Aacos

2a b c

1 cos Aa

2

a

= a acos A

2 a


=

R sin 2A12 4R sin A

=

sin A1 4.

A2 4 4 cos2

= 1 1 A.8 sin2 4 2

1 1 1 382 4 8 4 . (Using

A 1sin2 8

)

(iii) 1/ 3s a s b s c

s a3

3s

s a3

4

22

s s. s a3

2s

3 3

2

maxs

3 3 .

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