21. Properties of Triangles’ Solutions Part 4#

  1. The diagram is given below:

    Problem 151

    Let ABCABC be the triangle. Let OO be the circumcenter and I,I, the incenter.

    Clearly, OA=OB=OC=R,IE=r[IEAB]OA = OB = OC = R, IE=r[IE\perp AB]

    Let OMBCOM\perp BC then BOM=COM=A\angle BOM = \angle COM = A

    Now, OA=R,AI=rcosecA2=4RsinA2sinB2sinC2sinA2OA = R, AI = r\cosec\frac{A}{2} = \frac{4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}{\sin\frac{A}{2}}

    =4RsinB2sinC2= 4R\sin\frac{B}{2}\sin\frac{C}{2}

    OAB=OBA=B(90A)=A+B90=90C\angle OAB = \angle OBA = B - (90^\circ - A) = A + B - 90^\circ = 90^\circ - C

    OAI=BAIBAO=A2(90C)\therefore \angle OAI = \angle BAI - \angle BAO = \frac{A}{2} - (90^\circ - C)

    =CB2= \frac{C - B}{2}

    Applying cosine law in OAI,\triangle OAI,

    cosCB2=OA2+AI2OI22OA.AI\cos\frac{C - B}{2} = \frac{OA^2 + AI^2 - OI^2}{2OA.AI}

    =R2+16R2sin2B2sin2C2OI22.R.4RsinB2sinC2= \frac{R^2 + 16R^2\sin^2\frac{B}{2}\sin^2\frac{C}{2} - OI^2}{2.R.4R\sin\frac{B}{2}\sin\frac{C}{2}}

    OI2=R2[1+8sinB2sinC2{2sinB2sinC2cos(BC2)}]OI^2 = R^2\left[1 + 8\sin\frac{B}{2}\sin\frac{C}{2}\left\{2\sin\frac{B}{2}\sin\frac{C}{2} - \cos\left(\frac{B - C}{2}\right)\right\}\right]

    =R2[1+8sinB2sinC2{cosBC2cosB+C2cosBC2}]= R^2\left[1 + 8\sin\frac{B}{2}\sin\frac{C}{2}\left\{\cos\frac{B - C}{2} - \cos\frac{B + C}{2} - \cos\frac{B - C}{2}\right\}\right]

    =R2[18sinA2sinB2sinC2]= R^2\left[1 - 8\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\right]

    =R22Rr= R^2 - 2Rr

    Let us prove the necessary condition for the second part.

    Let bb be the A.M. of aa and cc i.e. 2b=a+c2b = a + c

    2sinB=sinA+sinC2.2sinB2cosB2=2sinA+C2cosAC2\Rightarrow 2\sin B = \sin A + \sin C \Rightarrow 2.2\sin\frac{B}{2}\cos\frac{B}{2} = 2\sin\frac{A + C}{2}\cos\frac{A - C}{2}

    2sinB2=cosAC22cosA+C2=cosAC2\Rightarrow 2\sin\frac{B}{2} = \cos\frac{A - C}{2} \Rightarrow 2\cos\frac{A + C}{2} = \cos\frac{A - C}{2}

    cosBIO=BI2+IO2BO22BI.IO\cos BIO = \frac{BI^2 + IO^2 - BO^2}{2BI.IO}

    Now BI2+IO2BO2=r2cosec2B2+R22rRR2BI^2 + IO^2 - BO^2 = r^2\cosec^2\frac{B}{2} + R^2 - 2rR - R^2

    =r2cosec2B22rR= r^2\cosec^2\frac{B}{2} - 2rR

    =r.4RsinA2sinB2sinC2sin2B22rR= \frac{r.4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}{\sin^2\frac{B}{2}} - 2rR

    =2rR.2sinA2sinC2sinB22rR= \frac{2rR.2\sin\frac{A}{2}\sin\frac{C}{2}}{\sin\frac{B}{2}} - 2rR

    =2rR[cosAC2cosA+C2]sinB22rR= \frac{2rR\left[\cos\frac{A - C}{2} - \cos\frac{A + C}{2}\right]}{\sin\frac{B}{2}} - 2rR

    =2rR2cosA+C2cosA+C2sinB22rR=0= 2rR\frac{2\cos\frac{A + C}{2}- \cos\frac{A + C}{2}}{\sin\frac{B}{2}} - 2rR = 0

    Thus, BIO\triangle BIO is a right angled triangle.

    Now the sufficient condition can be proved similarly.

  2. A2+B2=π2C2\frac{A}{2} + \frac{B}{2} = \frac{\pi}{2} - \frac{C}{2}

    cot(A2+B2)=cot(π2C2)\Rightarrow \cot\left(\frac{A}{2} + \frac{B}{2}\right) = \cot\left(\frac{\pi}{2} - \frac{C}{2}\right)

    cotA2cotB21cotA2+cotB2=tanC2=1cotC2\Rightarrow \frac{\cot\frac{A}{2}\cot\frac{B}{2} - 1}{\cot\frac{A}{2} + \cot\frac{B}{2}} = \tan\frac{C}{2} = \frac{1}{\cot\frac{C}{2}}

    cotA2+cotB2+cotC2=cotA2cotB2cotC2\Rightarrow \cot\frac{A}{2} + \cot \frac{B}{2} + \cot\frac{C}{2} = \cot\frac{A}{2}\cot\frac{B}{2}\cot\frac{C}{2}

  3. The diagram is given below:

    Problem 153

    HL=r2,ID=rIL=IDLD=IDHK=rr2HL = r_2, ID=r \therefore IL = ID - LD = ID - HK = r - r_2

    In IHL,\triangle IHL,

    cotB2=HLIL=r2rr2\cot\frac{B}{2} = \frac{HL}{IL} = \frac{r_2}{r - r_2}

    Similarly, cotA2=r1rr1\cot\frac{A}{2} = \frac{r_1}{r - r_1} and

    cotC2=r3rr3\cot\frac{C}{2} = \frac{r_3}{r - r_3}

    Now following the result of previous problem

    r1rr1+r2rr2+r3rr3=r1r2r3(rr1)(rr2)(rr3)\frac{r_1}{r - r_1} + \frac{r_2}{r - r_2} + \frac{r_3}{r - r_3} = \frac{r_1r_2r_3}{(r - r_1)(r - r_2)(r - r_3)}

  4. The diagram is given below:

    Problem 154

    Let OO be the circumcenter and PP the orthocenter of the ABC.\triangle ABC. From geometry,

    BOF=COF=A\angle BOF = \angle COF = A and AP=2OF=2RcosA,BF=RsinAAP = 2OF = 2R\cos A, BF = R\sin A

    From right angled ADB\triangle ADB

    BD=ccosB,AD=csinBBD = c\cos B, AD = c\sin B

    Now, PM=AD(AP+MD)=csinB(2RcosA+RcosA)=csinB3RcosAPM = AD - (AP + MD) = c\sin B - (2R\cos A + R\cos A) = c\sin B - 3R\cos A

    OM=FD=BFBD=RsinAccosBOM = FD = BF - BD = R\sin A - c\cos B

    tanθ=PMOM=csinB3RcosARsinAccosB\therefore \tan\theta = \frac{PM}{OM} = \frac{c\sin B - 3R\cos A}{R\sin A - c\cos B}

    =2RsinCsinB3RcosARsinA2RsinCcosB= \frac{2R\sin C\sin B - 3R\cos A}{R\sin A - 2R\sin C\cos B}

    =2sinBsinC2cos[π(B+C)]sin[π(B+C)2sinCcosB]= \frac{2\sin B\sin C - 2\cos[\pi - (B + C)]}{\sin[\pi - (B + C)- 2\sin C\cos B]}

    =2sinBsinC+3cos(B+C)sin(B+C)2sinCcosB= \frac{2\sin B\sin C + 3\cos(B + C)}{\sin(B + C) - 2\sin C\cos B}

    =3cosBcosCsinBsinCcosCsinBsinCcosB= \frac{3\cos B\cos C - \sin B\sin C}{\cos C\sin B - \sin C\cos B}

    =3tanBtanCtanBtanC= \frac{3 - \tan B\tan C}{\tan B - \tan C}

  5. The diagram is given below:

    Problem 155

    Let ABCABC be the triangle. Let OO and II be the circumcenter and in-center of the ABC.\triangle ABC. Let PP be the center and x,x, the radius of the circle drawn which touches the inscribed and circumscribed circle, of ABC\triangle ABC and the side BCBC externally. Let us join OPOP and extend up to Q.Q. Let IDBC.ID\perp BC. Clearly, PP will lie on the extended part of ID.ID. Draw the line ONON parallel to the line IP.IP. Join NP.NP.

    Clearly, OB=OCOQ=R,OP=OQPQ=RxOB = OC OQ = R, OP = OQ - PQ = R - x

    ON=OM+MN=RcosA+xON = OM + MN = R\cos A + x

    NP=MD=OMCD=a2rcotC2NP = MD = OM - CD = \frac{a}{2} - r\cot \frac{C}{2}

    =a2Δs.s(sc)Δ=a2(sc)= \frac{a}{2} - \frac{\Delta}{s}.\frac{s(s - c)}{\Delta} = \frac{a}{2} - (s - c)

    =cb2= \frac{c - b}{2}

    From right angled ONP,OP2=ON2+NP2\triangle ONP, OP^2 = ON^2 + NP^2

    (Rx)2=(RcosA+x)2+(cb)2(R - x)^2 = (R\cos A + x)^2 + \left(c - b\right)^2

    R2+x22Rx=R2cos2A+x2+2RxcosA+(cb2)2R^2 + x^2 - 2Rx = R^2\cos^2A + x^2 + 2Rx\cos A + \left(\frac{c - b}{2}\right)^2

    2Rx(1+cosA)=R2(1cos2A)(cb2)22Rx(1 + \cos A) = R^2(1 - \cos^2A) - \left(\frac{c - b}{2}\right)^2

    4Rxcos2A2=R2sin2A(bc2)2=a24(bc)244Rx\cos^2\frac{A}{2} = R^2\sin^2A - \left(\frac{b - c}{2}\right)^2 = \frac{a^2}{4} - \frac{(b - c)^2}{4}

    4Rxs(sa)bc=a+bc2.ab+c2=(sb)(sc)4Rx\frac{s(s - a)}{bc} = \frac{a +b - c}{2}.\frac{a - b + c}{2} = (s - b)(s - c)

    x=Δatan2A2x = \frac{\Delta}{a}\tan^2\frac{A}{2}

  6. The diagram is given below:

    Problem 156

    Since angles n the same segment of a circle are equal. BED=BAD=A2\therefore \angle BED= \angle BAD = \frac{A}{2}

    and BEF=BCF=C2\angle BEF = \angle BCF = \frac{C}{2}

    Now DEF=BEF+BED=C2+A2=C+A2=90B2\angle DEF = \angle BEF + \angle BED = \frac{C}{2} + \frac{A}{2} = \frac{C + A}{2} = 90^\circ - \frac{B}{2}

    Similarly, DFE=90C2,EDF=90A2DFE = 90^\circ - \frac{C}{2}, \angle EDF = 90^\circ - \frac{A}{2}

    Now area of DEF=12DE.DF.sinEDF\triangle DEF = \frac{1}{2}DE.DF.\sin \angle EDF

    Let RR be the circum-radius of ABC\triangle ABC then clearly RR is also the circum-radius of DEF.\triangle DEF.

    Applying sine rule in DEF,\triangle DEF, we have

    DEsinDFE=DFsinDEF=EFsinEDF=2R\frac{DE}{\sin DFE} = \frac{DF}{\sin DEF} = \frac{EF}{\sin EDF} = 2R

    DE=2RsinDFE=2Rsin(90C2)=2RcosC2\Rightarrow DE = 2R\sin DFE = 2R\sin\left(90^\circ - \frac{C}{2}\right) = 2R\cos\frac{C}{2}

    Similarly, DF=2RcosB2DF = 2R\cos\frac{B}{2}

    So, area of DEF=12.4R2cosB2cosC2.sin(90A2)\triangle DEF = \frac{1}{2}.4R^2\cos\frac{B}{2}\cos\frac{C}{2}.\sin\left(90^\circ - \frac{A}{2}\right)

    =2R2cosA2cosB2cosC2= 2R^2\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}

    =2R2s(sa)bcs(sb)cas(sc)ab= 2R^2\sqrt{\frac{s(s - a)}{bc}}\sqrt{\frac{s(s - b)}{ca}}\sqrt{\frac{s(s - c)}{ab}}

    =2R2ss(sa)(sb)(sc)abc=2R2sΔabc=R2.s.4Δ2abc=R2s2abc4Δ= \frac{2R^2s\sqrt{s(s - a)(s - b)(s - c)}}{abc} = \frac{2R^2s\Delta}{abc} = \frac{R^2.s.4\Delta}{2abc} = \frac{R^2s}{2\frac{abc}{4\Delta}}

    Here, are of ABC=Δ\triangle ABC = \Delta

    =R2s2R=R2s= \frac{R^2s}{2R} = \frac{R}{2}s

    ΔDEFΔABC=Rs2Δ=R2r\Rightarrow \frac{\Delta DEF}{\Delta ABC} = \frac{Rs}{2\Delta} = \frac{R}{2r}

  7. The diagram is given below:

    Problem 157

    ΔABC=ΔABC(ΔACB+ΔBAC+ΔCAB)\Delta A'B'C' = \Delta ABC - (\Delta AC'B' + \Delta BA'C' + \Delta CA'B')

    ΔACB=12AC.ABsinA\Delta AC'B' = \frac{1}{2}AC'.AB'\sin A

    CC\because CC' is the internal bisector of C\angle C

    ACCB=ACCB=ba\frac{AC'}{C'B} = \frac{AC}{CB} = \frac{b}{a}

    AC=bk,CB=ak,\Rightarrow AC' = bk, C'B = ak, where kk is some constant dependent on angles.

    AC+CB=ABc=ak+bkk=ca+bAC' + C'B = AB \Rightarrow c = ak + bk \Rightarrow k = \frac{c}{a + b}

    AC=bca+b\therefore AC' = \frac{bc}{a + b}

    Similarly, AB=bca+cAB' = \frac{bc}{a + c}

    ΔACB=12bca+bbca+csinA\Delta AC'B' = \frac{1}{2}\frac{bc}{a + b}\frac{bc}{a + c}\sin A

    Let Δ\Delta be the area of ABC,\triangle ABC, then Δ=12bcsinA\Delta = \frac{1}{2}bc\sin A

    ΔACB=bc(a+b)(a+c)Δ\therefore \Delta AC'B' = \frac{bc}{(a + b)(a + c)}\Delta

    Similalry, ΔBAC=ac(a+b)(b+c)Δ\Delta BA'C' = \frac{ac}{(a + b)(b + c)}\Delta

    and ΔCAB=ab(a+c)(b+c)Δ\Delta CA'B' = \frac{ab}{(a + c)(b + c)}\Delta

    ΔABC=Δ.2abc(a+b)(b+c)(c+a)\therefore \Delta A'B'C' = \Delta.\frac{2abc}{(a + b)(b + c)(c + a)}

    ΔABCΔABC=2sinAsinBsinC(sinA+sinB)(sinB+sinC)(sinC+sinA)\therefore \frac{\Delta A'B'C'}{\Delta ABC} = \frac{2\sin A\sin B\sin C}{(\sin A + \sin B)(\sin B + \sin C)(\sin C + \sin A)}

    =2sinA2sinB2sinC2cosAB2cosBC2cosCA2= \frac{2\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}{\cos\frac{A - B}{2}\cos\frac{B - C}{2}\cos\frac{C - A}{2}}

  8. The diagram is given below:

    Problem 158

    Let the given triangle be ABCABC and the similar triangle inscribed in triangle ABCABC be ABCA'B'C' such that

    A=A,B=B,C=CA=A', B=B', C=C'

    Let BC=λa,AC=λb,AB=λcB'C'=\lambda a, A'C'=\lambda b, A'B' = \lambda c

    According to the quuestion BOC=θ\angle B'OC=\theta

    Clearly, OCB=Bθ=ACB\angle OC'B = B - \theta = \angle AC'B'

    BCA=180(Bθ+C)=A+θ\angle BC'A' = 180^\circ - (B - \theta + C) = A + \theta

    ABC=180(Bθ+A)=C+θ\angle AB'C' = 180^\circ - (B - \theta + A) = C + \theta

    ABC=180(C+θ+B)=Aθ\angle A'B'C' = 180^\circ - (C + \theta + B) = A - \theta

    Applying sine rule in the triangle BCA,BC'A',

    BAsin(A+θ)=λbsinBBA=λbsinBsin(A+θ)\frac{BA'}{\sin(A +\theta)} = \frac{\lambda b}{\sin B}\Rightarrow BA' = \frac{\lambda b}{\sin B}\sin(A + \theta)

    BA=λ2Rsin(A+θ)\Rightarrow BA' = \lambda 2R\sin(A + \theta)

    Applying sine rule in ABC,A'B'C,

    ACsin(Aθ)=λcsinC\frac{A'C}{\sin(A - \theta)} = \frac{\lambda c}{\sin C}

    AC=λ2Rsin(Aθ)\Rightarrow A'C = \lambda 2R\sin(A - \theta)

    BC=BA+ACBC = BA' + A'C

    a=λ2Rsin(A+θ)+λ2Rsin(Aθ)\Rightarrow a = \lambda 2R\sin(A + \theta) + \lambda 2R\sin(A - \theta)

    a=2Rλ[sin(A+θ)+sin(Aθ)]=2Rλ2sinAcosθ\Rightarrow a = 2R\lambda[\sin(A + \theta) + \sin(A - \theta)] = 2R\lambda 2\sin A\cos\theta

    a=2λacosθ\Rightarrow a = 2\lambda a\cos\theta

    2λcosθ=1\Rightarrow 2\lambda\cos\theta = 1

  9. We have to prove that r1+r2+r3r=4Rr_1 + r_2 + r_3 - r = 4R

    L.H.S. =Δsa+Δsb+ΔscΔs= \frac{\Delta}{s - a} + \frac{\Delta}{s - b} + \frac{\Delta}{s - c} - \frac{\Delta}{s}

    =Δ[sa+sb(sa)(sb)+ss+cs(sc)]= \Delta\left[\frac{s - a + s - b}{(s - a)(s - b)} + \frac{s - s + c}{s(s - c)}\right]

    =Δ[c(sa)(sb)+cs(sc)]=\Delta\left[\frac{c}{(s - a)(s - b)} + \frac{c}{s(s - c)}\right]

    =Δ.c[s(sc)+(sa)(sb)s(sa)(sb)(sc)]=\Delta.c \left[\frac{s(s - c) + (s - a)(s - b)}{s(s - a)(s - b)(s - c)}\right]

    =Δ.c.1Δ2[s2sc+s2(a+b)s+ab]=\Delta.c.\frac{1}{\Delta^2}[s^2 - sc + s^2 - (a + b)s + ab]

    =cΔ[2s2s(a+b+c)+ab]=cΔ[2s22s2+ab]=\frac{c}{\Delta}[2s^2 - s(a + b + c) + ab] = \frac{c}{\Delta}[2s^2 - 2s^2 + ab]

    =abcΔ=4R== \frac{abc}{\Delta} = 4R = R.H.S.

  10. We have to prove that 1r1+1r2+1r3=1r\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3} = \frac{1}{r}

    L.H.S. =saΔ+sbΔ+scΔ= \frac{s - a}{\Delta} + \frac{s - b}{\Delta} + \frac{s - c}{\Delta}

    =3s(a+b+c)Δ=3s2sΔ[2s=a+b+c]= \frac{3s - (a + b + c)}{\Delta} = \frac{3s - 2s}{\Delta} [\because 2s = a + b + c]

    =sΔ=1r== \frac{s}{\Delta} = \frac{1}{r} = R.H.S.

  11. We have to prove that 1r12+1r22+1r32+1r2=a2+b2+c2Δ2\frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \frac{1}{r^2} = \frac{a^2 + b^2 + c^2}{\Delta^2}

    L.H.S. =1r12+1r22+1r32+1r2= \frac{1}{r_1^2} + \frac{1}{r_2^2} + \frac{1}{r_3^2} + \frac{1}{r^2}

    =(sa)2+(sb)2+(sc)2+s2Δ2= \frac{(s - a)^2 + (s - b)^2 + (s - c)^2 + s^2}{\Delta^2}

    =4s22s(a+b+c)+a2+b2+c2Δ2= \frac{4s^2 - 2s(a + b + c) + a^2 + b^2 + c^2}{\Delta^2}

    =4s22s.2s+a2+b2+c2Δ2=\frac{4s^2 - 2s.2s + a^2 + b^2 + c^2}{\Delta^2}

    =a2+b2+c2Δ2== \frac{a^2 + b^2 + c^2}{\Delta^2} = R.H.S.

  12. r=Δs=Δssasar = \frac{\Delta}{s} = \frac{\Delta}{s}\frac{s - a}{s - a}

    We know that tanA2=(sb)(sc)s(sa)\tan \frac{A}{2} = \sqrt{\frac{(s - b)(s - c)}{s(s - a)}} and Δ=s(sa)(sb)(sc)\Delta = \sqrt{s(s - a)(s - b)(s - c)}

    r=(sa)tanA3\Rightarrow r = (s - a)\tan\frac{A}{3}

    Similarly, r=(sb)tanB2,r=(sc)tanC2r = (s - b)\tan\frac{B}{2}, r = (s - c)\tan\frac{C}{2}

  13. We have to prove that 1A=1A1+1A2+1A3\frac{1}{\sqrt{A}} = \frac{1}{\sqrt{A_1}} + \frac{1}{\sqrt{A_2}} + \frac{1}{\sqrt{A_3}}

    R.H.S. =1A1+1A2+1A3= \frac{1}{\sqrt{A_1}} + \frac{1}{\sqrt{A_2}} + \frac{1}{\sqrt{A_3}}

    =1π(1r1+1r2+1r3)= \frac{1}{\sqrt{\pi}}\left(\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}\right)

    =1π(saΔ+sbΔ+scΔ)=\frac{1}{\sqrt{\pi}}\left(\frac{s - a}{\Delta} + \frac{s - b}{\Delta} + \frac{s - c}{\Delta}\right)

    =1π(3s(a+b+c)Δ)=1πsΔ=1π.1r= \frac{1}{\sqrt{\pi}}\left(\frac{3s - (a + b + c)}{\Delta}\right) = \frac{1}{\sqrt{\pi}}\frac{s}{\Delta} = \frac{1}{\sqrt{\pi}}.\frac{1}{r}

    =1A== \frac{1}{\sqrt{A}} = L.H.S.

  14. r1bc+r2ca+r3ab=stanA2bc+stanB2ca=stanC2ab\frac{r_1}{bc} + \frac{r_2}{ca} + \frac{r_3}{ab} = \frac{s\tan\frac{A}{2}}{bc} + \frac{s\tan\frac{B}{2}}{ca} = \frac{s\tan\frac{C}{2}}{ab}

    =sabc(atanA2+btanB2+ctanC2)= \frac{s}{abc}\left(a\tan\frac{A}{2} + b\tan\frac{B}{2} + c\tan\frac{C}{2}\right)

    =sabc(2RsinAtanA2+2RsinBtanB2+2RsinCtanC2)= \frac{s}{abc}\left(2R\sin A\tan\frac{A}{2} + 2R\sin B\tan\frac{B}{2} + 2R\sin C\tan\frac{C}{2}\right)

    =sabc.4R(sin2A2+sin2B2+sin2C2)= \frac{s}{abc}.4R\left(\sin^2\frac{A}{2} + \sin^2\frac{B}{2} + \sin^2\frac{C}{2}\right)

    =sΔ(1cosA2+1cosB2+1cosC2)= \frac{s}{\Delta}\left(\frac{1 - \cos A}{2} + \frac{1 - \cos B}{2} + \frac{1 - \cos C}{2}\right)

    =1r(32cosA+cosB+cosC2)= \frac{1}{r}\left(\frac{3}{2} - \frac{\cos A + \cos B + \cos C}{2}\right)

    We know that cosA+cosB+cosC=1+rR\cos A + \cos B + \cos C = 1 + \frac{r}{R}

    =1r(1r2R)=1r12R== \frac{1}{r}\left(1 - \frac{r}{2R}\right) = \frac{1}{r} - \frac{1}{2R} = R.H.S.

  15. Let DD be the point where perpendicular from AA meets BC.BC. Then AD=hAD = h

    The diagram is given below:

    Problem 165

    Clearly, OB=r,AD=h,OD=hrOB = r, AD = h, OD=h - r (If OO is below BDBD then OD=rhOD = r - h)

    BD=OB2OB2=r2(hr)2=2rhh2BD = \sqrt{OB^2 - OB^2} = \sqrt{r^2 - (h - r)^2} = \sqrt{2rh - h^2}

    Area of triangle =12.2.BD.h=h2rhh2= \frac{1}{2}.2.BD.h = h\sqrt{2rh - h^2}

  16. Let the sides be a,b,ca, b, c then Δ=12ap1=12bp2=12cp3\Delta = \frac{1}{2}ap_1 = \frac{1}{2}bp_2 = \frac{1}{2}cp_3

    p1=2Δa,p2=2Δb,p3=2Δc\Rightarrow p_1 = \frac{2\Delta}{a}, p_2 = \frac{2\Delta}{b}, p_3 = \frac{2\Delta}{c}

    L.H.S. =cosAp1+cosBp2+cosCp3= \frac{\cos A}{p_1} + \frac{\cos B}{p_2} + \frac{\cos C}{p_3}

    =12Δ[acosA+bcosB+ccosC]= \frac{1}{2\Delta}[a\cos A + b\cos B + c\cos C]

    =2R2Δ[sinAcosA+sinBcosB+sinCcosC]= \frac{2R}{2\Delta}[\sin A\cos A + \sin B\cos B + \sin C\cos C]

    =R2Δ[sin2A+sin2B+sin2C]=R2Δ4sinAsinBsinC= \frac{R}{2\Delta}[\sin 2A + \sin 2B + \sin 2C] = \frac{R}{2\Delta}4\sin A\sin B\sin C

    =abc4Δ.1R2=RR2=1R== \frac{abc}{4\Delta}.\frac{1}{R^2} = \frac{R}{R^2} = \frac{1}{R} = R.H.S.

  17. This has been already proved in 149.

  18. L.H.S. =r1r2r3=ΔsaΔsb.Δsc= r_1r_2r_3 = \frac{\Delta}{s - a}\frac{\Delta}{s - b}.\frac{\Delta}{s - c}

    =Δ3(sa)(sb))(sc)=Δ.s= \frac{\Delta^3}{(s - a)(s - b))(s - c)} = \Delta .s

    R.H.S. =r3cot2A2cot2B2cot2C2= r^3\cot^2\frac{A}{2}\cot^2\frac{B}{2}\cot^2\frac{C}{2}

    =Δ3s3.s2(sa)2Δ2.s2(sb)2Δ2.s2(sc)2Δ2= \frac{\Delta^3}{s^3}.\frac{s^2(s - a)^2}{\Delta^2}.\frac{s^2(s - b)^2}{\Delta^2}.\frac{s^2(s - c)^2}{\Delta^2}

    =s3(sa)2(sb)2(sc)2Δ3=Δ.s= \frac{s^3(s - a)^2(s - b)^2(s - c)^2}{\Delta^3} = \Delta .s

  19. We have to prove that a(rr1+r2r3)=b(rr2+r3r1)=c(rr3+r1r2)=abca(rr_1 + r2r_3) = b(rr_2 + r_3r_1) = c(rr_3 + r_1r_2) = abc

    a(rr1+r2r3)=a(Δs.Δsa+Δ(sb)Δsc)a(rr_1 + r_2r_3) = a\left(\frac{\Delta}{s}.\frac{\Delta}{s - a} + \frac{\Delta}{(s - b)}\frac{\Delta}{s - c}\right)

    =Δ2.a((sb)(sc)+s(sa)s(sa)(sb)(sc))= \Delta^2.a\left(\frac{(s - b)(s - c) + s(s - a)}{s(s - a)(s - b)(s - c)}\right)

    =Δ2.a(2s2s(a+b+c)+bcΔ2)= \Delta^2.a\left(\frac{2s^2 - s(a + b + c) + bc}{\Delta^2}\right)

    =abc= abc

    Similalry other terms can be evaluated to same value of abc.abc.

  20. We have to prove that (r1+r2)tanC2=(r3r)cotC2=c(r_1 + r_2)\tan\frac{C}{2} = (r_3 - r)\cot\frac{C}{2} = c

    (r1+r2)tanC2=(Δsa+Δsb)(sa)(sb)Δ(r_1 + r_2)\tan\frac{C}{2} = \left(\frac{\Delta}{s - a} + \frac{\Delta}{s - b}\right)\frac{(s - a)(s - b)}{\Delta}

    =sb+sa=c= s - b + s - a = c

    (r3r)cotC2=(ΔscΔs)Δ(sa)(sb)(r_3 - r)\cot\frac{C}{2} = \left(\frac{\Delta}{s - c} - \frac{\Delta}{s}\right)\frac{\Delta}{(s - a)( s - b)}

    =Δ2(ss+cs(sa)(sb)(sc))=c= \Delta^2\left(\frac{s - s + c}{s(s - a)(s - b)(s - c)}\right) = c

  21. We have to prove that 4RsinAsinBsinC=acosA+bcosB+ccosC4R\sin A\sin B\sin C = a\cos A + b\cos B + c\cos C

    R.H.S. =R(2sinAcosA+2sinBcosB+2sinCcosC)=R(sin2A+sin2B+sin2C)= R(2\sin A\cos A + 2\sin B\cos B + 2\sin C\cos C) = R(\sin 2A + \sin 2B + \sin 2C)

    =R(2sin(A+B)cos(AB)+2sinCcosC)=2R(sinCcos(AB)+sinCcosC)[sin(A+B)=sin(πC)=sinC]= R(2\sin(A + B)\cos(A - B) + 2\sin C\cos C) = 2R(\sin C\cos(A - B) + \sin C\cos C)[\because \sin(A + B) = \sin(\pi - C) = \sin C]

    =2RsinC[cos(AB)cos(A+B)][cosC=cos(πAB)=cos(A+B)]= 2R\sin C[\cos(A - B) - \cos(A + B)][\because \cos C = \cos(\pi - A - B) = -\cos(A + B)]

    =2RsinC.2sinAsinB=4RsinAsinBsinC== 2R\sin C. 2\sin A\sin B = 4R\sin A\sin B\sin C = L.H.S.

  22. We have to prove that (r1r)(r2r)(r3r)=4Rr2(r_1 - r)(r_2 - r)(r_3 - r) = 4Rr^2

    L.H.S. =(ΔsaΔs)(ΔsbΔS)(ΔscΔs)= \left(\frac{\Delta}{s - a} - \frac{\Delta}{s}\right)\left(\frac{\Delta}{s - b} - \frac{\Delta}{S}\right)\left(\frac{\Delta}{s - c} - \frac{\Delta}{s}\right)

    =Δ3(ss+as(sa))(ss+bs(sb))(ss+cs(sc))= \Delta^3\left(\frac{s - s + a}{s(s - a)}\right)\left(\frac{s - s + b}{s(s - b)}\right)\left(\frac{s - s + c}{s(s - c)}\right)

    =Δ3.abcs3(sa)(sb)(sc)=Δ3.abcs2Δ2=abc.Δs2= \Delta^3.\frac{abc}{s^3(s - a)(s - b)(s - c)} = \frac{\Delta^3.abc}{s^2\Delta^2} = \frac{abc.\Delta}{s^2}

    =abcΔ.Δ2s2=4Rr2== \frac{abc}{\Delta}.\frac{\Delta^2}{s^2} = 4Rr^2 = R.H.S.

  23. We have to prove that r2+r12+r22+r32=16R2a2b2c2r^2 + r_1^2 + r_2^2 + r_3^2 = 16R^2 - a^2 - b^2 - c^2

    (r1+r2+r3r)2=r12+r22+r32+r22r(r1+r2+r3)+2(r1r2+r2r3+r3r1)(r_1 + r_2 + r_3 - r)^2 = r_1^2 + r_2^2 + r_3^2 + r^2 - 2r(r_1 + r_2 + r_3) + 2(r_1r_2 + r_2r_3 + r_3r_1)

    We know that r1+r2+r3r=4Rr_1 + r_2 + r_3 - r = 4R and (r1r2+r2r3+r3r1)=s2(r_1r_2 + r_2r_3 + r_3r_1) = s^2

    r(r1+r2+r3)=Δs(Δsa+Δsb+Δsc)r(r_1 + r_2 + r_3) = \frac{\Delta}{s}\left(\frac{\Delta}{s -a} + \frac{\Delta}{s - b} + \frac{\Delta}{s - c}\right)

    =Δ2s(sa)+Δ2s(sb)+Δ2s(sc)=s2+(ab+bc+ca)= \frac{\Delta^2}{s(s - a)} + \frac{\Delta^2}{s(s - b)} + \frac{\Delta^2}{s(s - c)} = -s^2 + (ab + bc + ca)

    16R2=r12+r22+r32+r22[s2+(ab+bc+ca)]+2s216R^2 = r_1^2 + r_2^2 + r_3^2 + r^2 -2[-s^2 + (ab + bc + ca)] + 2s^2

    r2+r12+r22+r32=16R2a2b2c2\Rightarrow r^2 + r_1^2 + r_2^2 + r_3^2 = 16R^2 - a^2 - b^2 - c^2

  24. We know that IA=rcosecA2,IB=rcosecB2,IC=rcosecC2IA = r\cosec\frac{A}{2}, IB=r\cosec\frac{B}{2}, IC=r\cosec\frac{C}{2}

    L.H.S. =r3sinA2sinB2sinC2=r3.4R4RsinA2sinB2sinC2= \frac{r^3}{\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}} = \frac{r^3.4R}{4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}}

    =r3.4Rr=4R.r2=4R.Δ2s2=abcΔs2= \frac{r^3.4R}{r} = 4R.r^2 = 4R.\frac{\Delta^2}{s^2} = \frac{abc\Delta}{s^2}

    R.H.S. =abc.(sa)2(sb)2(sc)2Δ3=abcΔs2= abc.\frac{(s - a)^2(s - b)^2(s - c)^2}{\Delta^3} = \frac{abc\Delta}{s^2}

  25. From here we can say that AI1=r1cosecA2AI_1 = r_1\cosec\frac{A}{2}

  26. II1=AI1AI=r1cosecA2rcosecA2II_1 = AI_1 - AI = r_1\cosec\frac{A}{2} - r\cosec\frac{A}{2}

    =(ΔsaΔs)cosecA2= \left(\frac{\Delta}{s - a} - \frac{\Delta}{s}\right)\cosec\frac{A}{2}

    =Δ.as(sa)bc(sb)(sc)=asecA3= \Delta .\frac{a}{s(s - a)}\sqrt{\frac{bc}{(s - b)(s - c)}} = a\sec\frac{A}{3}

  27. If E2E_2 be the point of contact of the circle whose center is I2I_2 with the side ACAC of the triangle ABC,ABC, we have

    AI2=AE2secI2AE2=AE2sec(90A2)=(sb)cosecA2AI_2 = AE_2\sec I_2AE_2 = AE_2sec\left(90^\circ - \frac{A}{2}\right) = (s - b)\cosec \frac{A}{2}

    I2I3=AI2+AI3=(sb+sc)cosecA2=acosecA2I_2I_3 = AI_2 + AI_3 = (s - b + s - c)\cosec\frac{A}{2} = a\cosec\frac{A}{2}

  28. We have deduced that II1=asecA2II_1 = a\sec\frac{A}{2} in problem 176. So II2=bsecB2II_2 = b\sec\frac{B}{2} and II3=csecC2II_3 = c\sec\frac{C}{2}

    L.H.S. =II1.II2.II3=abcsecA2secB2secC2= II_1.II_2.II_3 = abc\sec\frac{A}{2}\sec\frac{B}{2}\sec\frac{C}{2}

    =8R32sinA2cosA2.2sinB2cosB2.2sinC2cosC2cosA2cosB2cosC2= 8R^3\frac{2\sin\frac{A}{2}\cos\frac{A}{2}.2\sin\frac{B}{2}\cos\frac{B}{2}.2\sin\frac{C}{2}\cos\frac{C}{2}}{\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}}

    =16R2.4RsinA2sinB2sinC2=16R2r== 16R^2.4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2} = 16R^2r = R.H.S.

  29. II1=asecA2,I2I3=acosecA2II_1 = a\sec\frac{A}{2}, I_2I_3 = a\cosec\frac{A}{2}

    II12+I2I32=a2(1sin2A2+1sin2A2)II_1^2 + I_2I_3^2 = a^2\left(\frac{1}{\sin^2\frac{A}{2}} + \frac{1}{\sin^2\frac{A}{2}}\right)

    =(asinA2cosA2)2=(2a2sinA2cosA2)2= \left(\frac{a}{\sin\frac{A}{2}\cos\frac{A}{2}}\right)^2 = \left(\frac{2a}{2\sin\frac{A}{2}\cos\frac{A}{2}}\right)^2

    =16R2[a=2RsinA]= 16R^2 [\because a = 2R\sin A]

    Similalrly other terms can be proven to be equal to 16R216R^2

  30. We know that OI2=R2(18sinA2sinB2sinC2)OI^2 = R^2\left(1 - 8\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}\right)

    =R2[14(cosAB2cosA+B2)sinC2]= R^2\left[1 - 4\left(\cos \frac{A - B}{2} - \cos\frac{A + B}{2}\right)\sin\frac{C}{2}\right]

    =R2[14cosAB2cosA+B2+4sin2C2][sinC2=cosA+B2]= R^2\left[1 - 4\cos\frac{A - B}{2}\cos\frac{A + B}{2} + 4\sin^2\frac{C}{2}\right][\because \sin \frac{C}{2} = \cos\frac{A + B}{2}]

    =R2[12(cosA+cosB)+2(1cosC)]= R^2\left[1 - 2(\cos A + \cos B) + 2(1 - \cos C)\right]

    =R2(32cosA2cosB2cosC)=R^2(3 - 2\cos A - 2\cos B - 2\cos C)

  31. We have, IH2=AH2+AI22.AH.AI.cosIAHIH^2 = AH^2 + AI^2 - 2.AH.AI.\cos IAH

    IAH=A2HAC=A2(90C)=CB2\angle IAH = \frac{A}{2} - \angle HAC = \frac{A}{2} - (90^\circ - C) = \frac{C - B}{2}

    IH2=4R2cos2A+16R2sin2B2sin2C216R2cosAsinB2sinC2cosCB2IH^2 = 4R^2\cos^2A + 16R^2\sin^2\frac{B}{2}\sin^2\frac{C}{2} - 16R^2\cos A\sin\frac{B}{2}\sin\frac{C}{2}\cos\frac{C - B}{2}

    =4R2[cos2A+4sin2B2sin2C24cosAsinB2sinC2cosC2cosB24cosAsin2B2sin2C2]= 4R^2\left[\cos^2A + 4\sin^2\frac{B}{2}\sin^2\frac{C}{2} - 4\cos A\sin\frac{B}{2}\sin\frac{C}{2}\cos\frac{C}{2}\cos\frac{B}{2} - 4\cos A\sin^2\frac{B}{2}\sin^2\frac{C}{2}\right]

    =4R2[cos2A+4sin2B2sin2C2(1cosA)cosAsinBsinC]= 4R^2\left[\cos^2A + 4\sin^2\frac{B}{2}\sin^2\frac{C}{2}(1 - \cos A) - \cos A\sin B\sin C\right]

    =4R2[8sin2A2sin2B2sin2C2+cos2AcosAsinBsinC]= 4R^2\left[8\sin^2\frac{A}{2}\sin^2\frac{B}{2}\sin^2\frac{C}{2} + \cos^2A - \cos A\sin B\sin C\right]

    =2r2+4R2cosA(cosAsinBsinC)= 2r^2 + 4R^2\cos A(\cos A - \sin B\sin C)

    =2r24R2cosAcosBcosC= 2r^2 - 4R^2\cos A\cos B\cos C

  32. We know that OG=13OHOG2=OH29OG = \frac{1}{3}OH \Rightarrow OG^2 = \frac{OH^2}{9}

    =19[R28R2cosAcosBcosC]=R29[14{cos(A+B)+cos(AB)}cosC]= \frac{1}{9}[R^2 - 8R^2\cos A\cos B\cos C] = \frac{R^2}{9}[1 - 4\{\cos(A + B) + \cos(A - B)\}\cos C]

    =R29[1+4cos2C+4cos(AB)cos(A+B)]= \frac{R^2}{9}[1 + 4\cos^2C + 4\cos(A - B)\cos(A + B)]

    =R29[1+2(1+cos2C)+2(cos2A+cos2C)]= \frac{R^2}{9}[1 + 2(1 + \cos 2C) +2(\cos 2A + \cos 2C)]

    =R29[3+cos2A+cos2B+cos2C]= \frac{R^2}{9}[3 + \cos 2A + \cos 2B + \cos 2C]

    =R29[92(1cos2A)2(1cos2B)2(1cos2C)]= \frac{R^2}{9}[9 - 2(1 - \cos 2A) - 2(1 - \cos 2B) - 2(1 - \cos 2C)]

    =R29[94(sin2A+sin2B+sin2C)]= \frac{R^2}{9}[9 - 4(\sin^2A + \sin^2B + \sin^2C)]

    =R219(2RsinA)219(2RsinB)219(2RsinC)2= R^2 - \frac{1}{9}(2R\sin A)^2 - \frac{1}{9}(2R\sin B)^2 - \frac{1}{9}(2R\sin C)^2

    =R219(a2+b2+c2)= R^2 - \frac{1}{9}(a^2 + b^2 + c^2)

  33. The diagram is given below:

    Problem 183

    Clearly, R=b2sinα2=bcosecα22R = \frac{b}{2\sin\frac{\alpha}{2}} = \frac{b\cosec\frac{\alpha}{2}}{2}

  34. We know that in a ABC,\triangle ABC, r=4RsinA2sinB2sinC2r = 4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}

    Let ADAD be the perpedicular bisector to BC.BC.

    Δ=BD.AD=bcosα.bsinα=12b2sin2α\Delta = BD.AD = b\cos\alpha.b\sin\alpha = \frac{1}{2}b^2\sin2\alpha

    r=Δs=12b2sin2α12(b+b+2bcosα)=bsin2α2(1+cosα)r = \frac{\Delta}{s} = \frac{\frac{1}{2}b^2\sin2\alpha}{\frac{1}{2}(b + b + 2b\cos\alpha)} = \frac{b\sin2\alpha}{2(1 + \cos\alpha)}

  35. OI=OD+DI=OD+rOI = |OD + DI| = |OD + r| because α<π/4,A>π/2\alpha < \pi/4, A>\pi/2 and OO lies on ADAD produced.

    From right-angled ODB,\triangle ODB, we get

    OD2=OB2BD2=R2b2cos2αOD^2 = OB^2 - BD^2 = R^2 - b^2\cos^2\alpha

    =14b2sin2αb2cos2α= \frac{1}{4}\frac{b^2}{\sin^2\alpha} - b^2\cos^2\alpha

    =b(14sin2αcos2α)4sin2α=b2(cos2αsin2α)4sin2α= \frac{b^(1 - 4\sin^2\alpha\cos^2\alpha)}{4\sin^2\alpha} = \frac{b^2(\cos^2\alpha - \sin^2\alpha)}{4\sin^2\alpha}

    =b2cos22α(2sinα)2= \frac{b^2\cos^22\alpha}{(2\sin\alpha)^2}

    OI=bsin2α2(1+cosα)+bcos2α2sinα\therefore OI = \left|\frac{b\sin2\alpha}{2(1 + \cos\alpha)} + \frac{b\cos2\alpha}{2\sin\alpha}\right|

    =bsin2α4cos2α2+bcos2α4sinα2cosα2= \left|\frac{b\sin2\alpha}{4\cos^2\frac{\alpha}{2}} + \frac{b\cos2\alpha}{4\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}\right|

    =b4cosα/2.sin2αsinα/2+cos2αcosα/2sinα/2cosα/2= \left|\frac{b}{4\cos\alpha/2}.\frac{\sin2\alpha\sin\alpha/2 + \cos2\alpha\cos\alpha/2}{\sin\alpha/2\cos\alpha/2}\right|

    =bcos3α/22sinαcosα/2= \left|\frac{b\cos3\alpha/2}{2\sin\alpha\cos\alpha/2}\right|

  36. L.H.S. =1ab+1bc+1ca=a+b+cabc= \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} = \frac{a + b + c}{abc}

    =2s4RS=12RS/s=12Rr== \frac{2s}{4RS} = \frac{1}{2RS/s} = \frac{1}{2Rr} = R.H.S.

  37. We know that r1=Δsa,r2=Δsb,r3=Δscr_1 = \frac{\Delta}{s - a}, r_2 = \frac{\Delta}{s - b}, r_3 = \frac{\Delta}{s - c} and r=Δsr = \frac{\Delta}{s}

    L.H.S. =3Δ(sa)(sb)(sc)=3sΔΔ2=3sΔ=3r== \frac{3\Delta}{(s - a)(s - b)(s - c)} = \frac{3s\Delta}{\Delta^2} = \frac{3s}{\Delta} = \frac{3}{r} = R.H.S.

  38. The diagram is given below:

    Problem 188

    2s=2(α+β+γ)s=α+β+γ2s = 2(\alpha + \beta + \gamma) \Rightarrow s = \alpha + \beta + \gamma

    We know that r=Δss2=(sa)(sb)(sc)sr = \frac{\Delta}{s} \Rightarrow s^2 = \frac{(s - a)(s - b)(s - c)}{s}

    Clearly, sa=γ,sb=β,sc=αs - a = \gamma, s - b = \beta, s - c = \alpha

    s=αβγα+β+γ\Rightarrow s = \frac{\alpha\beta\gamma}{\alpha + \beta + \gamma}

  39. The diagram is given below:

    Problem 189

    Let PQ=x,PQBC,RS=y,RSAC,TU=z,TUABPQ=x, PQ\parallel BC, RS = y, RS\parallel AC, TU = z, TU\parallel AB

    In APQ,xsinA=AQsinB=APsinC\triangle APQ, \frac{x}{\sin A} = \frac{AQ}{\sin B} = \frac{AP}{\sin C}

    AQ=bxa,AP=cxa\Rightarrow AQ = \frac{bx}{a}, AP = \frac{cx}{a}

    r=(x+AP+AQ2)tanA2=a+b+c2xtanA2r = \left(\frac{x + AP + AQ}{2}\right)\tan\frac{A}{2} = \frac{a + b + c}{2}x\tan\frac{A}{2}

    =sxatanA2=(sa)tanA2= \frac{sx}{a}\tan\frac{A}{2} = (s - a)\tan\frac{A}{2}

    sxa=sa\Rightarrow \frac{sx}{a} = s - a

    Similarly, syb=sb\frac{sy}{b} = s - b and szc=sc\frac{sz}{c} = s - c

    s(xa+yb+zc)=3s(a+b+c)\Rightarrow s\left(\frac{x}{a} + \frac{y}{b} + \frac{z}{c}\right) = 3s - (a + b + c)

    xa+xb+xc=1\Rightarrow \frac{x}{a} + \frac{x}{b} + \frac{x}{c} = 1

  40. The diagram is given below:

    Problem 190

    Since II is the incenter, AIAI will be angle bisector. Let AIAI cut circumcirlce at D.D.

    DBI=DBC+IBC=DAB+ABI=BID\angle DBI=\angle DBC+\angle IBC=\angle DAB+\angle ABI=\angle BID and then DB=DIDB=DI

    Likewise DC=DIDC = DI and then DB=BI=DCDB = BI = DC

    I1CI_1C bisects BCTICI1=90\angle BCT \Longrightarrow \angle ICI_1 = 90^\circ

    Let the perpendicular bisector of BCBC cut circumcircle at MM also.

    SAI1 BMD\triangle SAI_1 ~ \triangle BMD

    Power of I1I_1 w.r.t the circumcircle of ABC=OI12R2=I1D.I1A\triangle ABC = OI_1^2 - R^2 = I_1D.I_1A

    MDBD=I1ASI12Rr1=OI12R2\Rightarrow \frac{MD}{BD} = \frac{I_1A}{SI_1} \Rightarrow 2Rr_1 = OI_1^2 - R^2

    Thus, OI1=R2+2Rr1OI_1 = R^2 + 2Rr_1

    Thus length of tangent t12=OI12R2=2Rr1t_1^2 = OI_1^2 - R^2 = 2Rr_1

    1t12+1t22+1t32=12R(1r1+1r2+1r3)\frac{1}{t_12} + \frac{1}{t_2^2} + \frac{1}{t_3^2} = \frac{1}{2R}\left(\frac{1}{r_1} + \frac{1}{r_2} + \frac{1}{r_3}\right)

    =12R3s(a+b+c)Δ=4Δ2.sΔ=2sabc= \frac{1}{2R}\frac{3s - (a + b + c)}{\Delta} = \frac{4\Delta}{2}.\frac{s}{\Delta} = \frac{2s}{abc}

  41. r1=Δsar_1 = \frac{\Delta}{s - a} and so on. Given,

    (1r1r2)(1r1r3)=2\left(1 - \frac{r_1}{r_2}\right)\left(1 - \frac{r_1}{r_3}\right) = 2

    (1sbsa)(1scsa)=2\left(1 - \frac{s - b}{s - a}\right)\left(1 - \frac{s - c}{s - a}\right) = 2

    (ba)(ca)=2(sa)2(b - a)(c - a) = 2(s - a)^2

    bc+a2acab=(b+ca)2/2bc + a^2 - ac - ab = (b + c - a)^2/2

    b2+c2=a2b^2 + c^2 = a^2

    Thus the triangle is right-angled.

  42. Area of in-circleArea of triangle=πr2Δ=πΔ.Δ2s2\frac{\text{Area of in-circle}}{\text{Area of triangle}} = \frac{\pi r^2}{\Delta} = \frac{\pi}{\Delta}.\frac{\Delta^2}{s^2}

    =π.Δs2= \pi .\frac{\Delta}{s^2}

    cotA2cotB2cotC2=s(sa)(sb)(sc).s(sb)(sa)(sc).s(sc)(sa)(sb)\cot\frac{A}{2}\cot\frac{B}{2}\cot\frac{C}{2} = \sqrt{\frac{s(s - a)}{(s - b)(s - c)}.\frac{s(s - b)}{(s - a)(s - c)}.\frac{s(s - c)}{(s - a)(s - b)}}

    =s4s(sa)(sb)(sc)=s2Δ= \sqrt{\frac{s^4}{s(s - a)(s - b)(s - c)}} = \frac{s^2}{\Delta}

    πcotA2cotB2cotC2=πΔs2\Rightarrow \frac{\pi}{\cot\frac{A}{2}\cot\frac{B}{2}\cot\frac{C}{2}} = \frac{\pi\Delta}{s^2}

    Thus, we have the desired result by combining both the equations.

  43. The diagram is given below:

    Problem 193

    Let OO be the center of the regular polygon A1,A2,A3,,AnA_1, A_2, A_3, \ldots, A_n which has nn sides.

    Since it is a regular polyogn so A1OA2=A2OA3==AnOA1=2πn\angle A_1OA_2 = \angle A_2OA_3 = \ldots = \angle A_nOA_1 = \frac{2\pi}{n}

    Also, let OA1=OA2==OAn=rOA_1 = OA_2 = \ldots = OA_n = r

    Applying cosine rule in A1OA2,\triangle A_1OA_2,

    cos2πn=OA12+OA22A1A222OA1.OA2\cos\frac{2\pi}{n} = \frac{OA_1^2 + OA_2^2 - A_1A_2^2}{2OA_1.OA_2}

    A1A22=2r2(1cos2πn)\Rightarrow A_1A_2^2 = 2r^2\left(1 - \cos \frac{2\pi}{n}\right)

    A1A22=4r2sinπn\Rightarrow A_1A_2^2 = 4r^2\sin^\frac{\pi}{n}

    A1A2=2rsinπn\Rightarrow A_1A_2 = 2r\sin\frac{\pi}{n}

    Likewise A1A3=2rsin2πnA1A4=2rsin3πnA_1A_3 = 2r\sin\frac{2\pi}{n} A_1A_4 = 2r\sin\frac{3\pi}{n}

    Given, 1A1A2=1A1A3+1A1A4\frac{1}{A_1A_2} = \frac{1}{A_1A_3} + \frac{1}{A_1A_4}

    12rsinπn=12rsin2πn+12rsin3πn\Rightarrow \frac{1}{2r\sin\frac{\pi}{n}} = \frac{1}{2r\sin\frac{2\pi}{n}} + \frac{1}{2r\sin\frac{3\pi}{n}}

    1sinπn1sin3πn=1sin2πn\Rightarrow \frac{1}{\sin\frac{\pi}{n}} - \frac{1}{\sin\frac{3\pi}{n}} = \frac{1}{\sin\frac{2\pi}{n}}

    2cos2πnsinπnsinπnsin3πn=1sin2πn\Rightarrow \frac{2\cos\frac{2\pi}{n}\sin\frac{\pi}{n}}{\sin\frac{\pi}{n}\sin\frac{3\pi}{n}} = \frac{1}{\sin\frac{2\pi}{n}}

    2cos2πnsin2πn=sin3πn\Rightarrow 2\cos\frac{2\pi}{n}\sin\frac{2\pi}{n} = \sin\frac{3\pi}{n}

    sin4πn=sin3πn\Rightarrow \sin\frac{4\pi}{n} = \sin\frac{3\pi}{n}

    cos7π2n.sinπ2n=0\Rightarrow \cos\frac{7\pi}{2n}.\sin\frac{\pi}{2n} = 0

    cos7π2n=0[sinπ2n0]\Rightarrow \cos\frac{7\pi}{2n} = 0\left[\sin\frac{\pi}{2n} \neq 0\right]

    7π2n=odd integer×π2\Rightarrow \frac{7\pi}{2n} = \text{odd integer} \times \frac{\pi}{2}

    n=7odd integer=7[nI,n>1]\Rightarrow n = \frac{7}{\text{odd integer}} = 7[n\in I, n > 1]

  44. The diagram is given below:

    Problem 194

    Let OO be the center of the circumscribing circle regular polygon A1,A2,A3,,AnA_1, A_2, A_3, \ldots, A_n which has nn sides.

    Since the polygon is regular, therefore OO will also be the center of inscribing circle.

    Let ODA1A2.OD\perp A_1A_2. Now A1OA2=2πn\angle A_1OA_2 = \frac{2\pi}{n}

    AOD=A2OD=πn\angle A_OD = \angle A_2OD = \frac{\pi}{n}

    Also, A1D=A2D=a/2A_1D = A_2D = a/2 where aa is the length of a side of the polygon.

    Here, R=R = radius of the circumscribing circle =OA= OA

    and r=r = radius of the inscribing circle =OD= OD

    From right angled triangle ODA1ODA_1

    sinπn=a/2RR=a2cosecπn\sin\frac{\pi}{n} = \frac{a/2}{R} \Rightarrow R = \frac{a}{2}\cosec\frac{\pi}{n}

    and tanπn=a/2rr=a2cotπn\tan\frac{\pi}{n} = \frac{a/2}{r} \Rightarrow r = \frac{a}{2}\cot\frac{\pi}{n}

    R+r=a2[cosecπn+cotπn]\therefore R + r = \frac{a}{2}\left[\cosec\frac{\pi}{n} + \cot\frac{\pi}{n}\right]

    =a2[1+cosπnsinπn]= \frac{a}{2}\left[\frac{1 + \cos\frac{\pi}{n}}{\sin\frac{\pi}{n}}\right]

    =a2cotπ2n= \frac{a}{2}\cot\frac{\pi}{2n}

  45. Let ABCDABCD be a quadrilateral such that AB=3cm,BC=4cm,CD=5cmAB = 3 cm, BC = 4 cm, CD= 5 cm and AD=6cm.AD = 6 cm.

    Also, let that BAC=θ\angle BAC = \theta and BCD=120θ\angle BCD = 120^\circ - \theta as it is given that sum of pair of opposite angles is 120.120^\circ.

    Applying cosine law in ABD,\triangle ABD,

    cosθ=AB2+AD2BD22.AB.ADBD=4536cosθ\cos\theta = \frac{AB^2 + AD^2 - BD^2}{2.AB.AD} \Rightarrow BD = 45 - 36\cos\theta

    Applying cosine law in BCD,\triangle BCD,

    cos(120θ)=BC2+CD2BD22.BC.CDBD2=41+20cosθ203sinθ\cos(120^\circ - \theta) = \frac{BC^2 + CD^2 - BD^2}{2.BC.CD} \Rightarrow BD^2 = 41 + 20\cos \theta - 20\sqrt{3}\sin\theta

    Thus, 4536cosθ=41+20cosθ203sinθ45 - 36\cos\theta = 41 + 20\cos\theta - 20\sqrt{3}\sin\theta

    14cosθ53sinθ=1\Rightarrow 14\cos\theta - 5\sqrt{3}\sin\theta = 1

    Area of the quadrilateral =ΔABD+ΔBCD= \Delta ABD + \Delta BCD

    =123.6.sinθ+124..5sin(120θ)= \frac{1}{2}3.6.\sin\theta + \frac{1}{2}4..5\sin(120^\circ - \theta)

    =14sinθ+53sinθ=z= 14\sin\theta + 5\sqrt{3}\sin\theta = z (let)

    Solving the two equations thus obtaiined, we get

    196(sin2θ+cos2θ)+75(cos2θ+sin2θ)=z2+1196(\sin^2\theta + \cos^2\theta) + 75(\cos^2\theta + \sin^2\theta) = z^2 + 1

    z=230\Rightarrow z = 2\sqrt{30} sq.cm.

  46. Let ABCDABCD be a cyclic quadrilateral such that AD=2,AB=5,DAB=60AD = 2, AB = 5, \angle DAB = 60^\circ

    Since the quadrilateral is cyclic BCD=120\angle BCD = 120^\circ

    Area of quadrilateral ABD=12.2.5.sin60=532ABD = \frac{1}{2}.2.5.\sin60^\circ = \frac{5\sqrt{3}}{2}

    Area of BCD=\triangle BCD = Area of quadrilateral ABCDABCD - Area of ABD\triangle ABD

    =43=532=332= 4\sqrt{3} = \frac{5\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}

    Let CD=x,BC=yCD = x, BC = y

    Now area of BCD=12.x.y.sin120332=12xy32\triangle BCD = \frac{1}{2}.x.y.\sin120^\circ \Rightarrow \frac{3\sqrt{3}}{2} = \frac{1}{2}xy\frac{\sqrt{3}}{2}

    xy=6\Rightarrow xy = 6

    Applying cosine rule in ABD,\triangle ABD,

    cos60=AD2+AB2BD22.AD.ABBD2=19\cos60^\circ = \frac{AD^2 + AB^2 - BD^2}{2.AD.AB} \Rightarrow BD^2 = 19

    Applying cosine rule in BCD,\triangle BCD,

    cos120=x2+y2192xyx2+y2=13\cos120^\circ = \frac{x^2 + y^2 - 19}{2xy}\Rightarrow x^2 + y^2 = 13

    (x+y)2=25x+y=±5(x + y)^2 = 25 \Rightarrow x + y = \pm5

    (xy)2=1xy=±1(x - y)^2 = 1 \Rightarrow x - y = \pm1

    x=3,y=2\Rightarrow x = 3, y = 2 or x=2,y=3x = 2, y = 3

  47. From question, AB=1,BD=3.AB = 1, BD = \sqrt{3}. Let BAD=θ,AD=x,BC=yBAD=\theta, AD = x, BC = y and CD=z.CD = z. Since the given circle is also circum-circle of ABD,BDsinθ=2R\triangle ABD, \Rightarrow \frac{BD}{\sin\theta} = 2R

    sinθ=32θ=60\Rightarrow \sin\theta = \frac{\sqrt{3}}{2}\Rightarrow \theta = 60^\circ

    Now BCD=18060=120\angle BCD = 180^\circ - 60^\circ = 120^\circ

    Applying cosine rule in ABD,\triangle ABD,

    cos60=AB2+AD2BD22.AB.AD12=1+x232x\cos60^\circ = \frac{AB^2 + AD^2 - BD^2}{2.AB.AD} \Rightarrow \frac{1}{2} = \frac{1 + x^2 - 3}{2x}

    x2x2=0x=2\Rightarrow x^2 - x - 2 = 0 \Rightarrow x = 2

    Applying cosine law in BCD,\triangle BCD,

    cos120=y2+z232yzy2+z2+yz=3\cos120^\circ = \frac{y^2 + z^2 - 3}{2yz} \Rightarrow y^2 + z^2 + yz = 3

    Area of quadrilateral ABCD=ABCD = Area of BCD\triangle BCD + Area of ABD\triangle ABD

    332=121.x.sin60+12yzsin120\frac{3\sqrt{3}}{2} = \frac{1}{2}1.x.\sin60^\circ + \frac{1}{2}yz\sin 120^\circ

    yz=1\Rightarrow yz = 1

    y2+z2=2\Rightarrow y^2 + z^2 = 2

    (y+z)2=4,(yz)2=0\Rightarrow (y + z)^2 = 4, (y - z)^2 = 0

    y=z=1\Rightarrow y = z = 1

  48. Let ABCDABCD be the cyclic quadrilateral in which AB=a,BC=b,CD=c,DA=d.AB = a, BC = b, CD = c, DA =d.

    Applying cosine rule in ABC,\triangle ABC,

    cosB=a2+b2AC22abAC2=a2+b22abcosB\cos B = \frac{a^2 + b^2 - AC^2}{2ab} \Rightarrow AC^2 = a^2 + b^2 - 2ab\cos B

    Applying cosine rule in ADC,\triangle ADC,

    cos(πB)=c2+d2AC22cdAC2=c2+d2+2cdcosB\cos(\pi - B) = \frac{c^2 + d^2 - AC^2}{2cd} \Rightarrow AC^2 = c^2 + d^2 + 2cd\cos B

    cosB=a2+b2c2d22(ab+cd)\Rightarrow \cos B = \frac{a^2 + b^2 - c^2 - d^2}{2(ab + cd)}

    tan2B2=1cosB1+cosB\tan^2\frac{B}{2}= \frac{1 - \cos B}{1 + \cos B}

    tanB2=(Sa)(Sb)(Sc)(Sd)\Rightarrow \tan\frac{B}{2} = \sqrt{\frac{(S - a)(S - b)}{(S - c)(S - d)}}

  49. Let ABCDABCD be the quadrilateral for which Ab=a,BC=b,CD=c,DA=d.Ab = a, BC=b, CD=c, DA = d. Let diagonals be, AC=x,BD=y.AC=x, BD = y.

    Given, AOD=BOC=αAOB=COD=180α\angle AOD = \angle BOC = \alpha \therefore \angle AOB = \angle COD = 180^\circ - \alpha

    Area of the quadrilateral =12xysinα= \frac{1}{2}xy\sin\alpha

    Applying cosine rule in AOB,\triangle AOB,

    a2=AO2+BO22.AO.BO.cos(180α)a^2 = AO^2 + BO^2 - 2.AO.BO.\cos(180^\circ - \alpha)

    a2=AO2+BO2+2.AO.BO.cosαa^2 = AO^2 + BO^2 + 2.AO.BO.\cos\alpha

    Likewise in BOC,\triangle BOC,

    b2=BO2+CO2+2.BO.CO.cosαb^2 = BO^2 + CO^2 + 2.BO.CO.\cos \alpha

    And in COD\triangle COD

    c2=CO2+DO2+2.CO.DO.cosαc^2 = CO^2 + DO^2 + 2.CO.DO.\cos\alpha

    And in AOD,\triangle AOD,

    d2=AO2+DO2+2AO.DO.cosαd^2 = AO^2 + DO^2 + 2AO.DO.\cos\alpha

    a2+c2b2d2=2cosα.x.ya^2 + c ^2 - b^2 - d^2 = 2\cos\alpha.x.y

    Thus, area =12(a2+c2b2d2)= \frac{1}{2}(a^2 + c^2 - b^2 - d^2)

  50. If the quadrilateral ABCDABCD can have a circle inscribed such that it touches the quadrilateral on sides AB,BC,CD,DAAB, BC, CD, DA at points P,Q,R,SP,Q,R,S then we will have

    AP=AS,BP=BQ,CQ=CR,DR=DSAP = AS, BP = BQ, CQ = CR, DR = DS

    Since lengths of tnagents are equal,

    AP+BP+CR+DR=AS+BQ+CQ+DS\therefore AP + BP + CR + DR = AS + BQ + CQ + DS

    AB+CD=AD+BC\Rightarrow AB + CD = AD + BC

    a+c=b+d\Rightarrow a + c = b + d

    s=a+c=b+d\Rightarrow s = a + c = b + d

    Area of cyclic quadrilateral =(sa)(sb)(sc)(sd)=abcd=12r(a+b+c+d)= \sqrt{(s - a)(s - b)(s - c)(s - d)} = \sqrt{abcd} = \frac{1}{2}r(a + b + c + d) where rr is the in-radius.

    r=2abcda+b+c+dr = \frac{2\sqrt{abcd}}{a + b + c + d}