Properties of Triangle

For option (a) $\cos P = {{{q^2} + {r^2} - {p^2}} \over {2qr}}$ ..... (i)


$\because$ ${{{q^2} + {r^2}} \over 2} \ge \sqrt {{q^2}{r^2}} (AM \ge GM)$

$ \Rightarrow {q^2} + {r^2} \ge 2qr$

From Eq. (i), we get

$\cos P \ge {{2qr - {p^2}} \over {2qr}}$

$ \Rightarrow \cos P \ge 1 - {{{p^2}} \over {2qr}}$ $\to$ option (a) is correct.

For option (b)

${{(q - r)\cos P + (p - r)\cos Q} \over {p + q}}$$ = {{(q\cos P + p\cos Q) - r(\cos P + \cos Q)} \over {p + q}}$

$ \Rightarrow 4{\lambda ^2} - 2\lambda = 0$

$ \Rightarrow u\,.\,v\, = 0$ (Rejected)

or $u\,.\,v = {1 \over 2}$

$\therefore$ $u\,.\,v = {1 \over 2}$

$\therefore$ ${\left| {3u + 5v} \right|^2} = 9{\left| u \right|^2} + 25{\left| v \right|^2} + 3 \times 5 \times 2 \times u\,.\,v$

$ = 9 + 25 + 30 \times {1 \over 2}$

$ = 49$ ($\because$ $\left| u \right| = \left| v \right| = 1$, given)

$\therefore$ $\left| {3u + 5v} \right| = 7$

$ = {{r - r(\cos P + \cos Q)} \over {p + q}} = {{r - r\cos P - r\cos Q} \over {p + q}}$

$ = {{r - (q - \cos R) - (p - q\cos R)} \over {p + q}}$

$ = {{(r - p - q) + (p + q)\cos R} \over {p + q}}$

$ = \cos R + {{r - (p + q)} \over {p + q}} \le \cos R$ ($\because$ $r < p + q$)

Hence, option (b) is correct.

For option (c)

${{q + r} \over p} = {{\sin Q + \sin R} \over {\sin P}} \ge {{2\sqrt {\sin Q.\sin R} } \over {\sin P}}$

$\to$ option (c) is incorrect.

For option (d)

If p < q and p < r, then p is the smallest side and hence one of Q or R can be obtuse.

So, one of cos Q and cos R can be negative.

Therefore, $\cos Q > {p \over r}$ and $\cos R > {p \over q}$ cannot hold always.

Option (d) is incorrect.

For a $\Delta $XYZ, it is given that

$\tan {X \over 2} + \tan {Z \over 2} = {{2y} \over {x + y + z}}$

$ \Rightarrow {\Delta \over {s(s - x)}} + {\Delta \over {s(s - z)}} = {y \over s}$

$ \Rightarrow \Delta {{(s - z + s - x)} \over {(s - x)(s - z)}} = y$

$ \Rightarrow \Delta = (s - x)(s - z)$

$ \Rightarrow s(s - x)(s - y)(s - z) = {(s - x)^2}{(s - z)^2}$

$ \Rightarrow {s^2} - sy = {s^2} - (x + z)s + xz$

$ \Rightarrow s(x + z - y) = xz$

$ \Rightarrow {(x + z)^2} - {y^2} = 2xz$

$ \Rightarrow {x^2} + {z^2} = {y^2} \Rightarrow y = {\pi \over 2}$

$ \because $ $X + Y + Z = \pi \Rightarrow X + Z = {\pi \over 2} = Y$

$ \Rightarrow X + Z = Y$

$ \because $ $\tan {X \over 2} = \sqrt {{{1 - \cos x} \over {1 + \cos x}}} = \sqrt {{{1 - {z \over y}} \over {1 + {z \over y}}}} $



$ = \sqrt {{{y - z} \over {y + z}}} = {{\sqrt {{y^2} - {z^2}} } \over {y + z}}$

$\tan {X \over 2} = {x \over {y + z}}$ {$ \because $ x² + z² = y²}

Let a non-right angled $\Delta $PQR.

Now, by sine rule



${P \over {\sin P}} = {q \over {\sin Q}} = {r \over {\sin R}} = 2 \times $ circumradius

$ \Rightarrow {{\sqrt 3 } \over {\sin P}} = {1 \over {\sin Q}} = {r \over {\sin R}} = 2 \times 1$

[circumradius = 1 unit]

$ \Rightarrow $ ${\sin P}$ = ${{\sqrt 3 } \over 2}$ and sinQ = ${1 \over 2}$

$ \Rightarrow P = 120^\circ $ and $ \Rightarrow Q = 30^\circ $

($ \because $ $\Delta $PQR is non-right angled triangle)

So, R = 30$^\circ $

$ \Rightarrow $ r = 1, so $\Delta $PQR is an isosceles triangle. And, since RS and PE are the median of $\Delta $PQR, so 'O' is centroid of the $\Delta $PQR.

Now,

Option (a),

From Apollonius theorem,

$2(P{E^2} + Q{E^2}) = P{Q^2} + P{R^2}$

$ \Rightarrow 2\left( {P{E^2} + {3 \over 4}} \right) = 1 + 1$

$ \Rightarrow P{E^2} = 1 - {3 \over 4} \Rightarrow P{E^2} = {1 \over 4}$

$ \Rightarrow PE = {1 \over 2}$ units

and $OE = {1 \over 3}PE = {1 \over 6}$ units

[$ \because $ O divides PE is 2 : 1]

Option (b),

Again from Apollonius theorem,

2 (PS² + RS²) = PR² + QR²

$ \Rightarrow 2\left( {{1 \over 4} + R{S^2}} \right) = 1 + 3$

$ \Rightarrow R{S^2} = 2 - {1 \over 4} \Rightarrow R{S^2} = {7 \over 4}$

$ \Rightarrow RS = {{\sqrt 7 } \over 2}$ units

Option (c),

Area of $\Delta $SOE = ${1 \over 2}(OE)\,(ST)$

= ${1 \over 2} \times {1 \over 6}[(PS)sin60^\circ ]$

= ${1 \over {12}} \times {1 \over 2} \times {{\sqrt 3 } \over 2}$

= ${{\sqrt 3 } \over {48}}$ square units

Option (d),

$ \because $ Inradius of

$\Delta PQR = {\Delta \over s} = {{{1 \over 2}pq\sin R} \over {{1 \over 2}(p + q + r)}} = {{{1 \over 2}(\sqrt 2 )(1){1 \over 2}} \over {{1 \over 2}(\sqrt 3 + 1 + 1)}}$

$ = {{\sqrt 3 } \over 2}(2 - \sqrt 3 )$ units

Hence, options (a), (b) and (d) are correct.

It is given that

$ 2 s=x+y+z $

Let us consider

$ \frac{s-x}{4}=\frac{s-y}{3}=\frac{s-z}{2}=k $

That is, $s=4 k+x$

$ \begin{aligned} &s =3 k+y \\\\ &s =2 k+z \\\\ &3 s =9 k+(x+y+z)=9 k+2 s \\\\ &s =9 k \\\\ &x =9 k-4 k=5 k \\\\ &y =6 k, z=7 k \end{aligned} $


Area of in-circle of the triangle $X Y Z$ is

$ \begin{aligned} \pi r^2 & =\pi\left(\frac{\Delta}{s}\right)^2 \\\\ & =\frac{\pi}{s^2} \Delta^2=\frac{\pi}{s^2} s(s-x)(s-y)(s-z) \\\\ & =\frac{\pi}{81 k^2} \times 9 k \times 4 k \times 3 k \times 2 k \\\\ & =\frac{\pi}{9 k} \times 24 k^3=\frac{\pi}{9} 24 k^2 \end{aligned} $

$ \begin{aligned} & \text {Therefore, } \frac{\pi}{9} 24 k^2=\frac{8 \pi}{3} \\\\ & \Rightarrow \quad k^2=\frac{8 \pi}{3} \times \frac{9}{24 \pi}=1 \Rightarrow k=1 \end{aligned} $

The sides of the triangle is given by $x=5, y=6, z=7$. Now, the area of $\triangle X Y Z$ is

$ \sqrt{s(s-x)(s-y)(s-z)}=\sqrt{9 \times 4 \times 3 \times 2}=6 \sqrt{6} $

Hence, option $(\mathrm{A})$ is correct.

Now, $R=\frac{x y z}{4 \Delta}=\frac{5 \times 6 \times 7}{4 \times 6 \sqrt{6}}=\frac{35}{4 \sqrt{6}}$

$ \begin{aligned} & \sin \frac{X}{2} \sin \frac{Y}{2} \sin \frac{Z}{2} \\\\ & \left.\quad=\sqrt{\left[\frac{(s-y)(s-z)}{y z}\right]\left[\frac{(s-z)(s-z)}{x z}\right]\left[\frac{(s-y)(s-x)}{x y}\right]}\right. \\\\ & \quad=\frac{(s-z)(s-y)(s-z)}{x y z} \\\\ & \text {Now, } \frac{(s-z)(s-y)(s-z)}{x y z}=\frac{4 \times 3 \times 2}{5 \times 6 \times 7}=\frac{4}{25} \end{aligned} $

Hence, option (C) is correct.

Now, $\sin ^2\left(\frac{X+Y}{2}\right)=\sin ^2\left(\frac{\pi-Z}{2}\right)=\cos ^2 \frac{z}{2}=\frac{s(s-z)}{x y}$

$ =\frac{9 \times 2}{5 \times 6}=\frac{3}{5} $

Hence, option (D) is correct.

For the above circle PN and PM are its pair of tangents. So, its length are equal i.e. $|\mathrm{PN}|=|\mathrm{PM}| \text {. }$

Similarly $|\mathrm{QL}|=|\mathrm{QN}|$ and $|\mathrm{RL}|=|\mathrm{RM}|$

Let $|\mathrm{PN}|=|\mathrm{PM}|=x,|\mathrm{QL}|=|\mathrm{QN}|=y$ and $|R L|=|R M|=z$

Given $|\mathrm{PN}|,|\mathrm{QL}|$ and $|\mathrm{RM}|$ are consecutive even integer

$\therefore \quad x=2 n, y=2 n+2 \text { and } z=2 n+4$

Given, $\cos P=\frac{1}{3}$

$\begin{aligned} & \Rightarrow \quad \frac{(4 n+4)^2+(4 n+2)^2-(4 n+6)^2}{2(4 n+4)(4 n+2)}=\frac{1}{3} \\ & \left(16 n^2+16+32 n\right)+\left(16 n^2+4+16 n\right) \\ & \Rightarrow \frac{-\left(16 n^2+36+48 n\right)}{32 n^2+48 n+16}=\frac{1}{3} \\ & \Rightarrow \quad 48 n^2-48=32 n^2+48 n+16 \\ & \Rightarrow \quad 16 n^2-48 n-64=0 \\ & \Rightarrow \quad n^2-3 n-4=0 \\ & \Rightarrow \quad n^2-4 n+n-4=0 \\ & \Rightarrow \quad(n-4)(n+1)=0 \\ & \Rightarrow \quad n=4 \\ \end{aligned}$

$\begin{aligned} & \text { Now }|P Q|=4 n+2=18,|Q R|=4 n+6=22 \\ & \text { and }|R P|=4 n+4=20 \end{aligned}$

Hints :

(i) Recall cosine rule in a triangle $\triangle \mathrm{ABC}$,

$\begin{aligned} & \cos \mathrm{A}=\frac{b^2+c^2-a^2}{2 b c}, \cos \mathrm{B}=\frac{c^2+a^2-b^2}{2 c a} \\ \text{and} \quad & \cos \mathrm{C}=\frac{a^2+b^2-c^2}{2 a b}\end{aligned}$

(ii) Three consecutive even integers are $2 n, 2 n +2,2 n+4$ where $n$ is an integer.

$ \text { Area of } \triangle \mathrm{ABC} $

$ \begin{aligned} & \text { = area }(\Delta \mathrm{ABD})+\text { area }(\Delta \mathrm{ACD}) \\ & \text { Let } \mathrm{AB}=c, \mathrm{BC}=a \text { and } \mathrm{AC}=b \\ & \frac{1}{2} b c \sin \mathrm{~A}=\frac{1}{2} c \times \mathrm{AD} \sin \frac{\mathrm{~A}}{2} \\ & +\frac{1}{2} \times b \times \mathrm{AD} \sin \frac{\mathrm{~A}}{2} \\ & \Rightarrow b c \sin \mathrm{~A}=c \times \mathrm{AD} \sin \frac{\mathrm{~A}}{2}+b \mathrm{AD} \sin \frac{\mathrm{~A}}{2} \\ & \Rightarrow b c 2 \sin \frac{\mathrm{~A}}{2} \cdot \cos \frac{\mathrm{~A}}{2}=c \times \mathrm{AD} \\ & \sin \frac{\mathrm{~A}}{2}+b \mathrm{AD} \sin \frac{\mathrm{~A}}{2} \\ & \because \quad \left(\sin \mathrm{~A}=2 \sin \frac{\mathrm{~A}}{2} \cos \frac{\mathrm{~A}}{2}\right) \\ & \Rightarrow \quad 2 b c \cos \frac{\mathrm{~A}}{2}=c \times \mathrm{AD}+b \times \mathrm{AD} \end{aligned} $

$ \Rightarrow \quad \mathrm{AD}=\frac{2 b c \cos \frac{\mathrm{~A}}{2}}{b+c} $

Option (B) is correct.

In $\triangle \mathrm{ADE}$,

$ \begin{aligned} \cos \left(\frac{\mathrm{A}}{2}\right)= & \frac{\mathrm{AD}}{\mathrm{AE}} \\ \mathrm{AE}= & \frac{\mathrm{AD}}{\cos \left(\frac{\mathrm{~A}}{2}\right)}=\frac{2 b c \cos \left(\frac{\mathrm{~A}}{2}\right)}{(b+c)\left(\cos \frac{\mathrm{A}}{2}\right)} \\ & \text { (using above calculated value of } \mathrm{AD} \text { ) } \\ = & \frac{2 b c}{b+c} \\ \mathrm{AE}= & \frac{2}{\frac{1}{b}+\frac{1}{c}} \text { Harmonic mean of } b \text { and } c \end{aligned} $

Option (A) is correct.

$ \begin{aligned} &\begin{aligned} \mathrm{EF} & =\mathrm{ED}+\mathrm{DF} \\ & =2 \mathrm{DE} \end{aligned}\\ &\text { Since, } \Delta \mathrm{ADE} \text { and } \Delta \mathrm{ADF} \text { is congruent by ASA }\\ &\begin{aligned} & =2 \times \mathrm{AD} \tan \left(\frac{\mathrm{~A}}{2}\right) \\ & =2 \times \frac{2 b c}{b+c} \cos \left(\frac{\mathrm{~A}}{2}\right) \frac{\sin \left(\frac{\mathrm{A}}{2}\right)}{\cos \left(\frac{\mathrm{A}}{2}\right)} \end{aligned} \end{aligned} $

$ \begin{aligned} \angle \mathrm{DAE} & =\angle \mathrm{DAF}=\angle \frac{\mathrm{A}}{2} \\ \angle \mathrm{ADE} & =\angle \mathrm{ADF}=90^{\circ} \\ \text { and } \mathrm{AD} & =\mathrm{AD}(\text { common side }) \\ \Rightarrow \mathrm{DE} & =\mathrm{DF} \text { by } \mathrm{CPCT} \\ \mathrm{EF} & =\frac{4 b c}{b+c} \sin \left(\frac{\mathrm{~A}}{2}\right) \end{aligned} $

Option (C) is correct.

Since, AD is bisector and perpendicular on side EF

Therefore, $\triangle \mathrm{AEF}$ is an isosceles.