JEE Advanced · 2016 · 22 May 2016 · Paper 1 · Properties of Triangle

In a triangle

X Y Z

, let

x, y, z

be the lengths of sides opposite to the angles

X, Y, Z

, respectively, and

2 s = x + y + z

. If

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

and area of incircle of the triangle

X Y Z

\frac{8 π}{3}

, then

Answer: (A), (C), (D)

Solution

From

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

, we get

s - x = 4 k

s - y = 3 k

s - z = 2 k

.

Adding:

3 s - (x + y + z) = 9 k

, so

3 s - 2 s = 9 k

, thus

s = 9 k

.

Then

x = s - 4 k = 5 k

y = 6 k

z = 7 k

.

Area by Heron's formula:

Δ = s (s - x) (s - y) (s - z) = 9 k \cdot 4 k \cdot 3 k \cdot 2 k = 216 k^{4} = 66 k^{2}

.

Inradius:

r = \frac{Δ}{s} = \frac{6 6 k ^{2}}{9 k} = \frac{2 6 k}{3}

.

Given

π r^{2} = \frac{8 π}{3}

, so

r^{2} = \frac{8}{3}

, thus

r = \frac{2 2}{3} = \frac{2 6}{3}

.

This gives

k = 1

, so

s = 9

x = 5

y = 6

z = 7

, and

Δ = 66

.

(A) True.

(C) Using

r = 4 R sin \frac{X}{2} sin \frac{Y}{2} sin \frac{Z}{2}

where

R = \frac{x y z}{4Δ} = \frac{210}{24 6} = \frac{35 6}{24}

, we get

sin \frac{X}{2} sin \frac{Y}{2} sin \frac{Z}{2} = \frac{r}{4 R} = \frac{4}{35}

. True.

(D)

cos Z = \frac{x ^{2} + y ^{2} - z ^{2}}{2 x y} = \frac{25 + 36 - 49}{60} = \frac{1}{5}

. Then

sin^{2} (\frac{X + Y}{2}) = sin^{2} (\frac{π - Z}{2}) = cos^{2} \frac{Z}{2} = \frac{1 + c o s Z}{2} = \frac{3}{5}

. True.

JEE Advanced · 2016 · 22 May 2016 · Paper 1 · One or more correct options