JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Properties of Triangle

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Let x, y and z be positive real numbers. Suppose x, y and z are lengths of the sides of a triangle opposite to its angles X, Y and Z, respectively. If

tan \frac{X}{2} + tan \frac{Z}{2} = \frac{2 y}{x + y + z},

then which of the following statements is/are TRUE?

(A) $2 Y = X + Z$
(B) $Y = X + Z$ ✓ Correct
(C) $tan \frac{X}{2} = \frac{x}{y + z}$ ✓ Correct
(D) $x^{2} + z^{2} - y^{2} = x z$

Answer: (B), (C)

Solution

Using the formula

tan \frac{A}{2} = \frac{Δ}{s ( s - a )}

where

s

is the semi-perimeter and

Δ

is the area:

tan \frac{X}{2} + tan \frac{Z}{2} = \frac{Δ}{s ( s - x )} + \frac{Δ}{s ( s - z )} = \frac{Δ}{s} (\frac{2 s - ( x + z )}{( s - x ) ( s - z )}) = \frac{Δ y}{s ( s - x ) ( s - z )}

Given this equals

\frac{2 y}{2 s} = \frac{y}{s}

, we have:

\frac{Δ}{( s - x ) ( s - z )} = 1

This simplifies to

(x + y + z) (x + z - y) = (y + z - x) (x + y - z)

, which gives

x^{2} + z^{2} = y^{2}

.

Therefore,

∠ Y = 90°

, so

Y = \frac{π}{2}

and

X + Z = \frac{π}{2}

. (B) is TRUE, (A) is FALSE.

For (C):

tan \frac{X}{2} = \frac{Δ}{s ( s - x )} = \frac{\frac{1}{2} x z}{\frac{( y + z ) ^{2} - x ^{2}}{4}} = \frac{2 x z}{y ^{2} + z ^{2} + 2 y z - x ^{2}} = \frac{2 x z}{2 z ^{2} + 2 y z} = \frac{x}{y + z}

. (C) is TRUE.

For (D): From

x^{2} + z^{2} = y^{2}

, we have

x^{2} + z^{2} - y^{2} = 0 \neq = x z

in general. (D) is FALSE.

JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · One or more correct options

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JEE Advanced · 2020 · 27 Sep 2020 · Paper 1 · Properties of Triangle

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