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Properties of Triangle question

2014 · Shift 2 · Q35
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  5. /2014 · Shift 2 · Q35

Properties of Triangle question

2014 · Shift 2 · Q35

JEE AdvancedMathematicsProperties of TriangleMCQ+3 / −1
In a triangle the sum of two sides is xxx and the product of the same sides is yyy. If x2−c2=y{x^2} - {c^2} = yx2−c2=y, where ccc is the third side of the triangle, then the ratio of the in radius to the circum-radius of the triangle is
  1. A
    3y2x(x+c){{3y} \over {2x\left( {x + c} \right)}}2x(x+c)3y​
  2. B
    3y2c(x+c){{3y} \over {2c\left( {x + c} \right)}}2c(x+c)3y​
  3. C
    3y4x(x+c){{3y} \over {4x\left( {x + c} \right)}}4x(x+c)3y​
  4. D
    3y4c(x+c){{3y} \over {4c\left( {x + c} \right)}}4c(x+c)3y​
View written solutionFree

Correct answer: B

Let the two sides be aaa and bbb, and the third side be ccc.

We are given:

  1. a+b=xa+b=xa+b=x
  2. ab=yab=yab=y
  3. x2−c2=yx^2-c^2=yx2−c2=y

We need to find rR\dfrac{r}{R}Rr​, where rrr is the inradius and RRR is the circumradius.

1. Use the standard formula for rR\dfrac{r}{R}Rr​

For any triangle,

Δ=rs=abc4R\Delta = rs = \frac{abc}{4R}Δ=rs=4Rabc​

where sss is the semiperimeter.

So,

r=Δs,R=abc4Δr = \frac{\Delta}{s}, \qquad R = \frac{abc}{4\Delta}r=sΔ​,R=4Δabc​

Hence,

rR=Δ/sabc/(4Δ)=4Δ2sabc\frac{r}{R} = \frac{\Delta/s}{abc/(4\Delta)} = \frac{4\Delta^2}{sabc}Rr​=abc/(4Δ)Δ/s​=sabc4Δ2​

A more convenient identity is:

Δ2=s(s−a)(s−b)(s−c)\Delta^2 = s(s-a)(s-b)(s-c)Δ2=s(s−a)(s−b)(s−c)

Thus,

rR=4(s−a)(s−b)(s−c)abc\frac{r}{R} = \frac{4(s-a)(s-b)(s-c)}{abc}Rr​=abc4(s−a)(s−b)(s−c)​

2. Express semiperimeter terms in terms of xxx and ccc

Since a+b=xa+b=xa+b=x,

s=a+b+c2=x+c2s = \frac{a+b+c}{2} = \frac{x+c}{2}s=2a+b+c​=2x+c​

Now,

s−a=a+b+c2−a=b+c−a2s-a = \frac{a+b+c}{2}-a = \frac{b+c-a}{2}s−a=2a+b+c​−a=2b+c−a​ s−b=a+c−b2s-b = \frac{a+c-b}{2}s−b=2a+c−b​ s−c=a+b−c2=x−c2s-c = \frac{a+b-c}{2} = \frac{x-c}{2}s−c=2a+b−c​=2x−c​

Instead of expanding directly, use the identity:

(s−a)(s−b)=(b+c−a)(a+c−b)4(s-a)(s-b) = \frac{(b+c-a)(a+c-b)}{4}(s−a)(s−b)=4(b+c−a)(a+c−b)​

Now,

(b+c−a)(a+c−b)=c2−(a−b)2(b+c-a)(a+c-b) = c^2-(a-b)^2(b+c−a)(a+c−b)=c2−(a−b)2

So,

(s−a)(s−b)=c2−(a−b)24(s-a)(s-b) = \frac{c^2-(a-b)^2}{4}(s−a)(s−b)=4c2−(a−b)2​

But,

(a−b)2=(a+b)2−4ab=x2−4y(a-b)^2 = (a+b)^2-4ab = x^2-4y(a−b)2=(a+b)2−4ab=x2−4y

Hence,

c2−(a−b)2=c2−(x2−4y)=c2−x2+4yc^2-(a-b)^2 = c^2-(x^2-4y)=c^2-x^2+4yc2−(a−b)2=c2−(x2−4y)=c2−x2+4y

Given x2−c2=yx^2-c^2=yx2−c2=y, so

c2−x2=−yc^2-x^2=-yc2−x2=−y

Therefore,

c2−x2+4y=−y+4y=3yc^2-x^2+4y = -y+4y=3yc2−x2+4y=−y+4y=3y

Thus,

(s−a)(s−b)=3y4(s-a)(s-b)=\frac{3y}{4}(s−a)(s−b)=43y​

Also,

s−c=x−c2s-c=\frac{x-c}{2}s−c=2x−c​

Therefore,

(s−a)(s−b)(s−c)=3y4⋅x−c2=3y(x−c)8(s-a)(s-b)(s-c)=\frac{3y}{4}\cdot \frac{x-c}{2} = \frac{3y(x-c)}{8}(s−a)(s−b)(s−c)=43y​⋅2x−c​=83y(x−c)​

3. Compute rR\dfrac{r}{R}Rr​

Using

rR=4(s−a)(s−b)(s−c)abc\frac{r}{R} = \frac{4(s-a)(s-b)(s-c)}{abc}Rr​=abc4(s−a)(s−b)(s−c)​

and ab=yab=yab=y, we get

rR=4⋅3y(x−c)8yc\frac{r}{R} = \frac{4\cdot \frac{3y(x-c)}{8}}{yc}Rr​=yc4⋅83y(x−c)​​ =3(x−c)2c= \frac{3(x-c)}{2c}=2c3(x−c)​

Now use the given relation:

x2−c2=yx^2-c^2=yx2−c2=y

which factors as

(x−c)(x+c)=y(x-c)(x+c)=y(x−c)(x+c)=y

So,

x−c=yx+cx-c = \frac{y}{x+c}x−c=x+cy​

Substitute into rR\dfrac{r}{R}Rr​:

rR=32c⋅yx+c=3y2c(x+c)\frac{r}{R} = \frac{3}{2c}\cdot \frac{y}{x+c} = \frac{3y}{2c(x+c)}Rr​=2c3​⋅x+cy​=2c(x+c)3y​

4. Match with options

Thus,

rR=3y2c(x+c)\boxed{\frac{r}{R}=\frac{3y}{2c(x+c)}}Rr​=2c(x+c)3y​​

This matches Option B.

5. Verification with stored answer

Stored correct answer: B

Our derived answer: B

So the answer agrees with the stored answer.

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