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Find the sides

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area of a triangle

Heron's formula

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Source: Indian RMO 1994 Problem 2

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1592 posts

Rushil

#1 h Oct 25, 2005, 8:34 AM • 2 Y

Y by Adventure10, Mango247

In a triangle $ABC$ , the incircle touches the sides $BC, CA, AB$ at $D, E, F$ respectively. If the radius if the incircle is $4$ units and if $BD, CE , AF$ are consecutive integers, find the sides of the triangle $ABC$ .

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749 posts

#2 h Oct 25, 2005, 11:36 AM • 1 Y

Y by Adventure10

well known bd, ce, af = s-b, s-c, s-a.

So (s-a) - (s-c) = 1, (s-c) - (s-b) = 1. They imply c-a = 1 and b-c = 1.

So b = 1+c and a = c-1. So s = (a+b+c)/2 = 3c/2. Since s-c is integer, c is twice an integer. Put c =2k.

Finally (s-b, s-c, s-a) becomes (k-1, k, k+1). From area formula, $\sqrt{3k(k)(k+1)(k-1)/2}$ is area.

From $sr = area$ formula, area is $6k$ . So we must equate $36 = 3(k+1)(k-1)/2$ . Then $k = 5$ (has to be positive).

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#3 h Dec 8, 2005, 4:45 AM • 2 Y

Y by Adventure10, Mango247

Singular wrote:

From area formula, $\sqrt{3k(k)(k+1)(k-1)/2}$ is area.

From $sr = area$ formula, area is $6k$ . So we must equate $36 = 3(k+1)(k-1)/2$ . Then $k = 5$ (has to be positive).

Your solution is incorrect.

Area is wrong, using your notation, the area should be $\sqrt{3k(k)(k+1)(k-1)}$ because $s=3k$

Second area is also wrong, it should be $12k$

So $144k^2=3k^2(k+1)(k-1)\Rightarrow48=(k^2-1)\Rightarrow k=7$ The triangle is 13-14-15.

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madhusudan kale

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madhusudan kale

#4 h May 12, 2012, 9:49 AM • 2 Y

Y by Adventure10, Mango247

we equate the area of whole triangle (calculated using Heron's formula $\sqrt{s(s-a)(s-b)(s-c)}$ )
and the area of smaller triangles for example $\Delta ODC, \Delta AOF$ etc. SEE ATTACHMENT.
Take $AE=n, CF=n+1 , BD= n+2.$

and by that we get the sides as 13,14,15.

Attachments:

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7054 posts

#5 h May 16, 2012, 3:22 AM • 2 Y

Y by Adventure10, Mango247

$(s-a)(s-b)(s-c)=sr^2\iff (n-1)n(n+1)=3n\cdot 16\iff$

$n^2-1=48\iff$ $n^2=49\iff$ $n=7\iff$ $(6,7,8)\iff$ $\Delta (13,14,15)$ .

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sayantanchakraborty

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sayantanchakraborty

#6 h Sep 11, 2013, 4:33 PM • 2 Y

Y by Adventure10, Mango247

Let BD=x,CE=x+1 and AF=x+2.
Let I be the incentre s.t. ID=IE=IF=4
Area of ABC=area of AIB+area of BIC+area of CIA=12(x+1)
Applying heron's formula we get are of ABC=[(3x+3)(x+2)(x+1)x]^1/2
Equating these two areas we get x=6.
Thus the sides are 2x+1,2x+2,2x+3 i.e., 13,14 and 15 units.

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#7 h Jan 16, 2017, 3:41 PM • 2 Y

Y by Adventure10, Mango247

Let BD be $x$ , EC be $x+1$ , and AF be $x+2$ . Because of properties of in circles, AE will equal AF, CD will equal CE and BF will equal BD. Adding each side up, we get BC, AC, AB be $2x+1$ , $2x+3$ , and $2x+2$ respectively. Let $2x+2$ be $y$ , So the side lengths are in consecutive order with AB AC and BC be equal to $y$ , $y-1$ , and $y+1$

Note $y$ will have to be even.

By Heron's Formula $A=\sqrt{s(s-a)(s-b)(s-c)}$ , $A=\sqrt{(\frac{3y}{2})(\frac{y}{2})(\frac{y-2}{2})(\frac{y+2}{2})}=\frac{x\sqrt{3x^2-12}}{4}$

Since $A=sp$ , $A=\frac{12y}{2}$

Setting these equal, we get $\frac{x\sqrt{3x^2-12}}{4}=\frac{12y}{2}$ , or $y=14$ . This works, since $y$ has to be even.

Since this is one of the sides, we just add 1 and subtract 1, to find the other sides which are $\boxed{13, 14, 15}$ .

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#8 h Oct 30, 2024, 9:30 AM

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In short, calculate the $[ABC]$ in two ways, by $\Delta=rs$ and $\text{Heron's formula!}$

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