Question

If a and b are real numbers between 0 and 1 such that the point \[(a,1),(1,b)\] and \[(0,0)\] from an equilateral triangle, find a and b.

Answer 1

Hint: Here we are given the points from an equilateral triangle, we have to find the value of a and b. As all the sides are equal for an equilateral triangle, we get two different equations using a and b. Then solving them altogether will give us the needed value of a and b.

Complete step-by-step answer:
Given points \[(a,1),(1,b)\] and \[(0,0)\] from an equilateral triangle.

As all the sides are equal for an equilateral triangle, so we use the distance formula to find the distance between 2 points, i.e., for points \[(a,b)\] and \[(c,d)\] , the distance between them is \[\sqrt {{{(a - c)}^2} + {{(b - d)}^2}} \] .
We have the distance between \[(a,1)\] and \[(0,0)\]

\[ \Rightarrow \] \[\sqrt {{{(a - 0)}^2} + {{(1 - 0)}^2}} \] \[ = \sqrt {{a^2} + 1} \]

And, the distance between \[(0,0)\] and \[(1,b)\] is

 \[ \Rightarrow \] \[\sqrt {{{(0 - 1)}^2} + {{(0 - b)}^2}} = \sqrt {{b^2} + 1} \]

Also, the distance between \[(a,1)\] and \[(1,b)\] is

\[ \Rightarrow \] \[\sqrt {{{(a - 1)}^2} + {{(1 - b)}^2}} \]

As it forms an equilateral triangle,

So, distance between any 2 points are equal.

Hence we have, \[\sqrt {{{(a - 1)}^2} + {{(1 - b)}^2}} = \sqrt {{a^2} + 1} = \sqrt {{b^2} + 1} \]

On squaring, we get
\[{(a - 1)^2} + {(1 - b)^2} = {a^2} + 1 = {b^2} + 1\] ………….(1)
From the last two, we get
 \[{a^2} + 1 = {b^2} + 1\]
On subtracting 1 from both sides, we get
 \[ \Rightarrow {a^2} = {b^2}\]
On taking square root, we get
 \[ \Rightarrow a = \pm b\]
Now, again, from the first two terms of equation (1), we get
 \[{(a - 1)^2} + {(1 - b)^2} = {a^2} + 1\]
On taking, \[b = a\]
\[ \Rightarrow \]\[{(a - 1)^2} + {(1 - a)^2} = {a^2} + 1\]
As \[{(a - 1)^2} = {(1 - a)^2}\], we get
\[ \Rightarrow 2{(a - 1)^2} = {a^2} + 1\]
On using \[{(a - 1)^2} = {a^2} - 2a + 1\] , we get
 \[ \Rightarrow 2({a^2} - 2a + 1) = {a^2} + 1\]
On simplification, we get
 \[ \Rightarrow 2{a^2} - 4a + 2 = {a^2} + 1\]
 \[ \Rightarrow {a^2} - 4a + 1 = 0\]
For a quadratic equation \[a{x^2} + bx + c = 0\] the value of x is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\] ,
Using this, we get
 \[ \Rightarrow a = \dfrac{{ - ( - 4) \pm \sqrt {{{( - 4)}^2} - 4 \times 1 \times 1} }}{{2 \times 1}}\]
On simplification we get,
 \[ \Rightarrow a = \dfrac{{4 \pm \sqrt {16 - 4} }}{2}\]
On solving the value under the root we get,
 \[ \Rightarrow a = \dfrac{{4 \pm \sqrt {12} }}{2}\]
On further simplification we get,
 \[ \Rightarrow a = \dfrac{{4 \pm 2\sqrt 3 }}{2}\]
On dividing we get,
 \[ \Rightarrow a = 2 \pm \sqrt 3 \]
As, \[b = a\]
So, we will have,
 \[ \Rightarrow b = 2 \pm \sqrt 3 \]

Note: For a equilateral triangle,
1.Three kinds of cevians coincide, and are equal, for (and only for) equilateral triangles:
2.The three altitudes have equal lengths.
3.The three medians have equal lengths.
4.The three angle bisectors have equal lengths.