Question

If $a$ and $b$ are real numbers between $0$ and $1$ such that the points $(a,1),(1,b)$ and $(0,0)$ form an equilateral triangle, find a and b.

• A. $a=2-\sqrt{3},b=2-\sqrt{3}$
• B. $a=2+\sqrt{3},b=2-\sqrt{3}$
• C. $a=2-\sqrt{3},b=2+\sqrt{3}$
• D. $a=2+\sqrt{3},b=2+\sqrt{3}$

Answer 1

The correct option is A $a=2-\sqrt{3},b=2-\sqrt{3}$

$△ABCisanequilateraltriangleAB=BC=AC{\left(AB\right)}^2={\left(BC\right)}^2={\left(AC\right)}^{2}AB=\sqrt{{(1-a)}^2+{(b-1)}^2}BC=\sqrt{1+b^2}AC=\sqrt{a^2+1}Also,{(1-a)}^2+{(b-1)}^2=1+b^2=1+a^2=>a^2=b^{2}Also,{(1-a)}^2+{(b-1)}^2=1+a^2-2a+1+b^2-2b=>1+a^2-2a+1+b^2-2b=1+b^2=>a^2-2a-2b+2=1=>b^2-2b-2b+1=0=>b^2-4b+1=0=>b=\frac{4\pm \sqrt{16-4}}{2}=\frac{4\pm \sqrt{12}}{2}=>b=2\pm \sqrt{3}b=2-\sqrt{3}a=2-\sqrt{3}$