In the coordinate plane, let A be a point on the x-axis, and let B be a point on the y-axis, so that AB is tangent to the unit circle. Find the minimum value of AB

Guest Jun 1, 2019

#1**+1 **

In the coordinate plane, let A be a point on the x-axis, and let B be a point on the y-axis, so that AB is tangent to the unit circle. Find the minimum value of AB.

The point of contact let P be, the origin of the coordinate plane O.

Triangle AOB is similar to triangle \(PX_PO\).

\(A_{PX_PO}=\frac{x}{2}\cdot y\)

\(y=\sqrt{1-x^2}\)

\(A=\frac{x}{2}\sqrt{1-x^2}\\ A=\frac{1}{2}\sqrt{x^2-x^4}\)

\(A=\frac{1}{2} (x^2-x^4)^{\frac{1}{2}}\)

\(A'=\frac{1}{2}(2x-4x^3)\cdot \frac{1}{2}(x^2-x^4)^{-\frac{1}{2}}=0\\ A'=\frac{2x-4x^3}{4(x^2-x^4)^{\frac{1}{2}}}=0\\ A'=\frac{x-2x^3}{2(x^2-x^4)^{\frac{1}{2}}}=0\\ \color{blue}x=\frac{1}{\sqrt{2}}\)

\(x=0,7071067811865476\)

\(A=\frac{1}{2}\sqrt{x^2-x^4}\)

\(A_{X_PO}=0.5\)

\(A_{AOB}=4\cdot A_{X_PO}=2 \)

The smallest area of the triangle AOB is equal to 2.

!

asinus Jun 1, 2019

#2**+2 **

Denote the point of tangency as T.

To attain minimum, AO and OB must be as short as they can be. As the unit circle is symmetric, AO must be equal to OB when minimum of AB is attained.

\(\text{AO} = \text{OB} \wedge\angle \text{AOB} = 90^{\circ}\).

When minimum of AB is attained, \(\triangle \text{AOB}\) is a 90-45-45 triangle.

OT = 1 unit

As \(\triangle \text{AOB}\) is an isosceles triangle, the radius joining O to the point of tangency is a perpendicular bisector and it divides \(\triangle \text{AOB}\) into 2 equal halves, which are two 90-45-45 triangles.(If you can't understand this, go to the kitchen, grab a square piece of bread, cut it along the main diagonal, and then cut one of the two parts into half.)

As \(\triangle \text{AOT}\) is a 90-45-45 triangle with angle OTA = 90 degrees, TA = TO = 1 unit.

As the shape is symmetric along the line OT, TB = TA = 1 unit.

So when minimum is attained, AB = TA + TB = 1 + 1 = **2 units**

MaxWong Jun 1, 2019