Smallest area of triangle

Geometry Level 4

As shown in the figure, given two fixed points, $O=(0,0)$ and $A=(10,10)$ . $C$ is a point on $x$ -axis, while $D$ is a point on a straight line, which make a $60^\circ$ with $x$ -axis. Now the three points $C, A$ and $D$ lie on a straight line.

Let the smallest possible value of the area of triangle $OCD$ be $s$ . Find the value of $\lfloor 1000s\rfloor$ .

The answer is 84529.

1 solution

Mayank Chaturvedi
May 22, 2016

Let the slope of line L1 be m. As the line pass through (10,10) we can find the coordinates of C and D in terms of m (as shown in the figure). L3: $y-\sqrt{3}x=0$ ; length of perpendicular from C to L3 = $\left| \frac { 5\sqrt { 3 } (m-1) }{ m } \right|$

L1: $(y-10)=m(x-10)$

We look at base L3 and Perpendicular from C

Area = $\frac{1}{2} (base)(height)$ = $\frac { 1 }{ 2 } \left| \frac { 20(1-m) }{ \sqrt { 3 } -m } \right| \left| \frac { 5\sqrt { 3 } (m-1) }{ m } \right|$

=> $50\sqrt{3} \left| \frac { { (m-1) }^{ 2 } }{ m(\sqrt { 3 } -m) } \right|$ . This equation has its minima at m=1 and $\sqrt{3} (2+\sqrt{3})$ of which, second one is desirable as in m=1, O and C co-inside.

So, minimum area comes to be 84.52994.... at m= $\sqrt{3} (2+\sqrt{3})$

Nice solution. Up voted.

Niranjan Khanderia - 5 years ago

Smallest area of triangle

The answer is 84529.

1 solution

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