minimum size

To see the full image of a wall the minimum size of mirror required is n. Find n/a where "a" is the size of the wall? how can we prove it?

Easy Math Editor

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

ummm... I think that the answer is 0; It can be as small as you like Consider connecting all 4 corners of the wall to a single point, forming a solid. Now take a cross section parallel to the wall, and place the mirror at the intersection of that cross section and the solid. Then technically, somebody standing at the reflection of the original point through the mirror could see the whole wall. Provided they are transparent.

Brian Reinhart - 8 years, 2 months ago

The size of the wall = a Therefore the minimum size of mirror required to view the full wall = a/2 = n So, n/a = a/2a = 1/2.

Abhishek Mohapatra - 8 years, 2 months ago

the value of n/a=1/3............ but how can we prove??????????

Rahul Vernwal - 8 years, 2 months ago

Rahul the proof can be found ind a ICSE 9 book in chapter light.

Pratik Acharya - 8 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$