Try this beautiful Problem on Geometry based on Positive Integers and Quadrilateral from AMC 10 A, 2015. You may use sequential hints to solve the problem.

**Positive Integers and Quadrilateral** – AMC-10A, 2015- Problem 24

For some positive integers $p$, there is a quadrilateral $A B C D$ with positive integer side lengths, perimeter $p$, right angles at $B$ and $C, A B=2$, and $C D=A D$. How many different values of $p<2015$ are possible?

,

- $30$
- $31$
- $61$
- $62$
- $63$

**Key Concepts**

Geometry

Rectangle

Integer

## Suggested Book | Source | Answer

#### Suggested Reading

Pre College Mathematics

#### Source of the problem

AMC-10A, 2015 Problem-24

#### Check the answer here, but try the problem first

$31$

## Try with Hints

#### First Hint

Given that $ ABCD$ is a quadrilateral whose perimeter $p$, right angles at $B$ and $C, A B=2$, and $C D=A D$. we have to find out How many different values of $p<2015$ are possible.

Now draw a perpendicular $AE$ on $CD$ . Let us assume that $BC=AE=x$ Then $CE=2$ and $DE=y-2$

Now can you finish the problem?

#### Second Hint

Now from the \(\triangle ADE\) we can write $x^{2}+(y-2)^{2}=y^{2}$

\(\Rightarrow x^{2}-4 y+4=0\)

\(\Rightarrow x^2=4(y-1)\), Thus, $y$ is one more than a perfect square.

Therefore the perimeter will be $p=2+x+2 y=2 y+2 \sqrt{y-1}+2$

Now according to the problem $p<2015$

So, $p=2+x+2 y=2 y+2 \sqrt{y-1}+2 <2015$

Now Can you finish the Problem?

#### Third Hint

Now $y=31^{2}+1=962$ and $y=32^{2}+1=1025$

Here $y=31^{2}+1=962$ is valid but $y=32^{2}+1=1025$ is not. On the lower side, $y=1$ does not work (because $x>0$ ), but $y=1^{2}+1$ does work. Hence, there are $31$ valid $y$ (all $y$ such that $y=n^{2}+1$ for $1 \leq n \leq 31$ )

Therefore the correct answer is $31$

## Other useful links

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=0qSCPw0YhUY