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# Positive Integers and Quadrilateral | AMC 10A 2015 | Sum 24

Try this beautiful Problem on Rectangle and triangle from AMC 10A, 2015. Problem-24. You may use sequential hints to solve the problem.

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?

,

• $30$
• $31$
• $61$
• $62$
• $63$

Geometry

Rectangle

Integer

## Suggested Book | Source | Answer

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 $31$

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