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Explore the Back-StoryTry 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.

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

Pre College Mathematics

AMC-10A, 2015 Problem-24

$31$

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?

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?

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 $31$

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

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.

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

Pre College Mathematics

AMC-10A, 2015 Problem-24

$31$

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?

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?

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 $31$

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

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