Removed coreners from a square

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Removed coreners from a square

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#1 h Aug 11, 2019, 9:08 AM • 5 Y

Y by dudu27_ctc, polynomialian, Adventure10, Mango247, Mango247

Form a square with sides of length $5$ , triangular pieces from the four coreners are removed to form a regular octagonn. Find the area removed to the nearest integer.

This post has been edited 1 time. Last edited by duttaditya18, Aug 11, 2019, 9:09 AM
Reason: Proper Question

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meet18

#4 h Aug 11, 2019, 9:45 AM • 4 Y

Y by paragdey01, polynomialian, Adventure10, Mango247

Let $x$ be the side length of octagon. We have that $x*root(2) +x=5$ . It is easy now.

This post has been edited 2 times. Last edited by meet18, Aug 11, 2019, 9:47 AM

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#6 h Aug 11, 2019, 9:51 AM • 1 Y

Y by Adventure10

so we get the answer as 5 or 4?

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#10 h Aug 11, 2019, 9:59 AM • 1 Y

Y by Adventure10

it is 4.3, so 4

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VERYunfortunate

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VERYunfortunate

#11 h Aug 11, 2019, 10:39 AM • 1 Y

Y by Adventure10

4 using simple relation

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867 posts

#12 h Aug 14, 2019, 7:17 AM • 4 Y

Y by deadkettles, geometrer_enthu, Adventure10, Mango247

I got this one wrong :wallbash:

:wallbash:

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#13 h Oct 2, 2019, 6:26 AM • 3 Y

Y by Randomizeriitian, Adventure10, Mango247

4 is the answer

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#14 h Dec 26, 2019, 12:49 PM • 1 Y

Y by Adventure10

Can someone please explain me

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#15 h Jun 4, 2020, 4:07 PM

Y by

meet18 wrote:

Let $x$ be the side length of octagon. We have that $x*root(2) +x=5$ . It is easy now.

I got the right answer ( 4 after rounding 4.3) by solving the equation :
sqrt(2) * x + 2x = 5 (x is the length of the leg of the triangle being removed)

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#16 h Jun 4, 2020, 4:09 PM • 1 Y

Y by The_Conjecturer

PRMO 2019 p1 lol. (search it)

This post has been edited 1 time. Last edited by Dem0nfang, Jun 4, 2020, 4:09 PM

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ATGY

#17 h Jul 18, 2020, 12:32 PM • 1 Y

Y by polynomialian

We have:
$x\sqrt2 = 5 - 2x$ $\implies x\sqrt2 + 2x = 5$ $\implies x\sqrt2(1 + \sqrt2) = 5$ $\implies x = \frac{5(\sqrt2 - 1)}{\sqrt2} \approx \sqrt2$ $\implies \text{removed area approximate} = \boxed{4}$ $\blacksquare$

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Thayar

#18 h Nov 4, 2020, 12:16 PM

Y by

How do you get the first statement?

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#19 h Nov 6, 2020, 10:21 PM

Y by

Wasn't it trivial @above ? I mean its just pythagoras theorem to isosceles triangles. Try convincing it to yourself

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Thayar

#20 h Apr 9, 2021, 3:15 AM

Y by

Yeah, I got it now. Thanks

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#22 h Sep 2, 2023, 6:33 AM

Y by

Chinmaydagod wrote:

so we get the answer as 5 or 4?

5
55

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blob22

#23 h Feb 26, 2025, 6:02 AM

Y by

We know that to be a regular octagon, the triangle needs to be isosceles right triangle. Let $x$ be the short side of the triangle and $y$ as the side of the octagon. The equation that we can form is:
$\begin{align*} (2x+y)^2=5 \\ \end{align*}$
and also
$\begin{align*} y^2=2x^2\\ y=x\sqrt{2} \end{align*}$
Substitute $y=x\sqrt{2}$ into $(2x+y)^2=5$ thus we get.

$\begin{align*} (2x+x\sqrt{2})^2=5\\ 4x^2+4x^2\sqrt{2}+2x^2=5\\ 6x^2+4x^2\sqrt{2}=5\\ x^2(6+4\sqrt{2})=5\\ x^2=\frac{5}{6+4\sqrt{2}} \end{align*}$
We need to find the value of $5-(4\times\frac{x^2}{2})$
$\begin{align*} 5-(4\times\frac{x^2}{2}) = 5-(2x^2)=5-(2\times(\frac{5}{6+4\sqrt{2}})=5-(\frac{10}{6+4\sqrt{2}}) \approx \boxed{4} \end{align*}$

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#24 h Mar 6, 2025, 9:49 PM

Y by

Note that $5-2x=x\sqrt{2}\implies x=\frac{5\sqrt{2}-5}{\sqrt{2}}$ so approximate area is $\boxed{4}$

This post has been edited 1 time. Last edited by S_14159, Mar 6, 2025, 9:49 PM

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#25 h Mar 18, 2025, 9:21 AM

Y by

why can't it be like 4 equilateral triangles are removed from each corner having side length $5/3$ ?
Then the answer becomes $(5/3)^2$ √ $3$ and the answer becomes 4.811 or approximately 5?

This post has been edited 1 time. Last edited by Kaprekar6147, Mar 18, 2025, 9:21 AM

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