max equilateral inside a square (2001 Denmark MO - Mohr Contest p5)

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High School Math Grades 9-12, Ages 13-18

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max equilateral inside a square (2001 Denmark MO - Mohr Contest p5)

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#1 h Aug 16, 2020, 9:57 AM

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Is it possible to place within a square an equilateral triangle whose area is larger than $9/ 20$ of the area of the square?

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#2 h Aug 16, 2020, 2:08 PM

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#3 h Aug 16, 2020, 3:14 PM

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Area of the equilateral triangle $=a^{2}(2\sqrt{3}-3)$ , area of the square $=a^{2}$ .

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#4 h Aug 16, 2020, 3:17 PM

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Wait I thought area of equilateral triangle was $\frac{a^2\sqrt{3}}{4}$

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#5 h Aug 16, 2020, 3:29 PM

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@above it is. @2above was talking about the equilateral triangle which was placed inside the square.

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#6 h Aug 16, 2020, 4:04 PM

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If we labeled each vertex of the square as $A, B, C,$ and $D$ , then we probably could- The triangle $DMN$ , where $M$ is on side $AB$ and $N$ is on $BC$ , and $AN=CM$

And ofc, $DMN$ is equilateral, so there is one possibility

This post has been edited 1 time. Last edited by peace09, Aug 16, 2020, 4:04 PM

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#7 h Aug 16, 2020, 6:15 PM

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ilovepizza2020 wrote:

Wait I thought area of equilateral triangle was $\frac{a^2\sqrt{3}}{4}$

Funny thing about formulae they shape shift depending on inputs and restrictions:

$A=\pi r^2=\frac{\pi d^2}{4}=\frac{C^2}{4\pi}$ are all valid circle area formulae. each has an equivalent for regular polygons.

This post has been edited 1 time. Last edited by science_man_88, Aug 16, 2020, 6:23 PM
Reason: variable

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#8 h Aug 17, 2020, 4:10 PM

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https://lh6.googleusercontent.com/0pm8RYuxBltHkt4nni1K7y7a2soOiPJTz0tL4V_wLv5GR9pBfGjnx9vvpaZIoY85O6qJHE85KBu5n2TZtDqcUrjrENqLhfK2tWBRTHqi0I8eDC36levYy0YnJ-iKB8DjO3C0qu3T

We draw a 60 degree angle inside $\angle ADC$ in such a way that angles $FDC$ and $EDA$ are congruent ( $15^\circ$ angles). By LA, these triangles are congruent. We draw the diagonals of this square, $DB$ and $AC$ . Because $AE \cong CF$ , $EF \parallel AC$ (you know why) & $EBF$ is a isoceles right triangle.
If $DG = x$ , $GF =\frac{x}{\sqrt{3}}$ , $GB =\frac{x}{\sqrt{3}}$ , $DB = x\frac{\sqrt{3}+1}{\sqrt{3}}$ , $\text{side length of square}=x\frac{\sqrt{3}+1}{\sqrt{6}}$ , $\text{area of square}=x^2\frac{2+\sqrt{3}}{3}$ , $\text{area of equilateral triangle}=\frac{x^2}{\sqrt{3}}$ .
And the ratio of the triangle to the square is $\frac{3}{2\sqrt{3}+3}$ , which we know is greater than $\frac{9}{20}$ . $\blacksquare$

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HumanCalculator9

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HumanCalculator9

#9 h Aug 17, 2020, 4:14 PM

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Question:
What about placing a square in an equilateral triangle?
What's the largest percent of an equilateral triangle that can be covered by 1 square?

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#10 h Aug 17, 2020, 4:23 PM

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HumanCalculator9 wrote:

Question:
What about placing a square in an equilateral triangle?
What's the largest percent of an equilateral triangle that can be covered by 1 square?

Here's a related topic:
https://gis.stackexchange.com/questions/59215/how-to-find-the-maximum-area-rectangle-inside-a-convex-polygon

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#11 h Aug 17, 2020, 4:34 PM

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I think it's just this.

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HumanCalculator9

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HumanCalculator9

#12 h Aug 17, 2020, 4:44 PM

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thedodecagon wrote:

I think it's just this.

I think you're right
What percent of the triangle's area is inside the square?

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#14 h Aug 17, 2020, 5:23 PM

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here, you can do the rest of the calculating, labeled stuff.

This post has been edited 1 time. Last edited by thedodecagon, Aug 17, 2020, 5:24 PM
Reason: fixed link

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