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Perimeter of an equilateral triangle drawn with respect to a square.
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Here's a question that was asked in the International Kangaroo Math Contest 2016. The question goes like this:
If the perimeter of the square in the figure is 4 units, then what is the perimeter of the equilateral triangle?
What I did:
Well I tried something very naïve and it was the supposition that the equilateral triangle cuts the top side of the square at its midpoint. Hence giving the following result.
By Pythagoras' Theorem,
$$overlineAB = overlineMC = sqrtoverlineBC^2 + overlineBM^2 = sqrtleft(1right)^2 + left(frac12right)^2 = fracsqrt52$$
So perimeter of the triangle is:
$$beginalign
P&=overlineAF +overlineFM+overlineMC+overlineCD+overlineDE+overlineEA\
&= frac12+frac12+fracsqrt52+1+frac12+fracsqrt52\
&= frac52+sqrt5
endalign$$
However this is not the correct answer and I know that the problem is with the supposition that $M$ is the midpoint of $overlineAB$. So what is the correct method and answer?
Thanks for the attention.
geometry analytic-geometry
asked 4 hours ago
Faiq Irfan
512217
add a comment |Â
up vote
4
down vote
favorite
Here's a question that was asked in the International Kangaroo Math Contest 2016. The question goes like this:
If the perimeter of the square in the figure is 4 units, then what is the perimeter of the equilateral triangle?
What I did:
Well I tried something very naïve and it was the supposition that the equilateral triangle cuts the top side of the square at its midpoint. Hence giving the following result.
By Pythagoras' Theorem,
$$overlineAB = overlineMC = sqrtoverlineBC^2 + overlineBM^2 = sqrtleft(1right)^2 + left(frac12right)^2 = fracsqrt52$$
So perimeter of the triangle is:
$$beginalign
P&=overlineAF +overlineFM+overlineMC+overlineCD+overlineDE+overlineEA\
&= frac12+frac12+fracsqrt52+1+frac12+fracsqrt52\
&= frac52+sqrt5
endalign$$
However this is not the correct answer and I know that the problem is with the supposition that $M$ is the midpoint of $overlineAB$. So what is the correct method and answer?
Thanks for the attention.
geometry analytic-geometry
asked 4 hours ago
Faiq Irfan
512217
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Here's a question that was asked in the International Kangaroo Math Contest 2016. The question goes like this:
If the perimeter of the square in the figure is 4 units, then what is the perimeter of the equilateral triangle?
What I did:
Well I tried something very naïve and it was the supposition that the equilateral triangle cuts the top side of the square at its midpoint. Hence giving the following result.
By Pythagoras' Theorem,
$$overlineAB = overlineMC = sqrtoverlineBC^2 + overlineBM^2 = sqrtleft(1right)^2 + left(frac12right)^2 = fracsqrt52$$
So perimeter of the triangle is:
$$beginalign
P&=overlineAF +overlineFM+overlineMC+overlineCD+overlineDE+overlineEA\
&= frac12+frac12+fracsqrt52+1+frac12+fracsqrt52\
&= frac52+sqrt5
endalign$$
However this is not the correct answer and I know that the problem is with the supposition that $M$ is the midpoint of $overlineAB$. So what is the correct method and answer?
Thanks for the attention.
geometry analytic-geometry
asked 4 hours ago
Faiq Irfan
512217
Here's a question that was asked in the International Kangaroo Math Contest 2016. The question goes like this:
If the perimeter of the square in the figure is 4 units, then what is the perimeter of the equilateral triangle?
What I did:
Well I tried something very naïve and it was the supposition that the equilateral triangle cuts the top side of the square at its midpoint. Hence giving the following result.
By Pythagoras' Theorem,
$$overlineAB = overlineMC = sqrtoverlineBC^2 + overlineBM^2 = sqrtleft(1right)^2 + left(frac12right)^2 = fracsqrt52$$
So perimeter of the triangle is:
$$beginalign
P&=overlineAF +overlineFM+overlineMC+overlineCD+overlineDE+overlineEA\
&= frac12+frac12+fracsqrt52+1+frac12+fracsqrt52\
&= frac52+sqrt5
endalign$$
However this is not the correct answer and I know that the problem is with the supposition that $M$ is the midpoint of $overlineAB$. So what is the correct method and answer?
Thanks for the attention.
geometry analytic-geometry
geometry analytic-geometry
asked 4 hours ago
Faiq Irfan
512217
asked 4 hours ago
Faiq Irfan
512217
asked 4 hours ago
Faiq Irfan
512217
asked 4 hours ago
Faiq Irfan
512217
asked 4 hours ago
Faiq Irfan
512217
512217
add a comment |Â
add a comment |Â
5 Answers
5
active
oldest
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up vote
2
down vote
$M$ is NOT the midpoint of $AB$. Note that the angle $angle MCB$ is equal to $90^circ-60^circ=30^circ$, therefore
$$|MB|=|BC|tan(30^circ)=frac1sqrt3.$$
Moreover $|ED|=|MB|$ (why?).
Can you take it from here?
answered 4 hours ago
Robert Z
87.5k1056127
Wish I'd refreshed before posting my answer, I may have given too much away!
– Tartaglia's Stutter
4 hours ago
add a comment |Â
up vote
1
down vote
Note that the angle AED is $fracpi3$ radians (or 60 degrees if you prefer). Since the side opposite that angle is $1$, and $tan(fracpi3)$ is $sqrt3$, we know that the side ED must be of length $fracsqrt33$. The triangle MBC is similar to EAD, so the side MB is also $fracsqrt33$. You can use Pythagorean's Theorem and the fact that each side of the square is 1 to find the lengths of all the remaining sides needed.
answered 4 hours ago
Tartaglia's Stutter
944
New contributor
Tartaglia's Stutter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Nice answer. BTW I like your username.
– Robert Z
4 hours ago
add a comment |Â
up vote
1
down vote
Given a horizontal segment with length $L$ and two lines
$$
begincases
y_1 = x tan(fracpi3)\
y_2 = L-xtan(fracpi2)
endcases
$$
their intersection is at
$$
y_1=y_2Rightarrow x = fracLtan(fracpi3)+tan(fracpi2)
$$
hence the equilateral triangle has perimeter $3L$ and the square has perimeter $4 x = frac4Ltan(fracpi3)+tan(fracpi2) = frac4Lsqrt 3+1$
answered 5 mins ago
Cesareo
7,1072415
add a comment |Â
up vote
0
down vote
Referring to the figure, let $x$ and $h$ be half-side and height of equilateral triangle, resp.:
$hspace3cm$
For the equilateral triangle:
$$h^2+x^2=(2x)^2 Rightarrow h^2=3x^2 Rightarrow h=xsqrt3.$$
From the similarity of triangles:
$$frach4=fracx2x-4 Rightarrow fracxsqrt34=fracx2(x-2) Rightarrow x=frac2sqrt33+2 Rightarrow P=6x=4sqrt3+12.$$
Addendum: It was stated the perimeter of the square is $4$, not the side. So, the answer must be divided by $4$ to get $3+sqrt3$.
answered 4 hours ago
farruhota
16.8k2735
add a comment |Â
up vote
0
down vote
We only need to know that the $angle ABC=30°$, the rest is just straight forward.
answered 1 hour ago
Seyed
6,11531222
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
$M$ is NOT the midpoint of $AB$. Note that the angle $angle MCB$ is equal to $90^circ-60^circ=30^circ$, therefore
$$|MB|=|BC|tan(30^circ)=frac1sqrt3.$$
Moreover $|ED|=|MB|$ (why?).
Can you take it from here?
answered 4 hours ago
Robert Z
87.5k1056127
Wish I'd refreshed before posting my answer, I may have given too much away!
– Tartaglia's Stutter
4 hours ago
add a comment |Â
up vote
2
down vote
$M$ is NOT the midpoint of $AB$. Note that the angle $angle MCB$ is equal to $90^circ-60^circ=30^circ$, therefore
$$|MB|=|BC|tan(30^circ)=frac1sqrt3.$$
Moreover $|ED|=|MB|$ (why?).
Can you take it from here?
answered 4 hours ago
Robert Z
87.5k1056127
Wish I'd refreshed before posting my answer, I may have given too much away!
– Tartaglia's Stutter
4 hours ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$M$ is NOT the midpoint of $AB$. Note that the angle $angle MCB$ is equal to $90^circ-60^circ=30^circ$, therefore
$$|MB|=|BC|tan(30^circ)=frac1sqrt3.$$
Moreover $|ED|=|MB|$ (why?).
Can you take it from here?
answered 4 hours ago
Robert Z
87.5k1056127
$M$ is NOT the midpoint of $AB$. Note that the angle $angle MCB$ is equal to $90^circ-60^circ=30^circ$, therefore
$$|MB|=|BC|tan(30^circ)=frac1sqrt3.$$
Moreover $|ED|=|MB|$ (why?).
Can you take it from here?
answered 4 hours ago
Robert Z
87.5k1056127
edited 4 hours ago
answered 4 hours ago
Robert Z
87.5k1056127
answered 4 hours ago
Robert Z
87.5k1056127
answered 4 hours ago
Robert Z
87.5k1056127
87.5k1056127
Wish I'd refreshed before posting my answer, I may have given too much away!
– Tartaglia's Stutter
4 hours ago
add a comment |Â
Wish I'd refreshed before posting my answer, I may have given too much away!
– Tartaglia's Stutter
4 hours ago
Wish I'd refreshed before posting my answer, I may have given too much away!
– Tartaglia's Stutter
4 hours ago
Wish I'd refreshed before posting my answer, I may have given too much away!
– Tartaglia's Stutter
4 hours ago
add a comment |Â
up vote
1
down vote
Note that the angle AED is $fracpi3$ radians (or 60 degrees if you prefer). Since the side opposite that angle is $1$, and $tan(fracpi3)$ is $sqrt3$, we know that the side ED must be of length $fracsqrt33$. The triangle MBC is similar to EAD, so the side MB is also $fracsqrt33$. You can use Pythagorean's Theorem and the fact that each side of the square is 1 to find the lengths of all the remaining sides needed.
answered 4 hours ago
Tartaglia's Stutter
944
New contributor
Tartaglia's Stutter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Nice answer. BTW I like your username.
– Robert Z
4 hours ago
add a comment |Â
up vote
1
down vote
Note that the angle AED is $fracpi3$ radians (or 60 degrees if you prefer). Since the side opposite that angle is $1$, and $tan(fracpi3)$ is $sqrt3$, we know that the side ED must be of length $fracsqrt33$. The triangle MBC is similar to EAD, so the side MB is also $fracsqrt33$. You can use Pythagorean's Theorem and the fact that each side of the square is 1 to find the lengths of all the remaining sides needed.
answered 4 hours ago
Tartaglia's Stutter
944
New contributor
Tartaglia's Stutter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Nice answer. BTW I like your username.
– Robert Z
4 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Note that the angle AED is $fracpi3$ radians (or 60 degrees if you prefer). Since the side opposite that angle is $1$, and $tan(fracpi3)$ is $sqrt3$, we know that the side ED must be of length $fracsqrt33$. The triangle MBC is similar to EAD, so the side MB is also $fracsqrt33$. You can use Pythagorean's Theorem and the fact that each side of the square is 1 to find the lengths of all the remaining sides needed.
answered 4 hours ago
Tartaglia's Stutter
944
New contributor
Tartaglia's Stutter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Note that the angle AED is $fracpi3$ radians (or 60 degrees if you prefer). Since the side opposite that angle is $1$, and $tan(fracpi3)$ is $sqrt3$, we know that the side ED must be of length $fracsqrt33$. The triangle MBC is similar to EAD, so the side MB is also $fracsqrt33$. You can use Pythagorean's Theorem and the fact that each side of the square is 1 to find the lengths of all the remaining sides needed.
answered 4 hours ago
Tartaglia's Stutter
944
New contributor
Tartaglia's Stutter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 4 hours ago
Tartaglia's Stutter
944
New contributor
Tartaglia's Stutter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 4 hours ago
Tartaglia's Stutter
944
answered 4 hours ago
Tartaglia's Stutter
944
944
New contributor
Tartaglia's Stutter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Tartaglia's Stutter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Tartaglia's Stutter is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Nice answer. BTW I like your username.
– Robert Z
4 hours ago
add a comment |Â
Nice answer. BTW I like your username.
– Robert Z
4 hours ago
Nice answer. BTW I like your username.
– Robert Z
4 hours ago
Nice answer. BTW I like your username.
– Robert Z
4 hours ago
add a comment |Â
up vote
1
down vote
Given a horizontal segment with length $L$ and two lines
$$
begincases
y_1 = x tan(fracpi3)\
y_2 = L-xtan(fracpi2)
endcases
$$
their intersection is at
$$
y_1=y_2Rightarrow x = fracLtan(fracpi3)+tan(fracpi2)
$$
hence the equilateral triangle has perimeter $3L$ and the square has perimeter $4 x = frac4Ltan(fracpi3)+tan(fracpi2) = frac4Lsqrt 3+1$
answered 5 mins ago
Cesareo
7,1072415
add a comment |Â
up vote
1
down vote
Given a horizontal segment with length $L$ and two lines
$$
begincases
y_1 = x tan(fracpi3)\
y_2 = L-xtan(fracpi2)
endcases
$$
their intersection is at
$$
y_1=y_2Rightarrow x = fracLtan(fracpi3)+tan(fracpi2)
$$
hence the equilateral triangle has perimeter $3L$ and the square has perimeter $4 x = frac4Ltan(fracpi3)+tan(fracpi2) = frac4Lsqrt 3+1$
answered 5 mins ago
Cesareo
7,1072415
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Given a horizontal segment with length $L$ and two lines
$$
begincases
y_1 = x tan(fracpi3)\
y_2 = L-xtan(fracpi2)
endcases
$$
their intersection is at
$$
y_1=y_2Rightarrow x = fracLtan(fracpi3)+tan(fracpi2)
$$
hence the equilateral triangle has perimeter $3L$ and the square has perimeter $4 x = frac4Ltan(fracpi3)+tan(fracpi2) = frac4Lsqrt 3+1$
answered 5 mins ago
Cesareo
7,1072415
Given a horizontal segment with length $L$ and two lines
$$
begincases
y_1 = x tan(fracpi3)\
y_2 = L-xtan(fracpi2)
endcases
$$
their intersection is at
$$
y_1=y_2Rightarrow x = fracLtan(fracpi3)+tan(fracpi2)
$$
hence the equilateral triangle has perimeter $3L$ and the square has perimeter $4 x = frac4Ltan(fracpi3)+tan(fracpi2) = frac4Lsqrt 3+1$
answered 5 mins ago
Cesareo
7,1072415
answered 5 mins ago
Cesareo
7,1072415
answered 5 mins ago
Cesareo
7,1072415
answered 5 mins ago
Cesareo
7,1072415
7,1072415
add a comment |Â
add a comment |Â
up vote
0
down vote
Referring to the figure, let $x$ and $h$ be half-side and height of equilateral triangle, resp.:
$hspace3cm$
For the equilateral triangle:
$$h^2+x^2=(2x)^2 Rightarrow h^2=3x^2 Rightarrow h=xsqrt3.$$
From the similarity of triangles:
$$frach4=fracx2x-4 Rightarrow fracxsqrt34=fracx2(x-2) Rightarrow x=frac2sqrt33+2 Rightarrow P=6x=4sqrt3+12.$$
Addendum: It was stated the perimeter of the square is $4$, not the side. So, the answer must be divided by $4$ to get $3+sqrt3$.
answered 4 hours ago
farruhota
16.8k2735
add a comment |Â
up vote
0
down vote
Referring to the figure, let $x$ and $h$ be half-side and height of equilateral triangle, resp.:
$hspace3cm$
For the equilateral triangle:
$$h^2+x^2=(2x)^2 Rightarrow h^2=3x^2 Rightarrow h=xsqrt3.$$
From the similarity of triangles:
$$frach4=fracx2x-4 Rightarrow fracxsqrt34=fracx2(x-2) Rightarrow x=frac2sqrt33+2 Rightarrow P=6x=4sqrt3+12.$$
Addendum: It was stated the perimeter of the square is $4$, not the side. So, the answer must be divided by $4$ to get $3+sqrt3$.
answered 4 hours ago
farruhota
16.8k2735
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Referring to the figure, let $x$ and $h$ be half-side and height of equilateral triangle, resp.:
$hspace3cm$
For the equilateral triangle:
$$h^2+x^2=(2x)^2 Rightarrow h^2=3x^2 Rightarrow h=xsqrt3.$$
From the similarity of triangles:
$$frach4=fracx2x-4 Rightarrow fracxsqrt34=fracx2(x-2) Rightarrow x=frac2sqrt33+2 Rightarrow P=6x=4sqrt3+12.$$
Addendum: It was stated the perimeter of the square is $4$, not the side. So, the answer must be divided by $4$ to get $3+sqrt3$.
answered 4 hours ago
farruhota
16.8k2735
Referring to the figure, let $x$ and $h$ be half-side and height of equilateral triangle, resp.:
$hspace3cm$
For the equilateral triangle:
$$h^2+x^2=(2x)^2 Rightarrow h^2=3x^2 Rightarrow h=xsqrt3.$$
From the similarity of triangles:
$$frach4=fracx2x-4 Rightarrow fracxsqrt34=fracx2(x-2) Rightarrow x=frac2sqrt33+2 Rightarrow P=6x=4sqrt3+12.$$
Addendum: It was stated the perimeter of the square is $4$, not the side. So, the answer must be divided by $4$ to get $3+sqrt3$.
answered 4 hours ago
farruhota
16.8k2735
edited 4 hours ago
answered 4 hours ago
farruhota
16.8k2735
answered 4 hours ago
farruhota
16.8k2735
answered 4 hours ago
farruhota
16.8k2735
16.8k2735
add a comment |Â
add a comment |Â
up vote
0
down vote
We only need to know that the $angle ABC=30°$, the rest is just straight forward.
answered 1 hour ago
Seyed
6,11531222
add a comment |Â
up vote
0
down vote
We only need to know that the $angle ABC=30°$, the rest is just straight forward.
answered 1 hour ago
Seyed
6,11531222
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We only need to know that the $angle ABC=30°$, the rest is just straight forward.
answered 1 hour ago
Seyed
6,11531222
We only need to know that the $angle ABC=30°$, the rest is just straight forward.
answered 1 hour ago
Seyed
6,11531222
answered 1 hour ago
Seyed
6,11531222
answered 1 hour ago
Seyed
6,11531222
answered 1 hour ago
Seyed
6,11531222
6,11531222
add a comment |Â
add a comment |Â
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