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Stuck on a Geometry Puzzle

Geometry question with rectangles purely out of curiosityMaximizing the perimeter of a triangle inside a squareEuclidean geometry and irrational numbers.Geometry and natural numbersGeometry: Circle inscribed in squareGeometry proof given diagramFinding the unknown area.Geometry High School OlympiadArea of a square inside a square created by connecting point-opposite midpointOff Centre Square Geometry Problem

9

3

$begingroup$

($ABCD$ is a square. $|BE|=|EC|, |AF|=3cm, |GC|=4cm$. Determine the length of $|FG|$.)

How can I approach this problem, preferably without trigonometry?

(Except proving otherwise, is there a way to know that a given problem cannot be solved without trigonometry?)

geometry euclidean-geometry puzzle

share|cite|improve this question

edited 1 hour ago

blackened

asked 4 hours ago

blackenedblackened

377211

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Well, this is harder than it looks !
  $endgroup$
  – Statistic Dean
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @StatisticDean yup, sitting here for like 25 minutes now and drawing the 50th picture
  $endgroup$
  – Vinyl_coat_jawa
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, I think it is quite easy with trigonometry : you can write the law of cosines in triangles $ECG$, $EFC$ and $EFG$. From this you get a relation between $x$ and the length $L$ of the side of your square. Then you have also $(7+x)^2 = 2L^2$, and the two relations together give you $x$. But without trigonometry, I have no idea...
  $endgroup$
  – TheSilverDoe
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One reflection proves that G is on AD. A second reflection proves that AF=AC/4.
  $endgroup$
  – amI
  41 mins ago
  

  
  
  
  

add a comment | 

9

3

$begingroup$

($ABCD$ is a square. $|BE|=|EC|, |AF|=3cm, |GC|=4cm$. Determine the length of $|FG|$.)

How can I approach this problem, preferably without trigonometry?

(Except proving otherwise, is there a way to know that a given problem cannot be solved without trigonometry?)

geometry euclidean-geometry puzzle

share|cite|improve this question

edited 1 hour ago

blackened

asked 4 hours ago

blackenedblackened

377211

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Well, this is harder than it looks !
  $endgroup$
  – Statistic Dean
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @StatisticDean yup, sitting here for like 25 minutes now and drawing the 50th picture
  $endgroup$
  – Vinyl_coat_jawa
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Actually, I think it is quite easy with trigonometry : you can write the law of cosines in triangles $ECG$, $EFC$ and $EFG$. From this you get a relation between $x$ and the length $L$ of the side of your square. Then you have also $(7+x)^2 = 2L^2$, and the two relations together give you $x$. But without trigonometry, I have no idea...
  $endgroup$
  – TheSilverDoe
  4 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  One reflection proves that G is on AD. A second reflection proves that AF=AC/4.
  $endgroup$
  – amI
  41 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

($ABCD$ is a square. $|BE|=|EC|, |AF|=3cm, |GC|=4cm$. Determine the length of $|FG|$.)

How can I approach this problem, preferably without trigonometry?

(Except proving otherwise, is there a way to know that a given problem cannot be solved without trigonometry?)

geometry euclidean-geometry puzzle

share|cite|improve this question

edited 1 hour ago

blackened

asked 4 hours ago

blackenedblackened

377211

$endgroup$

share|cite|improve this question

edited 1 hour ago

blackened

asked 4 hours ago

blackenedblackened

377211

share|cite|improve this question

edited 1 hour ago

blackened

asked 4 hours ago

blackenedblackened

377211

asked 4 hours ago

blackenedblackened

377211

asked 4 hours ago

blackenedblackened

377211

3 Answers
3

active

oldest

votes

10

$begingroup$

Let $H$ be the midpoint of $AC$ and $angle EIC= 90^{circ}$. We can observe that $$FH+3=HG+4,quad FH+HG=x.$$ So we obtain $HG=frac{x-1}2$. If we let $angle HEG=alpha$, it follows that $$angle EFG=180^{circ}-angle FHE-angle FEH=180^{circ}-135^{circ}-(45^{circ}-alpha)=alpha.$$ Since $angle EHG=45^{circ}$, we have that
$Delta EHG$ and $Delta FEG$ are similar to each other (they also share an angle $angle EGF$). This gives $$EG:FG=HG:EGimplies EG^2 = FGcdot HG=frac{x(x-1)}2.$$ Now, note that $EI=frac14 AC=frac{x+7}4$ and $IG=IH-GH=frac{x+7}{4}-frac{x-1}2=frac{9-x}{4}$. Since $Delta EIG$ is a right triangle, by Pythagorean theorem, we find that
$$
EG^2=frac{x(x-1)}{2}=EI^2+IG^2=frac{(x+7)^2}{16}+frac{(9-x)^2}{16},
$$ which implies $x=5 $ or $x=-frac{13}3$. Since $x>0$, we get $x=5$.

share|cite|improve this answer

edited 2 hours ago

answered 3 hours ago

SongSong

15k1636

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe a typo. In the first line, I guess you meant $angle FEG=alpha$.
  $endgroup$
  – OnoL
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @OnoL Oh, you're right. It is fixed now. Thank you for the correction!
  $endgroup$
  – Song
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user Ah, I didn't know that. I'll fix it soon ... Thank you for your comment.
  $endgroup$
  – Song
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Song How did you come up with this so quickly, as if you were waiting with the solution in hand?
  $endgroup$
  – blackened
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @blackened If I was quick, I think it is only because I was lucky to see this post earlier than most users. (Of course there is no way I was waiting for you to post this question :) At first, I tried drawing some lines and adding points to get some information, and then I was able to notice similiarity of triangles. It gave me a hint and I pursued how far I could go with it.
  $endgroup$
  – Song
  1 hour ago
  
  
  

  
  
  
  

 | 
show 3 more comments

0

$begingroup$

Ok after some calculus, I figured out to solve this problem. Let's draw the vertical V that goes through E, and call $alpha$ and $frac{pi}{4}-alpha$ the angles we get on the left and right of the 45° angle cut by V and call c the length of the side of the square. We can then do some trigonometry to get :

• 
  $c cdot tan(alpha) - (c-frac{3}{sqrt{2}}) = c cdot tan(alpha)$ (Continue the line EF to cut AD)


• 
  $ tan(frac{pi}{4}-alpha) = c cdot frac{sqrt{2}}{8}-1$ (G is $frac{4}{sqrt{2}}$ away vertically from the right side and $frac{c}{2}-frac{4}{sqrt{2}}$ away horizontally from the middle)

With a little trigo, we have 2 equations in $c$ and $tan(alpha)$ that we can solve, which allow us to solve the whole problem. This makes me think that there is no "simple" answer without trigo :(

share|cite|improve this answer

answered 3 hours ago

Statistic DeanStatistic Dean

1263

New contributor

Statistic Dean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

-2

$begingroup$

Solution without trigonometry:
Point G and F can be used to partition your figure into trapezoids by drawing a horizontal line through them. Then you can add up the areas of the figures formed and setting them equal to the area of the square which is side times side.

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

John AdmasJohn Admas

294

New contributor

John Admas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I still do not understand. Unless I am missing something, three areas you mention do not add up. Can you be more explicit. (And rotate the image?)
  $endgroup$
  – blackened
  2 hours ago
  

  
  
  
  

add a comment | 

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10

$begingroup$

Let $H$ be the midpoint of $AC$ and $angle EIC= 90^{circ}$. We can observe that $$FH+3=HG+4,quad FH+HG=x.$$ So we obtain $HG=frac{x-1}2$. If we let $angle HEG=alpha$, it follows that $$angle EFG=180^{circ}-angle FHE-angle FEH=180^{circ}-135^{circ}-(45^{circ}-alpha)=alpha.$$ Since $angle EHG=45^{circ}$, we have that
$Delta EHG$ and $Delta FEG$ are similar to each other (they also share an angle $angle EGF$). This gives $$EG:FG=HG:EGimplies EG^2 = FGcdot HG=frac{x(x-1)}2.$$ Now, note that $EI=frac14 AC=frac{x+7}4$ and $IG=IH-GH=frac{x+7}{4}-frac{x-1}2=frac{9-x}{4}$. Since $Delta EIG$ is a right triangle, by Pythagorean theorem, we find that
$$
EG^2=frac{x(x-1)}{2}=EI^2+IG^2=frac{(x+7)^2}{16}+frac{(9-x)^2}{16},
$$ which implies $x=5 $ or $x=-frac{13}3$. Since $x>0$, we get $x=5$.

share|cite|improve this answer

edited 2 hours ago

answered 3 hours ago

SongSong

15k1636

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe a typo. In the first line, I guess you meant $angle FEG=alpha$.
  $endgroup$
  – OnoL
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @OnoL Oh, you're right. It is fixed now. Thank you for the correction!
  $endgroup$
  – Song
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user Ah, I didn't know that. I'll fix it soon ... Thank you for your comment.
  $endgroup$
  – Song
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Song How did you come up with this so quickly, as if you were waiting with the solution in hand?
  $endgroup$
  – blackened
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @blackened If I was quick, I think it is only because I was lucky to see this post earlier than most users. (Of course there is no way I was waiting for you to post this question :) At first, I tried drawing some lines and adding points to get some information, and then I was able to notice similiarity of triangles. It gave me a hint and I pursued how far I could go with it.
  $endgroup$
  – Song
  1 hour ago
  
  
  

  
  
  
  

 | 
show 3 more comments

10

$begingroup$

Let $H$ be the midpoint of $AC$ and $angle EIC= 90^{circ}$. We can observe that $$FH+3=HG+4,quad FH+HG=x.$$ So we obtain $HG=frac{x-1}2$. If we let $angle HEG=alpha$, it follows that $$angle EFG=180^{circ}-angle FHE-angle FEH=180^{circ}-135^{circ}-(45^{circ}-alpha)=alpha.$$ Since $angle EHG=45^{circ}$, we have that
$Delta EHG$ and $Delta FEG$ are similar to each other (they also share an angle $angle EGF$). This gives $$EG:FG=HG:EGimplies EG^2 = FGcdot HG=frac{x(x-1)}2.$$ Now, note that $EI=frac14 AC=frac{x+7}4$ and $IG=IH-GH=frac{x+7}{4}-frac{x-1}2=frac{9-x}{4}$. Since $Delta EIG$ is a right triangle, by Pythagorean theorem, we find that
$$
EG^2=frac{x(x-1)}{2}=EI^2+IG^2=frac{(x+7)^2}{16}+frac{(9-x)^2}{16},
$$ which implies $x=5 $ or $x=-frac{13}3$. Since $x>0$, we get $x=5$.

share|cite|improve this answer

edited 2 hours ago

answered 3 hours ago

SongSong

15k1636

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Maybe a typo. In the first line, I guess you meant $angle FEG=alpha$.
  $endgroup$
  – OnoL
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @OnoL Oh, you're right. It is fixed now. Thank you for the correction!
  $endgroup$
  – Song
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @user Ah, I didn't know that. I'll fix it soon ... Thank you for your comment.
  $endgroup$
  – Song
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Song How did you come up with this so quickly, as if you were waiting with the solution in hand?
  $endgroup$
  – blackened
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @blackened If I was quick, I think it is only because I was lucky to see this post earlier than most users. (Of course there is no way I was waiting for you to post this question :) At first, I tried drawing some lines and adding points to get some information, and then I was able to notice similiarity of triangles. It gave me a hint and I pursued how far I could go with it.
  $endgroup$
  – Song
  1 hour ago
  
  
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

Let $H$ be the midpoint of $AC$ and $angle EIC= 90^{circ}$. We can observe that $$FH+3=HG+4,quad FH+HG=x.$$ So we obtain $HG=frac{x-1}2$. If we let $angle HEG=alpha$, it follows that $$angle EFG=180^{circ}-angle FHE-angle FEH=180^{circ}-135^{circ}-(45^{circ}-alpha)=alpha.$$ Since $angle EHG=45^{circ}$, we have that
$Delta EHG$ and $Delta FEG$ are similar to each other (they also share an angle $angle EGF$). This gives $$EG:FG=HG:EGimplies EG^2 = FGcdot HG=frac{x(x-1)}2.$$ Now, note that $EI=frac14 AC=frac{x+7}4$ and $IG=IH-GH=frac{x+7}{4}-frac{x-1}2=frac{9-x}{4}$. Since $Delta EIG$ is a right triangle, by Pythagorean theorem, we find that
$$
EG^2=frac{x(x-1)}{2}=EI^2+IG^2=frac{(x+7)^2}{16}+frac{(9-x)^2}{16},
$$ which implies $x=5 $ or $x=-frac{13}3$. Since $x>0$, we get $x=5$.

share|cite|improve this answer

edited 2 hours ago

answered 3 hours ago

SongSong

15k1636

$endgroup$

share|cite|improve this answer

edited 2 hours ago

answered 3 hours ago

SongSong

15k1636

answered 3 hours ago

SongSong

15k1636

answered 3 hours ago

SongSong

15k1636

0

$begingroup$

Ok after some calculus, I figured out to solve this problem. Let's draw the vertical V that goes through E, and call $alpha$ and $frac{pi}{4}-alpha$ the angles we get on the left and right of the 45° angle cut by V and call c the length of the side of the square. We can then do some trigonometry to get :

• 
  $c cdot tan(alpha) - (c-frac{3}{sqrt{2}}) = c cdot tan(alpha)$ (Continue the line EF to cut AD)


• 
  $ tan(frac{pi}{4}-alpha) = c cdot frac{sqrt{2}}{8}-1$ (G is $frac{4}{sqrt{2}}$ away vertically from the right side and $frac{c}{2}-frac{4}{sqrt{2}}$ away horizontally from the middle)

With a little trigo, we have 2 equations in $c$ and $tan(alpha)$ that we can solve, which allow us to solve the whole problem. This makes me think that there is no "simple" answer without trigo :(

share|cite|improve this answer

answered 3 hours ago

Statistic DeanStatistic Dean

1263

New contributor

Statistic Dean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

0

$begingroup$

Ok after some calculus, I figured out to solve this problem. Let's draw the vertical V that goes through E, and call $alpha$ and $frac{pi}{4}-alpha$ the angles we get on the left and right of the 45° angle cut by V and call c the length of the side of the square. We can then do some trigonometry to get :

• 
  $c cdot tan(alpha) - (c-frac{3}{sqrt{2}}) = c cdot tan(alpha)$ (Continue the line EF to cut AD)


• 
  $ tan(frac{pi}{4}-alpha) = c cdot frac{sqrt{2}}{8}-1$ (G is $frac{4}{sqrt{2}}$ away vertically from the right side and $frac{c}{2}-frac{4}{sqrt{2}}$ away horizontally from the middle)

With a little trigo, we have 2 equations in $c$ and $tan(alpha)$ that we can solve, which allow us to solve the whole problem. This makes me think that there is no "simple" answer without trigo :(

share|cite|improve this answer

answered 3 hours ago

Statistic DeanStatistic Dean

1263

New contributor

Statistic Dean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

add a comment | 

$begingroup$

Ok after some calculus, I figured out to solve this problem. Let's draw the vertical V that goes through E, and call $alpha$ and $frac{pi}{4}-alpha$ the angles we get on the left and right of the 45° angle cut by V and call c the length of the side of the square. We can then do some trigonometry to get :

• 
  $c cdot tan(alpha) - (c-frac{3}{sqrt{2}}) = c cdot tan(alpha)$ (Continue the line EF to cut AD)


• 
  $ tan(frac{pi}{4}-alpha) = c cdot frac{sqrt{2}}{8}-1$ (G is $frac{4}{sqrt{2}}$ away vertically from the right side and $frac{c}{2}-frac{4}{sqrt{2}}$ away horizontally from the middle)

With a little trigo, we have 2 equations in $c$ and $tan(alpha)$ that we can solve, which allow us to solve the whole problem. This makes me think that there is no "simple" answer without trigo :(

share|cite|improve this answer

answered 3 hours ago

Statistic DeanStatistic Dean

1263

New contributor

Statistic Dean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Statistic DeanStatistic Dean

1263

New contributor

Statistic Dean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 3 hours ago

Statistic DeanStatistic Dean

1263

New contributor

Statistic Dean is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 3 hours ago

Statistic DeanStatistic Dean

1263

-2

$begingroup$

Solution without trigonometry:
Point G and F can be used to partition your figure into trapezoids by drawing a horizontal line through them. Then you can add up the areas of the figures formed and setting them equal to the area of the square which is side times side.

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

John AdmasJohn Admas

294

New contributor

John Admas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I still do not understand. Unless I am missing something, three areas you mention do not add up. Can you be more explicit. (And rotate the image?)
  $endgroup$
  – blackened
  2 hours ago
  

  
  
  
  

add a comment | 

-2

$begingroup$

Solution without trigonometry:
Point G and F can be used to partition your figure into trapezoids by drawing a horizontal line through them. Then you can add up the areas of the figures formed and setting them equal to the area of the square which is side times side.

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

John AdmasJohn Admas

294

New contributor

John Admas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  I still do not understand. Unless I am missing something, three areas you mention do not add up. Can you be more explicit. (And rotate the image?)
  $endgroup$
  – blackened
  2 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Solution without trigonometry:
Point G and F can be used to partition your figure into trapezoids by drawing a horizontal line through them. Then you can add up the areas of the figures formed and setting them equal to the area of the square which is side times side.

share|cite|improve this answer

edited 3 hours ago

answered 3 hours ago

John AdmasJohn Admas

294

New contributor

John Admas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

$endgroup$

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edited 3 hours ago

answered 3 hours ago

John AdmasJohn Admas

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answered 3 hours ago

John AdmasJohn Admas

294

New contributor

John Admas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

answered 3 hours ago

John AdmasJohn Admas

294