Find the side of the square.Right Triangles and AltitudesFind all positive integer solutions $(x,y,z)$ that satisfy $5^x cdot 7^y +4= 3^z$?Find the sum of the lengths of line segments $BD$ and $CE$To find two sides of a triangle when it is circumscribed a circlea square inscribed right triangleGiven medians of a right triangle, find the length of one side of the triangleHeron's formula when side lengths include radicalsHow to find a minimal path without using a differential equation?Finding the area of an isosceles triangle inscribed in a squareGiven two angles, exact location of $E$ and $F$ in a square $ABCD$ and diagonal $BD$ intersects $AE$ at $P$. What is the value of $angle PFC$?
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Find the side of the square.
Right Triangles and AltitudesFind all positive integer solutions $(x,y,z)$ that satisfy $5^x cdot 7^y +4= 3^z$?Find the sum of the lengths of line segments $BD$ and $CE$To find two sides of a triangle when it is circumscribed a circlea square inscribed right triangleGiven medians of a right triangle, find the length of one side of the triangleHeron's formula when side lengths include radicalsHow to find a minimal path without using a differential equation?Finding the area of an isosceles triangle inscribed in a squareGiven two angles, exact location of $E$ and $F$ in a square $ABCD$ and diagonal $BD$ intersects $AE$ at $P$. What is the value of $angle PFC$?
$begingroup$
The problem I am proposing to solve was posed in a math contest for students of 17-18 years old, this month.
With the data in the picture, find the side of the square.
I did find the solution but it involves a long way across calculating the combined length of the red segments (not drawn in the original problem) and minimizing it for points on the segment of length $1$. With the diagonal we can get the side. I think there is a better and quicker way to solve it. I bring it here, maybe somebody could find this solution.
real-analysis geometry contest-math
asked Mar 22 at 9:47
Rafa BudríaRafa Budría
5,9721825
$endgroup$
add a comment |
$begingroup$
The problem I am proposing to solve was posed in a math contest for students of 17-18 years old, this month.
With the data in the picture, find the side of the square.
I did find the solution but it involves a long way across calculating the combined length of the red segments (not drawn in the original problem) and minimizing it for points on the segment of length $1$. With the diagonal we can get the side. I think there is a better and quicker way to solve it. I bring it here, maybe somebody could find this solution.
real-analysis geometry contest-math
asked Mar 22 at 9:47
Rafa BudríaRafa Budría
5,9721825
$endgroup$
1
$begingroup$
Does the red "line" mean anything?
$endgroup$
– Arthur
Mar 22 at 9:57
$begingroup$
Not sure the labels are clear. does the "$1$" refer to one leg of the small right triangle (which would then have hypotenuse $sqrt 5$) or does the $1$ refer to the longer segment which connects the two right angled vertices of your two triangles? If the latter, then what does the red "line" signify?
$endgroup$
– lulu
Mar 22 at 9:57
$begingroup$
@Arthur. The red line means nothing ("not drawn in the original problem") but my try to solve it, as mentioned in the post.
$endgroup$
– Rafa Budría
Mar 22 at 14:29
add a comment |
$begingroup$
The problem I am proposing to solve was posed in a math contest for students of 17-18 years old, this month.
With the data in the picture, find the side of the square.
I did find the solution but it involves a long way across calculating the combined length of the red segments (not drawn in the original problem) and minimizing it for points on the segment of length $1$. With the diagonal we can get the side. I think there is a better and quicker way to solve it. I bring it here, maybe somebody could find this solution.
real-analysis geometry contest-math
asked Mar 22 at 9:47
Rafa BudríaRafa Budría
5,9721825
$endgroup$
The problem I am proposing to solve was posed in a math contest for students of 17-18 years old, this month.
With the data in the picture, find the side of the square.
I did find the solution but it involves a long way across calculating the combined length of the red segments (not drawn in the original problem) and minimizing it for points on the segment of length $1$. With the diagonal we can get the side. I think there is a better and quicker way to solve it. I bring it here, maybe somebody could find this solution.
real-analysis geometry contest-math
real-analysis geometry contest-math
asked Mar 22 at 9:47
Rafa BudríaRafa Budría
5,9721825
asked Mar 22 at 9:47
Rafa BudríaRafa Budría
5,9721825
asked Mar 22 at 9:47
Rafa BudríaRafa Budría
5,9721825
asked Mar 22 at 9:47
Rafa BudríaRafa Budría
5,9721825
asked Mar 22 at 9:47
Rafa BudríaRafa Budría
5,9721825
5,9721825
1
$begingroup$
Does the red "line" mean anything?
$endgroup$
– Arthur
Mar 22 at 9:57
$begingroup$
Not sure the labels are clear. does the "$1$" refer to one leg of the small right triangle (which would then have hypotenuse $sqrt 5$) or does the $1$ refer to the longer segment which connects the two right angled vertices of your two triangles? If the latter, then what does the red "line" signify?
$endgroup$
– lulu
Mar 22 at 9:57
$begingroup$
@Arthur. The red line means nothing ("not drawn in the original problem") but my try to solve it, as mentioned in the post.
$endgroup$
– Rafa Budría
Mar 22 at 14:29
add a comment |
1
$begingroup$
Does the red "line" mean anything?
$endgroup$
– Arthur
Mar 22 at 9:57
$begingroup$
Not sure the labels are clear. does the "$1$" refer to one leg of the small right triangle (which would then have hypotenuse $sqrt 5$) or does the $1$ refer to the longer segment which connects the two right angled vertices of your two triangles? If the latter, then what does the red "line" signify?
$endgroup$
– lulu
Mar 22 at 9:57
$begingroup$
@Arthur. The red line means nothing ("not drawn in the original problem") but my try to solve it, as mentioned in the post.
$endgroup$
– Rafa Budría
Mar 22 at 14:29
1
1
$begingroup$
Does the red "line" mean anything?
$endgroup$
– Arthur
Mar 22 at 9:57
$begingroup$
Does the red "line" mean anything?
$endgroup$
– Arthur
Mar 22 at 9:57
$begingroup$
Not sure the labels are clear. does the "$1$" refer to one leg of the small right triangle (which would then have hypotenuse $sqrt 5$) or does the $1$ refer to the longer segment which connects the two right angled vertices of your two triangles? If the latter, then what does the red "line" signify?
$endgroup$
– lulu
Mar 22 at 9:57
$begingroup$
Not sure the labels are clear. does the "$1$" refer to one leg of the small right triangle (which would then have hypotenuse $sqrt 5$) or does the $1$ refer to the longer segment which connects the two right angled vertices of your two triangles? If the latter, then what does the red "line" signify?
$endgroup$
– lulu
Mar 22 at 9:57
$begingroup$
@Arthur. The red line means nothing ("not drawn in the original problem") but my try to solve it, as mentioned in the post.
$endgroup$
– Rafa Budría
Mar 22 at 14:29
$begingroup$
@Arthur. The red line means nothing ("not drawn in the original problem") but my try to solve it, as mentioned in the post.
$endgroup$
– Rafa Budría
Mar 22 at 14:29
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
It is just a simple pythagoras theorem. You find the diagonal of the square $=sqrt 50$
Therefore the side of the square is $=5$
edited Mar 22 at 11:03
Oscar Lanzi
13.6k12136
answered Mar 22 at 10:07
SeyedSeyed
7,17641526
$endgroup$
$begingroup$
Do you mean diagonal of the square instead of diameter of the square @Seyed?
$endgroup$
– Dbchatto67
Mar 22 at 10:28
$begingroup$
@Dbchatto67, You are right, I fixed it. Thank you
$endgroup$
– Seyed
Mar 22 at 10:39
$begingroup$
When rendering a square root, place braces around the argument to get the radical sign to cover all characters. Thus sqrt50 properly gives $sqrt50$.
$endgroup$
– Oscar Lanzi
Mar 22 at 11:05
add a comment |
$begingroup$
Assuming the red line means nothing, and $1$ is the length of the gray segment between the $2$ segment and the $5$ segment:
Turn the square (almost) $45^circ$ around, so that the $5$ segment becomes horizontal, and place the corner of the square where the $5$ segment starts at the origin. The opposite corner of the square will be at $(7, 1)$.
What is the length of the diagonal of the square? Then what is its side length?
answered Mar 22 at 10:01
ArthurArthur
122k7122211
$endgroup$
add a comment |
$begingroup$
Rotate the square so that the $5$ and $2$ segments are horizontal:
Label the left most corner as $(0,0)$. Now hopefully you can see that the right most corner is at $(7,1)$.
From this, we can calculate the distance from the left most to the right most as
beginalignd^2&=7^2+1^2\
&=49+1\
&=50\
d&=sqrt50endalign
Now we can calculate the side length of the square as we know its diagonal length
beginaligns^2+s^2&=d^2\
2s^2&=sqrt50^2\
2s^2&=50\
s^2&=25\
s&=sqrt25\
s&=5endalign
answered Mar 22 at 10:17
lioness99alioness99a
3,9012727
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is just a simple pythagoras theorem. You find the diagonal of the square $=sqrt 50$
Therefore the side of the square is $=5$
edited Mar 22 at 11:03
Oscar Lanzi
13.6k12136
answered Mar 22 at 10:07
SeyedSeyed
7,17641526
$endgroup$
$begingroup$
Do you mean diagonal of the square instead of diameter of the square @Seyed?
$endgroup$
– Dbchatto67
Mar 22 at 10:28
$begingroup$
@Dbchatto67, You are right, I fixed it. Thank you
$endgroup$
– Seyed
Mar 22 at 10:39
$begingroup$
When rendering a square root, place braces around the argument to get the radical sign to cover all characters. Thus sqrt50 properly gives $sqrt50$.
$endgroup$
– Oscar Lanzi
Mar 22 at 11:05
add a comment |
$begingroup$
It is just a simple pythagoras theorem. You find the diagonal of the square $=sqrt 50$
Therefore the side of the square is $=5$
edited Mar 22 at 11:03
Oscar Lanzi
13.6k12136
answered Mar 22 at 10:07
SeyedSeyed
7,17641526
$endgroup$
$begingroup$
Do you mean diagonal of the square instead of diameter of the square @Seyed?
$endgroup$
– Dbchatto67
Mar 22 at 10:28
$begingroup$
@Dbchatto67, You are right, I fixed it. Thank you
$endgroup$
– Seyed
Mar 22 at 10:39
$begingroup$
When rendering a square root, place braces around the argument to get the radical sign to cover all characters. Thus sqrt50 properly gives $sqrt50$.
$endgroup$
– Oscar Lanzi
Mar 22 at 11:05
add a comment |
$begingroup$
It is just a simple pythagoras theorem. You find the diagonal of the square $=sqrt 50$
Therefore the side of the square is $=5$
edited Mar 22 at 11:03
Oscar Lanzi
13.6k12136
answered Mar 22 at 10:07
SeyedSeyed
7,17641526
$endgroup$
It is just a simple pythagoras theorem. You find the diagonal of the square $=sqrt 50$
Therefore the side of the square is $=5$
edited Mar 22 at 11:03
Oscar Lanzi
13.6k12136
answered Mar 22 at 10:07
SeyedSeyed
7,17641526
edited Mar 22 at 11:03
Oscar Lanzi
13.6k12136
edited Mar 22 at 11:03
Oscar Lanzi
13.6k12136
edited Mar 22 at 11:03
Oscar Lanzi
13.6k12136
13.6k12136
answered Mar 22 at 10:07
SeyedSeyed
7,17641526
answered Mar 22 at 10:07
SeyedSeyed
7,17641526
answered Mar 22 at 10:07
SeyedSeyed
7,17641526
7,17641526
$begingroup$
Do you mean diagonal of the square instead of diameter of the square @Seyed?
$endgroup$
– Dbchatto67
Mar 22 at 10:28
$begingroup$
@Dbchatto67, You are right, I fixed it. Thank you
$endgroup$
– Seyed
Mar 22 at 10:39
$begingroup$
When rendering a square root, place braces around the argument to get the radical sign to cover all characters. Thus sqrt50 properly gives $sqrt50$.
$endgroup$
– Oscar Lanzi
Mar 22 at 11:05
add a comment |
$begingroup$
Do you mean diagonal of the square instead of diameter of the square @Seyed?
$endgroup$
– Dbchatto67
Mar 22 at 10:28
$begingroup$
@Dbchatto67, You are right, I fixed it. Thank you
$endgroup$
– Seyed
Mar 22 at 10:39
$begingroup$
When rendering a square root, place braces around the argument to get the radical sign to cover all characters. Thus sqrt50 properly gives $sqrt50$.
$endgroup$
– Oscar Lanzi
Mar 22 at 11:05
$begingroup$
Do you mean diagonal of the square instead of diameter of the square @Seyed?
$endgroup$
– Dbchatto67
Mar 22 at 10:28
$begingroup$
Do you mean diagonal of the square instead of diameter of the square @Seyed?
$endgroup$
– Dbchatto67
Mar 22 at 10:28
$begingroup$
@Dbchatto67, You are right, I fixed it. Thank you
$endgroup$
– Seyed
Mar 22 at 10:39
$begingroup$
@Dbchatto67, You are right, I fixed it. Thank you
$endgroup$
– Seyed
Mar 22 at 10:39
$begingroup$
When rendering a square root, place braces around the argument to get the radical sign to cover all characters. Thus sqrt50 properly gives $sqrt50$.
$endgroup$
– Oscar Lanzi
Mar 22 at 11:05
$begingroup$
When rendering a square root, place braces around the argument to get the radical sign to cover all characters. Thus sqrt50 properly gives $sqrt50$.
$endgroup$
– Oscar Lanzi
Mar 22 at 11:05
add a comment |
$begingroup$
Assuming the red line means nothing, and $1$ is the length of the gray segment between the $2$ segment and the $5$ segment:
Turn the square (almost) $45^circ$ around, so that the $5$ segment becomes horizontal, and place the corner of the square where the $5$ segment starts at the origin. The opposite corner of the square will be at $(7, 1)$.
What is the length of the diagonal of the square? Then what is its side length?
answered Mar 22 at 10:01
ArthurArthur
122k7122211
$endgroup$
add a comment |
$begingroup$
Assuming the red line means nothing, and $1$ is the length of the gray segment between the $2$ segment and the $5$ segment:
Turn the square (almost) $45^circ$ around, so that the $5$ segment becomes horizontal, and place the corner of the square where the $5$ segment starts at the origin. The opposite corner of the square will be at $(7, 1)$.
What is the length of the diagonal of the square? Then what is its side length?
answered Mar 22 at 10:01
ArthurArthur
122k7122211
$endgroup$
add a comment |
$begingroup$
Assuming the red line means nothing, and $1$ is the length of the gray segment between the $2$ segment and the $5$ segment:
Turn the square (almost) $45^circ$ around, so that the $5$ segment becomes horizontal, and place the corner of the square where the $5$ segment starts at the origin. The opposite corner of the square will be at $(7, 1)$.
What is the length of the diagonal of the square? Then what is its side length?
answered Mar 22 at 10:01
ArthurArthur
122k7122211
$endgroup$
Assuming the red line means nothing, and $1$ is the length of the gray segment between the $2$ segment and the $5$ segment:
Turn the square (almost) $45^circ$ around, so that the $5$ segment becomes horizontal, and place the corner of the square where the $5$ segment starts at the origin. The opposite corner of the square will be at $(7, 1)$.
What is the length of the diagonal of the square? Then what is its side length?
answered Mar 22 at 10:01
ArthurArthur
122k7122211
answered Mar 22 at 10:01
ArthurArthur
122k7122211
answered Mar 22 at 10:01
ArthurArthur
122k7122211
answered Mar 22 at 10:01
ArthurArthur
122k7122211
122k7122211
add a comment |
add a comment |
$begingroup$
Rotate the square so that the $5$ and $2$ segments are horizontal:
Label the left most corner as $(0,0)$. Now hopefully you can see that the right most corner is at $(7,1)$.
From this, we can calculate the distance from the left most to the right most as
beginalignd^2&=7^2+1^2\
&=49+1\
&=50\
d&=sqrt50endalign
Now we can calculate the side length of the square as we know its diagonal length
beginaligns^2+s^2&=d^2\
2s^2&=sqrt50^2\
2s^2&=50\
s^2&=25\
s&=sqrt25\
s&=5endalign
answered Mar 22 at 10:17
lioness99alioness99a
3,9012727
$endgroup$
add a comment |
$begingroup$
Rotate the square so that the $5$ and $2$ segments are horizontal:
Label the left most corner as $(0,0)$. Now hopefully you can see that the right most corner is at $(7,1)$.
From this, we can calculate the distance from the left most to the right most as
beginalignd^2&=7^2+1^2\
&=49+1\
&=50\
d&=sqrt50endalign
Now we can calculate the side length of the square as we know its diagonal length
beginaligns^2+s^2&=d^2\
2s^2&=sqrt50^2\
2s^2&=50\
s^2&=25\
s&=sqrt25\
s&=5endalign
answered Mar 22 at 10:17
lioness99alioness99a
3,9012727
$endgroup$
add a comment |
$begingroup$
Rotate the square so that the $5$ and $2$ segments are horizontal:
Label the left most corner as $(0,0)$. Now hopefully you can see that the right most corner is at $(7,1)$.
From this, we can calculate the distance from the left most to the right most as
beginalignd^2&=7^2+1^2\
&=49+1\
&=50\
d&=sqrt50endalign
Now we can calculate the side length of the square as we know its diagonal length
beginaligns^2+s^2&=d^2\
2s^2&=sqrt50^2\
2s^2&=50\
s^2&=25\
s&=sqrt25\
s&=5endalign
answered Mar 22 at 10:17
lioness99alioness99a
3,9012727
$endgroup$
Rotate the square so that the $5$ and $2$ segments are horizontal:
Label the left most corner as $(0,0)$. Now hopefully you can see that the right most corner is at $(7,1)$.
From this, we can calculate the distance from the left most to the right most as
beginalignd^2&=7^2+1^2\
&=49+1\
&=50\
d&=sqrt50endalign
Now we can calculate the side length of the square as we know its diagonal length
beginaligns^2+s^2&=d^2\
2s^2&=sqrt50^2\
2s^2&=50\
s^2&=25\
s&=sqrt25\
s&=5endalign
answered Mar 22 at 10:17
lioness99alioness99a
3,9012727
answered Mar 22 at 10:17
lioness99alioness99a
3,9012727
answered Mar 22 at 10:17
lioness99alioness99a
3,9012727
answered Mar 22 at 10:17
lioness99alioness99a
3,9012727
3,9012727
add a comment |
add a comment |
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1
$begingroup$
Does the red "line" mean anything?
$endgroup$
– Arthur
Mar 22 at 9:57
$begingroup$
Not sure the labels are clear. does the "$1$" refer to one leg of the small right triangle (which would then have hypotenuse $sqrt 5$) or does the $1$ refer to the longer segment which connects the two right angled vertices of your two triangles? If the latter, then what does the red "line" signify?
$endgroup$
– lulu
Mar 22 at 9:57
$begingroup$
@Arthur. The red line means nothing ("not drawn in the original problem") but my try to solve it, as mentioned in the post.
$endgroup$
– Rafa Budría
Mar 22 at 14:29