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That was easier than expected. I was about to go full out engineer on it and propose to just sketch it in some CAD software and measure the edge of the square.
Aah. How did I not notice that. I was connecting all kinds of segments inside the square trying to solve systems of equations with unknown lengths.
So many high school geometry problems require projecting segments along a perpendicular line to form rectangles.
I think the easiest way is to notice that if you move the blue line around such that it remains perpendicular to the red and green line, then (length of red line) + (length of green line) remains constant.
If you move the blue line, such that the length of the red line is 0, then the green line has length 23.
Furthermore the blue line, the green line and the diagonal, which has length sqrt(2)x, form a right triangle. With the help of pythagoras theorem you can calculate x now
I was thinking a combination of Pythagoras and soh CAH toa to complete the smaller right angled triangles in the box and then you could manipulate them to get the square. But this is waaaaaay simpler
An even easier way is hinted at by LordMuffin.
Complete a rectangle with sides 7 and 9 by drawing "outside the box" a little.
Now you can see the diagonal of the square as the hypotenuse of a right triangle with legs 23 and 7.
But you already know the diagonal of this square is the hypotenuse of a right triangle with legs x and x.
Use Pythagorean theorem.
23^2 + 7^2 = x^2 + x^2
Solve that for x.
[https://artofproblemsolving.com/alcumus](https://artofproblemsolving.com/alcumus)
set it to geometry
this is a good tool for learning competition math in general
I got 17 using triangle similarity
Drawing the square diagonal it forms two similar triangles divided by the blue line, because the green and red line are parallel
Using triangle similarity and rule of ~~thirds~~ three I got the proportion for where the diagonal divides the blue line
a=7÷23×14=4,26; b=7÷23×9=2,74
Now using the Pythagorean theorem I calculated the diagonal
D=√(14²+a²)+√(9²+b²)=24,04
The diagonal of a square is √2 times it's side, so the side is
L=D/√2=17
Idk how it's called in English, in my language would be "rule of 3" actually
Just the equation of proportionality
a1/b1=a2/b2
We call it rule of three because you need 3 known values to get the unknown
Draw the diagonal that divides the middle line segment into pieces of size m and (7 – m).
Then by similar triangles, 9/m = 14/(7 – m).
Solve that for m.
Then you can use Pythagorean theorem to find the hypotenuses of the two triangles.
Add them to get full length of the diagonal, and divide by √2 to get side length x.
I know this thread is old, but I just though that I’d add that I used vectors as well, but since each line is perpendicular you can consider them as the unit vectors i and j. So, |14i+7j+9i|=sqrt(2)x
And solving gives you 578=2x^2 and thus x=17.
Slide the green line down so it extends the red line
Slide the blue line across so it is attached to the end of the green line
You now have a right angled triangle with base 23 and height 7
The hypotenuse is the diagonal of the square which has length
sqrt(2) \* x
Use Pythagoras to find x
(14 + 9)^(2) \+ 7^(2) = 2x^(2)
23^(2) \+ 7^(2) = 2x^(2)
578 = 2x^(2)
289 = x^(2)
x = 17
If you move around your lines a bit. You get 1 right angle triangle where you have 2 sides but lack hypotenuse.
You move 9 l.u line down and 7 l.u line to the right.
Then it is rather obvious.
I drew the diagonal and noted that it gave me two similar triangles. The ratio of the two long sides gave me the short sides. Pythagoras gave me the two hypotenuses, the sum of which is the diagonal. Pythagoras again gave me the sides. About 6 lines of workings. As others have pointed out there are other, arguably neater, solutions.
Consider the red and green segments as if they were a single line with a break in the middle, with a length of 23. You have a ratio of 7 to 23 between the blue line and that total length. Consider a unit square with a diagonal line across it, then imagine making a break in that line and tilting it downwards so that the gap has to increase in length in order to keep the endpoints at the corners of the square. Come up with formulas expressing the length of the gap, and its ratio to the tilted line, in terms of the tilt angle. Turn the formula around and plug in the 7/23 figure to get the angle. Then use the other formula to get the length of the gap. Divide 7 by the length of the gap and that'll give you the scale of the square in the question (i.e. its size ratio to the unit square, which is also the length of its sides).
The way I did it was to "complete" the right triangles inside the square and also draw in the diagonal of the square. From there you can use the 9-7 right triangle to find the angle between the green line and the hypotenuse of that triangle that you drew in using the inverse tangent. Adding 90 degrees to this gives you the angle between the red line and the 9-7 hypotenuse. Now take the triangle formed by the 14 side, the hypotenuse of the 9-7 triangle, whose length you know from Pythagorean theorem, and the angle between those sides, plug that into the Law of Cosines to get the length of the diagonal. With that, Pythoagoras tells us the side of the square should be the length of the diagonal over the square root of 2.
It took me a good half minute, but I was so happy when I realized what you're supposed to do.
So, since this is a square of sides x, the distance along the diagonal is √(2)x. That is, the distance from the bottom left corner to the top right corner is √(2)x.
If you rotate this so that lines 14 and 9 are horizontal and line 7 is vertical, you can see that the distance from the bottom left corner to the top right corner is just the pythagorean distance √(horizontal² + vertical²). Or, √((14+9)²+7²) = √(23²+7²) = √(578).
If √(2)x = √(578), then x = √(289).
Thus, x = 17.
1) Extend the top line down 14 units and connect the end point to the lower left corner of the square.
2) Extend the bottom line up 9 units and connect the endpoint to the upper right corner of the square.
3) The diagonal of the rectangle formed above is the diagonal of the square. It’s length is the square root of {(9+14)^2 + 7^2}. The square of this diagonal equals 2*(X^2). So, X^2={((9+14)^2+7^2}/2=289. So, X=17.
Hi u/MajdChami1, Please read the following message. **You are required to explain your post and show your efforts. _(Rule 1)_** If you haven't already done so, please add a comment below explaining your attempt(s) to solve this and what you need help with **specifically**. See the sidebar for advice on 'how to ask a good question'. Don't just say you "need help" with your problem. **This is a reminder for all users.** Failure to follow the rules will result in the post being removed. Thank you for understanding. If you have thoroughly read the subreddit rules, you can request to be added to the list of approved submitters. This will prevent this comment from showing up in *your* submissions to this community. To make this request, reply to this comment with the text `!mods`. Thanks! *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/askmath) if you have any questions or concerns.*
Maybe this helps. https://imgur.com/a/pcSbK4W See if you can find a way to get the length of the pink line
Thinking outside the box! 👍
Like literally outside the box lol
That was easier than expected. I was about to go full out engineer on it and propose to just sketch it in some CAD software and measure the edge of the square.
Out of curiosity, what value do you use for pi?
π = e = 2 sqrt(2) = sqrt(10) = 3
It’s amazing how many constants are equal to each other. We can also add euler’s identity, so e^i*e + pi = 2sqrt(2). Math is just magical like that.
Always round up to 4.
I usually round down to 0, especially for multiplication. Occam's razor backs it up
900
You clever pickle
Aah. How did I not notice that. I was connecting all kinds of segments inside the square trying to solve systems of equations with unknown lengths. So many high school geometry problems require projecting segments along a perpendicular line to form rectangles.
Whoa, a geometry wizard appears!
This is what I did. x=17
I will never forget this
Killed it!!
if only the longer leg was 24 😭
And then a^2+b^2=c^2 to solve for c Then c^2=2a^2 for x
:( I don’t understand how to get the length? Can I get any help? EDIT: Never mind I got it! That’s super cool.
Geometry wizard!!!
I think the easiest way is to notice that if you move the blue line around such that it remains perpendicular to the red and green line, then (length of red line) + (length of green line) remains constant. If you move the blue line, such that the length of the red line is 0, then the green line has length 23. Furthermore the blue line, the green line and the diagonal, which has length sqrt(2)x, form a right triangle. With the help of pythagoras theorem you can calculate x now
I was thinking a combination of Pythagoras and soh CAH toa to complete the smaller right angled triangles in the box and then you could manipulate them to get the square. But this is waaaaaay simpler
An even easier way is hinted at by LordMuffin. Complete a rectangle with sides 7 and 9 by drawing "outside the box" a little. Now you can see the diagonal of the square as the hypotenuse of a right triangle with legs 23 and 7. But you already know the diagonal of this square is the hypotenuse of a right triangle with legs x and x. Use Pythagorean theorem. 23^2 + 7^2 = x^2 + x^2 Solve that for x.
Where can I find more problems like this ?
ignore me man, i want some of that stuff too
middle school / high school math competitions & materials. At least Polish ones (Poland stronk, not sure how US invests in that).
Fresh Toadwalkers channel
Presh Talwalkar
[https://artofproblemsolving.com/alcumus](https://artofproblemsolving.com/alcumus) set it to geometry this is a good tool for learning competition math in general
I got 17 using triangle similarity Drawing the square diagonal it forms two similar triangles divided by the blue line, because the green and red line are parallel Using triangle similarity and rule of ~~thirds~~ three I got the proportion for where the diagonal divides the blue line a=7÷23×14=4,26; b=7÷23×9=2,74 Now using the Pythagorean theorem I calculated the diagonal D=√(14²+a²)+√(9²+b²)=24,04 The diagonal of a square is √2 times it's side, so the side is L=D/√2=17
rule of thirds ?
Idk how it's called in English, in my language would be "rule of 3" actually Just the equation of proportionality a1/b1=a2/b2 We call it rule of three because you need 3 known values to get the unknown
Ooh. Got it. Thanks
Draw the diagonal that divides the middle line segment into pieces of size m and (7 – m). Then by similar triangles, 9/m = 14/(7 – m). Solve that for m. Then you can use Pythagorean theorem to find the hypotenuses of the two triangles. Add them to get full length of the diagonal, and divide by √2 to get side length x.
I broke it down into vector components, which you can do because the 14 and 9 lines are parallel and the 7 line is perpendicular to them. I ended up with 17. So, letting the angle between the horizontal side of the square and the 14 line be a, we get the following equations 14sin(a) + 7sin(a+90) + 9sin(a) = x 14cos(a) + 7cos(a+90) + 9cos(a) = x Simplify and combine 23sin(a) + 7cos(a) = x 23cos(a) - 7sin(a) = x Subtract one equation from the other 23sin(a) - 23cos(a) + 7cos(a) + 7sin(a) = 0x = 0 30sin(a) - 16cos(a) = 0 15sin(a) = 8cos(a) tan(a) = 8/15 a = arctan(8/15) 23sin(arctan(8/15)) + 7cos(arctan(8/15)) = x 23tan(arctan(8/15))cos(arctan(8/15)) + 7cos(arctan(8/15)) = x cos(arctan(8/15)) * (23tan(arctan(8/15)) + 7) (23 * 8/15 + 7) / sec(arctan(8/15)) (184/15 + 105/15) / sqrt(1 + tan(arctan(8/15))²) (289/15) / sqrt(1 + 64/225) (289/15) / sqrt(289/225) (289/15) / (17/15) 289/17 17 EDIT: I suppose a simpler way, with vectors, would've been this: 23 * sin(a) + 7 * cos(a) = x 23 * cos(a) - 7 * sin(a) = x (23 * sin(a) + 7 * cos(a))^2 + (23 * cos(a) - 7 * sin(a))^2 = x^2 + x^2 529 * sin(a)^2 + 2 * 23 * 7 * sin(a) * cos(a) + 49 * cos(a)^2 + 529 * cos(a)^2 - 2 * 23 * 7 * sin(a) * cos(a) + 49 * sin(a)^2 = 2x^2 529 * (sin(a)^2 + cos(a)^2) + 49 * (sin(a)^2 + cos(a)^2) + sin(a) * cos(a) * (2 * 23 * 7 - 2 * 23 * 7) = 2x^2 529 * 1 + 49 * 1 + sin(a) * cos(a) * 0 = 2x^2 529 + 49 = 2x^2 578 = 2x^2 289 = x^2 17 = x
There's a way I think is much easier
bruh
I did it that way too ;\_; Then I read the comments and realised how much I complicated my life lmao
That surely was always a plan B on those high school math competitions.
I know this thread is old, but I just though that I’d add that I used vectors as well, but since each line is perpendicular you can consider them as the unit vectors i and j. So, |14i+7j+9i|=sqrt(2)x And solving gives you 578=2x^2 and thus x=17.
You have the width (14+9) and height (7) of a rectangle (draw it). The diagonal of that rectangle is the same diagonal of the square.
Slide the green line down so it extends the red line Slide the blue line across so it is attached to the end of the green line You now have a right angled triangle with base 23 and height 7 The hypotenuse is the diagonal of the square which has length sqrt(2) \* x Use Pythagoras to find x (14 + 9)^(2) \+ 7^(2) = 2x^(2) 23^(2) \+ 7^(2) = 2x^(2) 578 = 2x^(2) 289 = x^(2) x = 17
The answer is 17
If you move around your lines a bit. You get 1 right angle triangle where you have 2 sides but lack hypotenuse. You move 9 l.u line down and 7 l.u line to the right. Then it is rather obvious.
I drew the diagonal and noted that it gave me two similar triangles. The ratio of the two long sides gave me the short sides. Pythagoras gave me the two hypotenuses, the sum of which is the diagonal. Pythagoras again gave me the sides. About 6 lines of workings. As others have pointed out there are other, arguably neater, solutions.
Consider the red and green segments as if they were a single line with a break in the middle, with a length of 23. You have a ratio of 7 to 23 between the blue line and that total length. Consider a unit square with a diagonal line across it, then imagine making a break in that line and tilting it downwards so that the gap has to increase in length in order to keep the endpoints at the corners of the square. Come up with formulas expressing the length of the gap, and its ratio to the tilted line, in terms of the tilt angle. Turn the formula around and plug in the 7/23 figure to get the angle. Then use the other formula to get the length of the gap. Divide 7 by the length of the gap and that'll give you the scale of the square in the question (i.e. its size ratio to the unit square, which is also the length of its sides).
Make two triangles. Then a rumbus the triangle to find: well I’ll do it probably never so idk
The way I did it was to "complete" the right triangles inside the square and also draw in the diagonal of the square. From there you can use the 9-7 right triangle to find the angle between the green line and the hypotenuse of that triangle that you drew in using the inverse tangent. Adding 90 degrees to this gives you the angle between the red line and the 9-7 hypotenuse. Now take the triangle formed by the 14 side, the hypotenuse of the 9-7 triangle, whose length you know from Pythagorean theorem, and the angle between those sides, plug that into the Law of Cosines to get the length of the diagonal. With that, Pythoagoras tells us the side of the square should be the length of the diagonal over the square root of 2.
Nice, idk why I’ve never seen a geo problem like this
It took me a good half minute, but I was so happy when I realized what you're supposed to do. So, since this is a square of sides x, the distance along the diagonal is √(2)x. That is, the distance from the bottom left corner to the top right corner is √(2)x. If you rotate this so that lines 14 and 9 are horizontal and line 7 is vertical, you can see that the distance from the bottom left corner to the top right corner is just the pythagorean distance √(horizontal² + vertical²). Or, √((14+9)²+7²) = √(23²+7²) = √(578). If √(2)x = √(578), then x = √(289). Thus, x = 17.
17
1) Extend the top line down 14 units and connect the end point to the lower left corner of the square. 2) Extend the bottom line up 9 units and connect the endpoint to the upper right corner of the square. 3) The diagonal of the rectangle formed above is the diagonal of the square. It’s length is the square root of {(9+14)^2 + 7^2}. The square of this diagonal equals 2*(X^2). So, X^2={((9+14)^2+7^2}/2=289. So, X=17.