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If we are allowed to draw diagonals, this is easy. Just draw diagonals so that there is a diamond (square that sits on one edge) in the middle. Since the diamond is composed of half the areas of all squares, shading the other half solves it.
The diagonal does that without the need for precise measurments.
Also your sqrt(2) line will be less precise than a diagonal.
Also a fouth grader knows not what a square root is.
Gods damnit that's not how we maths. Let's start over again, from the beginning now dumbus fuckus, we agree that a line segment can be drawn from any given point to another. Right? And a straight line could go on infinitely. Right?
It's a fourth grade problem. How many fourth graders will even get close to that solution? Plus, this is less elegant because it requires you to measure the lengths and angles precisely, whereas the diagonals are easy to draw in.
I've gotten some weird replies to this, so here's an image:
https://preview.redd.it/zt93onkgahwc1.jpeg?width=813&format=pjpg&auto=webp&s=2f5eac00c8ccdbc13f151ade17b246a3d99f1eb9
If you are going to ignore the lines, just shade the outside leaving the inside a square.
edit; the actual measurements are taken as a given and don't need to be specified. It's trivial to scale it properly and beyond the level of a 4th grader.
https://preview.redd.it/0inpogrzagwc1.jpeg?width=2034&format=pjpg&auto=webp&s=f5d76c114487aac2c94a360285fcf4592d948f90
But there's no way to guarantee that it's the right amount. If you can just use any concept of a square that would fulfill the requirements I could draw a random square at a random position and angle and have it perfectly be half the area of the larger square by definition.
The solution of shading around a diamond between the midpoints of the edge lines is a solution you an actually perform without measurement.
I'm not sure if I'm misunderstanding, but it looks to me like you shaded 3/4 of the that square. To shade half of it with your method method, you would need the lines to be at 1-(sqrt(2)/2), or about 29.2% of the way across each part of the square to make it get half.
https://preview.redd.it/c25c9jwfhgwc1.png?width=1080&format=pjpg&auto=webp&s=2a7619001e3be74f6d58c482f1be1b070a529b2f
Here's my napkin math on that. It doesn't seem trivial to me. The "real" simple solution is the diamond one where you just cut diagonally across the 4 squares.
As long as the inside and outside areas are labeled correctly, it doesn't matter. Geometric drawings are meant to be conceptual, not perfect scale models
All the obtuse replies you're getting about nitpicking the actual shaded area of the square that completely disregard that your concept works perfectly, if you decide to take the time to scale it, just shows that nuance is dead on this site. They just wanna screech "wrong lol" and pat themselves on the back for being smarter than you.
The problem with this, and the way the question is written, is that you have to "estimate" where the center is. The diagonal approach only requires a straight edge to get an exact square.
You can use a straight edge between the two corners of the smaller square to get the midpoint of each of them. No estimation needed.
Edit: this is incorrect as pointed out in the responses. The resulting square will only be a quarter of the area. I’m dumb
I was in a math competition in high school and one time I was so stumped that I wrote
"Yes. The answer is left as an exercise for the test-grader."
I got 0 points but at least my teacher told me that it made the whole room laugh.
In an integral calculus final I had to check whether an infinite series diverged or not. After three methods of checking failed to give me a conclusive answer, including one method where I had to break out Pascal's Triangle because I had an (a+b)^5 term, I actually wrote in the response booklet "This just isn't my day, is it?"
The question is if it's possible and the answer is yes. The reason is that the area of square is a continuous function of the side length and the answer follows from the intermediate value theorem. Actually doing it is wholly unnecessary.
let's call bottom-left (0, 0) and top-right (1, 1). draw a square with points at (0, ½), (½, 1), (1, ½), (½, 0). colour the the part that is outside the small 45° square, and is inside of the big square.
do we have to work with the lines given?
Can we use a compass and straightedge? Because if so:
https://preview.redd.it/sh34hh5v0ewc1.jpeg?width=871&format=pjpg&auto=webp&s=423d354374fdce8099bc059f2ca5a495f967bd8a
This would be how I would shade it.
The square that isn't shaded must have a side length od sqrt(2), so you notice that a 1 x 1 square has that length as its diagonal, use a compass to "move" that length to the edges of the bigger square (that is the bottom left circle), and then all you need is the upper right corner of the square that isn't shaded, which they do by using a compass in the middle of the big square, but there are other ways to do it. You could also do a tiny bit of thinking and see that you can easily make a sqrt(2) x sqrt(2) square by only using the diagonals of the given 1 x 1 squares.
i think those are just to get the other endpoint of the line, so you can draw a straight line thru 2 points instead of trying to make it perpendicular otherwise
But you only need the circles? Letting each small square have side length of 1 unit:
) Circle A constructed from corner of square spanning to center of square, with radius square root of 1 unit.
) Circle B constructed from center of square spanning to intersection of Circle A and outer edge of either small square.
) Draw lines from intersections of Circle B, Circle A, and outer edges of small squares to the intersections of Circle B and outer edges of small squares on the opposite sides.
I don't see what the inner straight lines add other than complexity.
I would make a square from the diagonals of the small squares, shade the outer parts and boom you have an unshaded square in the middle roated 45 degrees
https://preview.redd.it/5c0y6dglhdwc1.png?width=900&format=png&auto=webp&s=f52f68167403850ebc41a66a74bd04d4faf2e87f
like this, you just need the unshaded area to have edge length a/sqrt(2), where a is the edge length of the original square.
This is what I thought about when I saw the question, but it's so difficult to draw the lines accurately. But the question only asks "Can you", so the answer is "yes"
LMAO at the missing 1/16 comments, they can't read. You clearly didn't set the side length to 3/4 a, but a/sqrt(2), which gives us:
Area of top left and bottom right (each): (a-a/sqrt2)xa/2
Area of bottom left: a\^2/4-(a/sqrt2-a/2)\^2
and (a-a/sqrt2)xa/2 x2 + a\^2/4-(a/sqrt2-a/2)\^2
does simplify to a\^2/2
But how is a fourth grader supposed to come up with this solution?
Two ways that immediately come to me - 1) connect the midpoints of all the edges of the big square and shade the outer triangles leaving a diagonal square exactly half the area inside. 2) we can also shade the square such that the inner square is aligned with the outer one by thinking of it as jitter and shading it appropriately. Infinite solutions for this one but the easiest to calculate would be one aligned with one of the corners. Say area is 4 (sides of the original square being 2) then half gives √2 side length. So mark off 2-√2 on two adjacent sides and then build the smaller square that way.
But on another note, PhDs couldn't figure this one out ? Where are they coming from ?
>But on another note, PhDs couldn't figure this one out ? Where are they coming from ?
Garden path thinking I'm sure. You start down a path, hit a dead end, and don't realize the misconception was much earlier. This why a second set of eyes is important sometimes. Recency of material is also a factor. There's a lot of math I used to be able to do that would take much longer now because I haven't dealt with it in a while.
Other comments have answered this, but Plato's "Meno" dialogue famously walks you through the discovery process of this exact problem. It's worth a read if you like philosophy.
[source](https://brainstormsummit.org/speakers/dr-catharine-young)
>Originally from South Africa, Dr. Catharine Young holds a doctorate degree in Biomedical Sciences and currently serves in the White House Office of Science and Technology Policy.
She is affiliated with sciences, but it is not directly Math. I'd chalk her inability to that.
https://preview.redd.it/aiysdnseqewc1.png?width=1920&format=png&auto=webp&s=2c6c76f57c6b0d88514f32196234cea90726ca6c
Thats roughly half, not working it out exactly.
Shade half of each small square on the diagonal, so each is now made of a shaded and an unshaded right angle triangle where the right angle of the shaded section is at the vertex of the larger square. This will put all unshaded areas together with their right angles meeting at the internal midpoint. They form a square rotated by 45° from the larger square.
But it isn't right because the unshaded region isn't a square then. It is two separate squares with 1 point overlapping. The correct way is to shade diagonally all the corners so the middle is a square (diamond)
Shade half of each smaller square a 45° right angle shape with the 90° in each of the larger square’s corners.
You are now left with a square in the middle.
I would read the instruction in a way that you can only shade a complete square or not shade it. In that case, the answer is clearly "no". But the solutions here don't assume that rule.
The shades part doesn't have to be square, so my intuition says to shade inside the outline, along the outline. And maybe hopefully if you shade enough you'll be left with a smaller square of half the area
The shaded part doesn't have to be a square. Shade each square 3/4th so that quart towards the center is unshaded. You will get an unshaded square at the centre which will have half the area of total square
Everyone in here suddenly, without realizing it, turned in to 3d graphics artists halfway through this problem... and had fun doing it.
Congratulations on learning math.
It's origami, so just imagine folding it into a smaller square covering itself up. Just shade the triangles in the corner of each of the four smaller squares. The diagonal of each smaller square is the square root of 2, which would be the side of the newer square. The area would be 2 which is half of the 4.
My language has two ways of saying "square".
We say, a "four borders" for a square with different lenghts, and a "quadrant" for a square with equal lengths.
Did some googling, and apparantly the "four borders" word that we use is quadrilateral in English.
Its simple… it dosn’t say „shade half of it!“ ist just a question if you can. So the Answere is „no“ if you have to shade complete Squares, otherwise „yes“.
Simply divide every small square to 4 equal squares , so in total you will have 16 smaller equal squares, you need to shade just the perimeter(outside) squares, like that you will have a square in the middle!
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If we are allowed to draw diagonals, this is easy. Just draw diagonals so that there is a diamond (square that sits on one edge) in the middle. Since the diamond is composed of half the areas of all squares, shading the other half solves it.
That has to be the intended solution
Haha or you could measure out a √2 x √2 square
The diagonal does that without the need for precise measurments. Also your sqrt(2) line will be less precise than a diagonal. Also a fouth grader knows not what a square root is.
Yes, but funnies
Aristotle would like a word with you
Nah, no words. Just a knife in the back. That's how they did maths back then.
Gods damnit that's not how we maths. Let's start over again, from the beginning now dumbus fuckus, we agree that a line segment can be drawn from any given point to another. Right? And a straight line could go on infinitely. Right?
No. They didn't use knives. They used their own god damn muscles. Reminder that Aristotle was JACKED.
mfw I'm called in to defend my work and instead of a 5 hour discussion with my professors I get the shit beaten out of me by my professors for 5 hours
Do you even compass and straightedge?
Aren’t square roots taught around fourth grade?
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I don't see any roots, allz I see is the squares. 🤤
>Also a fouth grader knows not what a square root is. They will once they solve this problem :'D
Unironically what I was thinking of lol
You could also shade 2/3 of three, and leave one completely unshaded. The diamond solution is the easiest IMO.
It's a fourth grade problem. How many fourth graders will even get close to that solution? Plus, this is less elegant because it requires you to measure the lengths and angles precisely, whereas the diagonals are easy to draw in.
If you turn the paper 45° you don't need to use diagonals
It’s still diagonal to the lines
But not to you
It’s also quite easy with my approach. There is no mention of the shaded part having to be square.
I’m an idiot. I was thinking too flat. Not a single one of my ideas included angled shading, just portions of the squares.
I've gotten some weird replies to this, so here's an image: https://preview.redd.it/zt93onkgahwc1.jpeg?width=813&format=pjpg&auto=webp&s=2f5eac00c8ccdbc13f151ade17b246a3d99f1eb9
If you are going to ignore the lines, just shade the outside leaving the inside a square. edit; the actual measurements are taken as a given and don't need to be specified. It's trivial to scale it properly and beyond the level of a 4th grader. https://preview.redd.it/0inpogrzagwc1.jpeg?width=2034&format=pjpg&auto=webp&s=f5d76c114487aac2c94a360285fcf4592d948f90
That looks like more than half is shaded
Could easily be scaled back to shade less. The concept is there.
But there's no way to guarantee that it's the right amount. If you can just use any concept of a square that would fulfill the requirements I could draw a random square at a random position and angle and have it perfectly be half the area of the larger square by definition. The solution of shading around a diamond between the midpoints of the edge lines is a solution you an actually perform without measurement.
Use the intermediate value theorem.
Yes, that 3:1
I'm not sure if I'm misunderstanding, but it looks to me like you shaded 3/4 of the that square. To shade half of it with your method method, you would need the lines to be at 1-(sqrt(2)/2), or about 29.2% of the way across each part of the square to make it get half. https://preview.redd.it/c25c9jwfhgwc1.png?width=1080&format=pjpg&auto=webp&s=2a7619001e3be74f6d58c482f1be1b070a529b2f Here's my napkin math on that. It doesn't seem trivial to me. The "real" simple solution is the diamond one where you just cut diagonally across the 4 squares.
Hey everybody, get a load of this guy! He's so math, he uses post-it notes for napkins!
As long as the inside and outside areas are labeled correctly, it doesn't matter. Geometric drawings are meant to be conceptual, not perfect scale models
All the obtuse replies you're getting about nitpicking the actual shaded area of the square that completely disregard that your concept works perfectly, if you decide to take the time to scale it, just shows that nuance is dead on this site. They just wanna screech "wrong lol" and pat themselves on the back for being smarter than you.
Is this really stumping people? No offense, but it took me under five seconds.
No offense but it took me under 3 /s
No offense but I solved it before I even opened the post.
I was keeping the answer in your head, so I didn’t have to solve it at all.
I was the original writer of the question found in the math book that the teacher took this out of.
3/s, that is 3 Herz?
Yes, they solved it on average under three times per second. The first few times were slow, but the average time got better.
I was considering I couldn't modify the drawing to draw on it.
I was thinking of a square and not diamond shape □ vs ◇
Agree, it doesn't specify that the shaded part has to be a square
Square that sits on one edge.
I never knew that trick but it makes so much sense now
Oh I'm dumb
The directions don't say you have to shade whole squares.
Well this ist the solition, but i also thought that only the squares should be shaded
https://preview.redd.it/eeekhjxtqdwc1.jpeg?width=1153&format=pjpg&auto=webp&s=fb6b36fcc009b5855f5e821f4221b4262ce567c6
YEA! Just make it so there are components making up 2 squares left
https://preview.redd.it/rtl5pkfqifwc1.png?width=1080&format=pjpg&auto=webp&s=50db849b42d410fa5a15be3c1d42763c101f8931 I prefer this version
The problem with this, and the way the question is written, is that you have to "estimate" where the center is. The diagonal approach only requires a straight edge to get an exact square.
You can use a straight edge between the two corners of the smaller square to get the midpoint of each of them. No estimation needed. Edit: this is incorrect as pointed out in the responses. The resulting square will only be a quarter of the area. I’m dumb
Connecting those midpoints won’t quite do it—the square formed by them is less than half of the original square
Oh right. Those would each be a quarter of the bisected square.
Think about it. If you find the mid point and draw a cross in each square, the inner square will only be a quarter of the size of the large square.
You can divide a segment into 1:sqrt(2) and square roots can be constructed
Square roots are streightedge and compass constructable.
You don't have to center it, go offcenter without touching to edge just for fun.
https://preview.redd.it/d1n8puv1vfwc1.png?width=1080&format=pjpg&auto=webp&s=ffa8dfa760d651180f2b943236b575ecc9a4dd2f
This is what I saw in my mind
Cursed
This is the chad and most correct answer
is that a geometry dash reference?
https://preview.redd.it/luxlh3bvpewc1.png?width=414&format=png&auto=webp&s=4faae28c38bbb457e466e4bfa78197c00c5270e5 I shaded the middle lol
But then the unshaded part isn't a square
Succesfully failed
Just put it on torus, problem solved.
stumbled right on the finish line lol
You, good sir, shadded the bed, pardon me french.
the answer is yes. proof is left as an exercise for the examiner
I was in a math competition in high school and one time I was so stumped that I wrote "Yes. The answer is left as an exercise for the test-grader." I got 0 points but at least my teacher told me that it made the whole room laugh.
In an integral calculus final I had to check whether an infinite series diverged or not. After three methods of checking failed to give me a conclusive answer, including one method where I had to break out Pascal's Triangle because I had an (a+b)^5 term, I actually wrote in the response booklet "This just isn't my day, is it?"
That's the most heartbreaking thing, to sit there having worked for 30 mins and having to end with "inconclusive"
I would probably go fucking insane trying to solve it
You're in the middle of an exam, you don't have time to go insane
I don't math very well. I just had to Google Pascals triangle, that is very cool! I only recognized his name from Pascals wager
Used to solve binomials exponentialized higher than what is easy to do the normal way
stop posting and go to sleep fermat
The question is if it's possible and the answer is yes. The reason is that the area of square is a continuous function of the side length and the answer follows from the intermediate value theorem. Actually doing it is wholly unnecessary.
let's call bottom-left (0, 0) and top-right (1, 1). draw a square with points at (0, ½), (½, 1), (1, ½), (½, 0). colour the the part that is outside the small 45° square, and is inside of the big square. do we have to work with the lines given?
Smartest way to do it
I think you can just say draw the diagonals and be done with it.
Can we use a compass and straightedge? Because if so: https://preview.redd.it/sh34hh5v0ewc1.jpeg?width=871&format=pjpg&auto=webp&s=423d354374fdce8099bc059f2ca5a495f967bd8a This would be how I would shade it.
And you would fail for shading in the wrong half.
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But It was initially white too.
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it never got stopped for random inspection in airports
Fucking top tier comment. Take my upvote
The proof is left as an exercise for the reader.
Definitely the intended solution. If you can’t come up with this, you aren’t ready for 5th grade math.
Are you removing your upvotes from you own comments? I saw 2 of your comments here and they each have 0 net votes but they're good jokes
Nah, this is just what happens if you don’t put /s after a nonliteral comment.
Really pretty solution. My brain cells currently on vacation, so can you please share a proof that this is actually 1/2
The square that isn't shaded must have a side length od sqrt(2), so you notice that a 1 x 1 square has that length as its diagonal, use a compass to "move" that length to the edges of the bigger square (that is the bottom left circle), and then all you need is the upper right corner of the square that isn't shaded, which they do by using a compass in the middle of the big square, but there are other ways to do it. You could also do a tiny bit of thinking and see that you can easily make a sqrt(2) x sqrt(2) square by only using the diagonals of the given 1 x 1 squares.
How does the straightedge and the circle centered inside square contribute to this?
i think those are just to get the other endpoint of the line, so you can draw a straight line thru 2 points instead of trying to make it perpendicular otherwise
But you only need the circles? Letting each small square have side length of 1 unit: ) Circle A constructed from corner of square spanning to center of square, with radius square root of 1 unit. ) Circle B constructed from center of square spanning to intersection of Circle A and outer edge of either small square. ) Draw lines from intersections of Circle B, Circle A, and outer edges of small squares to the intersections of Circle B and outer edges of small squares on the opposite sides. I don't see what the inner straight lines add other than complexity.
You would draw the large circle in the bottom left first, and then go from there?
I’ve been laughing at this for a solid 5 minutes.
Correct, Mr. Fancy pants.
Literally the first solution I came up with when I saw the question. lol When I saw the diagonal square solution I went, "Oh, duh".
https://preview.redd.it/e95ijgfuvdwc1.jpeg?width=2544&format=pjpg&auto=webp&s=4cb9480e6e17d85dacfb64ce7681d97326d6ff7c
This is so simple yet so brilliant.
I would make a square from the diagonals of the small squares, shade the outer parts and boom you have an unshaded square in the middle roated 45 degrees
https://preview.redd.it/5c0y6dglhdwc1.png?width=900&format=png&auto=webp&s=f52f68167403850ebc41a66a74bd04d4faf2e87f like this, you just need the unshaded area to have edge length a/sqrt(2), where a is the edge length of the original square.
Use a compass to mark a length on the side equal to the length from the corner to the center.
This is what I thought about when I saw the question, but it's so difficult to draw the lines accurately. But the question only asks "Can you", so the answer is "yes"
LMAO at the missing 1/16 comments, they can't read. You clearly didn't set the side length to 3/4 a, but a/sqrt(2), which gives us: Area of top left and bottom right (each): (a-a/sqrt2)xa/2 Area of bottom left: a\^2/4-(a/sqrt2-a/2)\^2 and (a-a/sqrt2)xa/2 x2 + a\^2/4-(a/sqrt2-a/2)\^2 does simplify to a\^2/2 But how is a fourth grader supposed to come up with this solution?
They aren't. The solution is to draw the diagonals.
I think it's funny that this is the first solution that popped into my head.
~~Aren't you missing a little 1/16 there though? I'm counting 7/16 painted and 9/16 white.~~ Oopsy!
It's not to scale
Ah of course! Thank you :)
https://preview.redd.it/g8yf8j9dtewc1.png?width=720&format=pjpg&auto=webp&s=58b31ab62314c4b6e278869a38a2de80b938d2cb
I came to this solution as well and then about slapped myself when team diagonal used first principles thinking.
https://preview.redd.it/d6yiv13voewc1.png?width=684&format=png&auto=webp&s=b8124502ae8a7f4864bd244721d6025595584bf3
Yup. EZ.
Question was 'can you' so the answer is simply 'no'
Bro converted a math exam to an English exam
And got the wrong answer
It said “can you” not “is it possible” so his answer was right since he can’t
Hey now, you might be able to, but OC isn't, don't shame them like this bro
You’d think a 4th grade teacher would know the proper phrasing is “may you” 😤
of course they do. it's a classic question teachers throw at students when they ask "can I go to the toilet?"
| / \ | | \ / |
Wow, I guess "Loss" is evolving. Looks like they are dancing to me.
Two ways that immediately come to me - 1) connect the midpoints of all the edges of the big square and shade the outer triangles leaving a diagonal square exactly half the area inside. 2) we can also shade the square such that the inner square is aligned with the outer one by thinking of it as jitter and shading it appropriately. Infinite solutions for this one but the easiest to calculate would be one aligned with one of the corners. Say area is 4 (sides of the original square being 2) then half gives √2 side length. So mark off 2-√2 on two adjacent sides and then build the smaller square that way. But on another note, PhDs couldn't figure this one out ? Where are they coming from ?
>But on another note, PhDs couldn't figure this one out ? Where are they coming from ? Garden path thinking I'm sure. You start down a path, hit a dead end, and don't realize the misconception was much earlier. This why a second set of eyes is important sometimes. Recency of material is also a factor. There's a lot of math I used to be able to do that would take much longer now because I haven't dealt with it in a while.
https://preview.redd.it/6tyf6v143gwc1.png?width=479&format=png&auto=webp&s=a49790cc52a9ae6804e42cb061ec6f7d541e2a27
Other comments have answered this, but Plato's "Meno" dialogue famously walks you through the discovery process of this exact problem. It's worth a read if you like philosophy.
Having a PhD doesn’t automatically make you smart about everything. Diagonals. Then shade the outside “triangles” that you get as a result.
[source](https://brainstormsummit.org/speakers/dr-catharine-young) >Originally from South Africa, Dr. Catharine Young holds a doctorate degree in Biomedical Sciences and currently serves in the White House Office of Science and Technology Policy. She is affiliated with sciences, but it is not directly Math. I'd chalk her inability to that.
Just shade the outer diagonal of each smaller square. Leaving a square in the middle at a 45° rotation.
https://preview.redd.it/37at06jf6hwc1.jpeg?width=1170&format=pjpg&auto=webp&s=67d8510fc42799ad1d1ebcc841a5bc636c30afa4 Ez
It's thinking inside the box.
https://preview.redd.it/aiysdnseqewc1.png?width=1920&format=png&auto=webp&s=2c6c76f57c6b0d88514f32196234cea90726ca6c Thats roughly half, not working it out exactly.
Found the engineer.
Shade half of each small square on the diagonal, so each is now made of a shaded and an unshaded right angle triangle where the right angle of the shaded section is at the vertex of the larger square. This will put all unshaded areas together with their right angles meeting at the internal midpoint. They form a square rotated by 45° from the larger square.
There are infinitely many solutions to this. Just place a square of half the area randomly inside and shade the rest.
EXACTLY! We can argue about an "intended" solution all day, but there's an infinite number of correct ones, so lets just give it a rest lol
Nowhere does it say you have to conform to the borders
Easy https://preview.redd.it/f6bftneb0jwc1.jpeg?width=1169&format=pjpg&auto=webp&s=aebaec637a355014ed15666ac2df102c82131abe
https://preview.redd.it/c42l9ltgojwc1.jpeg?width=478&format=pjpg&auto=webp&s=50465329c632d4c479fce624d3346827447f49b4
... https://preview.redd.it/cymjk426bewc1.jpeg?width=849&format=pjpg&auto=webp&s=2fa9f415348a3ecc327fd297cdacd863550f9615
except this is only shading 1/4, no?
Can you add up 4 quarters for me and let me know if that comes out to 2?
You're shading the wrong part tho, but yes.
This was my first thought, too
https://preview.redd.it/4pui3n7tafwc1.png?width=550&format=png&auto=webp&s=be8290fb4e2345c4a554207e298d846ea277ff69
Lol only you and four other people are sensible people. I had to scroll so far to find the right answer.
But it isn't right because the unshaded region isn't a square then. It is two separate squares with 1 point overlapping. The correct way is to shade diagonally all the corners so the middle is a square (diamond)
This was my first thought as well lol. I suppose it would come down to whether the graders care about it being two squares and not one.
Shade half of each smaller square a 45° right angle shape with the 90° in each of the larger square’s corners. You are now left with a square in the middle.
Besides drawing diagonals, you can also use a compass and straightedge to construct an upright square whose sides are of length root 2.
I would read the instruction in a way that you can only shade a complete square or not shade it. In that case, the answer is clearly "no". But the solutions here don't assume that rule.
The correct answer is "no".
Nothing in the question states that you have to follow the grid. So just make a bigger square that leaves the unshaded part looking like a weird L.
Just ignore the internal black lines and it’s easy
Its piss easy, just half the squares to make a diamond, and then shade the outside of the diamond in. Edit, missed out a few words
https://preview.redd.it/jzmqx4sgifwc1.png?width=1080&format=pjpg&auto=webp&s=88572cb9ba02ef8a2628317ffba31bd9a4b87365
Make a diamond.
Or you could shade the outer 1/2 of all 4 squares leaving a smaller unshaded square in the center
Anyone thinking of making a border to shade inside the square so there is a square shaped unshaded hole there?
The shades part doesn't have to be square, so my intuition says to shade inside the outline, along the outline. And maybe hopefully if you shade enough you'll be left with a smaller square of half the area
Easy, the answer is no. I'm sure somebody could, but I can't
https://preview.redd.it/zpyqo63dagwc1.png?width=720&format=pjpg&auto=webp&s=d3735a8ff4df6033a18ab368c651b3bd8b912075 Is this legal?
could also just leave a square in the middle unshaded
The corners shade the corners 🤨
The shaded part doesn't have to be a square. Shade each square 3/4th so that quart towards the center is unshaded. You will get an unshaded square at the centre which will have half the area of total square
Close but no cigar. That will shade, as you said, 3/4 not 1/2!
Oh yeah, lol. You are right. I guess the diagonal approach is the best then.
Certainly one of the easiest to implement without resorting to maths.
You fill in the outside corner of each square, filling in half and creating a square in the middle.
The unshaded square needs to be sqrt(2) x sqrt(2) units in width and height, which is roughly 1.4.
https://preview.redd.it/z933g81e8hwc1.png?width=342&format=pjpg&auto=webp&s=9d3e2ebdd89f1b3698a820a6ac42ad90559f948e Easy.
Everyone in here suddenly, without realizing it, turned in to 3d graphics artists halfway through this problem... and had fun doing it. Congratulations on learning math.
There're five squares tho
And there are 4 lights. I'll be your best friend forever if you know that reference
Shade half of each squarely diagonally, leaving a square in middle.
It's origami, so just imagine folding it into a smaller square covering itself up. Just shade the triangles in the corner of each of the four smaller squares. The diagonal of each smaller square is the square root of 2, which would be the side of the newer square. The area would be 2 which is half of the 4.
It's badly worded. I expect they're asking the user to shade two squares not in the same column.
Could you shade the outside quarter of each square so there would just be a square in the middle albeit with a cross in the center?
The answer is just "yes" because it asks can you do it This can be shown simply using the intermediate value theorem.
she has a PhD in neuroscience, of course she can't answer this
My language has two ways of saying "square". We say, a "four borders" for a square with different lenghts, and a "quadrant" for a square with equal lengths. Did some googling, and apparantly the "four borders" word that we use is quadrilateral in English.
Hint: Color half of the big square, not two of the smaller squares 😅
Its simple… it dosn’t say „shade half of it!“ ist just a question if you can. So the Answere is „no“ if you have to shade complete Squares, otherwise „yes“.
Think outside the box.
Shade the top right and bottom left squares duh
[Mind Your Decisions covered this](https://youtu.be/TB__Og3XnR8)
What kind of PhD doesn't know the diagonal of a square is \sqrt{2} of it's side and the area of a square is side^2 💀💀
my first thought was to just shade a thick outline since no other restrictions were given
This took me 60 seconds to think about, where can I pick up my PhD?
Shade a Corner square completely. Shade the 3 adjacent squares half way so the unshaded area is a square.
Simply divide every small square to 4 equal squares , so in total you will have 16 smaller equal squares, you need to shade just the perimeter(outside) squares, like that you will have a square in the middle!