I love how elegant the second one is. (Though the practicality of the third being contained within the original triangle is the most practical.) Thank you Dr Lin.
I don't think the third is that practical, it relies too much on the ability to draw perfect squares when the others take already existing lines to base on the construction of those. Not trying to argue, just saying.
The third one isn't practical at all. Have you ever actually constructed polygons using the straight edge and compass method? A modification of the first method is the most practical, and doesn't even need a ruled edge. The second method's scale becomes too large to be practical, and the third requires way too many construction lines to be practical. What's shown here are very approximated polygons that only use a straight edge, but when you consider accurate angles, you also need a compass to do this.
In the first one, you have to draw two rectangles and a whole graph with perfect orthogonal axis. In the third one, you just have to draw one square, of course it's the most practical.
@@adrianoaraujo1692 All you need to do it in the first is already provided by the triangle and trace a 45° angle line. The guy is just trying to make a clean example drawing to the side. In the final, you have to construct your own squares from zero, which completely mocks on the objective of the exercise in the first place. That is more like a way to make sure that the squares are well constructed. If you already had a way to construct perfect squares, why would you need to use any of these techniques to put them into the tringle?
I love this channel because it covers math ideas rather than making you solve equations on my math lab and getting the whole equation wrong because you forgot to include the negative sign somewhere
Try a game called Euclidea in Google store, it has tons of puzzles similar to this, just pure geometry puzzles, it starts pretty easy but ends up with super complicated stuff.
There is a way to do the first way with compass and straight edge. Here is how 1) label the traingle ABC with A on top. 2) extend BC through C to D such that BC=CD 3) draw perpendicular to AD through C. 4) draw a line parallel to BD through A and mark its intersection with the perpendicular through C as E. 5) draw segment ED 6) draw a line parallel to AB through C and mark its intersection with ED as G. 7) draw a line parallel to BD through G mark its intersections with AB and AC as H and I respectively. 8) drop perpendiculars through H and I to BC. You now have created the square inside the triangle
@@stuartryerse you are correct, the problem was I typed this up on my cell phone from memory of the construction I had done earlier but unfortunately got the point names messed up. I have fixed it
If you wanna find the side length of the square that’s inside the triangle you can use If angle is less or equal to 45 a = base(tan x)/2+(tan x) If angle is greater than 45 a = base(tan x)/1+ (tan x) Base being the entire base of the triangle X being one of the two congruent angles of the triangle And y being the side length of the square
Then you might like the app "Euclidia" (definitely available for Android, probably for iPhone as well, but not sure about that). It gives you tons of different tasks that can be constructed with only these two items, making it kind of a game where you have to look for the solution requiring the fewest steps (there are even bonus points for additional, symmetrical solutions in some tasks...). There are a lot of solution videos here on youtube as well, for every single task, in case one gets stuck.
@@Shadow81989 You're right, I love euclidia, though I haven't played it in forever. Definitely worth going back to now that I've forgotten all the solutions!
Me: Studying hard for my Calculus exams. RUclips recommended: 3 Ways to Draw Squares Inside Triangles - Numberphile. Me: Exams never were that important anyways.
How are you going to an exam? Aren't exams postponed because of Covid-19? I know online education is a thing but I can't imagine they would let anyone do their exam online as it is easy to cheat. Maybe exams are still going just with precautions but wouldn't think so
I really encourage these kind of construction videos. Simple enough to get kids hooked on maths. Had good fun for a few hours trying it out. The third method was too good. We need more Dr Lin. Very infectious enthusiasm.
This is one of the challenges in the Euclidea app. My approach was to skew the original triangle until it's a right-angled triangle, and bisect the right angle, to find the correct height. It's similar to approach 1 but keeping everything drawn in-place.
This is why I always loved geometry in a school. Most of the time everything you need to solve the problem is right in front of you; you just need some cleverness to connect all the dots and make deductions to each step to reach the desired end. Plus it's visual, which I was always better at. You can see how everything interacts.
4th Way: raise the perpendiculars of the base to form a square then draw lines from the top corners of that square down to the center of the base. Didn't actually try yet, just thought of it right after seeing the second method. Edit: As john osullivan has pointed out, this only works for isosceles triangles. I originally thought this was a requirement, but it's not, and the other 3 ways work on ANY triangle that has acute angles on both sides of its base, not just isosceles triangles.
This is actually really useful for me- I'm doing an artists book of geometric constructions and I'm doing the Magnum Opus symbol (square within a triangle within a circle). Hitherto I've been constructing the shape by starting with the square and the constructing an equilateral triangle from its top edge, but that doesn't give me control over the size of the final circle (important when laying out a book). Using the third method I should be able to construct the symbol in reverse order- big circle, then triangle, then square. Thanks Dr Lin! More geometry videos please!
I never used to like geometry. Even when I had a really good geometry teacher, I didn't like it. I had finally started to understand algebra, and then they put me in geometry. I have a much better appreciation for, and even love of, geometry now. Partially because of that teacher, partially because of physics (you really need all kinds of math in order to do the problems on physics tests), and partially because of videos like this. One of the really neat things about geometry in general, is so much of it can be done without any numbers, kind of like what Calvin Lin does in this video. I will definitely be showing some of my students these methods (I tutor part-time).
Aaah. I truly feel the fun of childhood, when I was figuring out by myself how to build cubes and other 3D objects out of paper. Playful geometry is the best.
You can also do the first method contained almost in the same footprint as the original triangle: Construct a triangle t' with the same base as the original triangle t, but with the top vertex shifted horizontally to the left such that the bottom left corner is 90°. Then construct the angle bisector of the bottom left corner and the intersection with the right edge of t' gives you the height of the square.
Great suggestion! I'd use that the next time. I originally kept the approaches distinct, but decided to combine it as we started filming. The glee at the end was pure joy.
You can indeed use method 1 with a compass and straight edge. construct a perpendicular line to the base in the bottom left point of the triangle. mark the length of the top left edge of the triangle on the base. mark the length of the base on our perpendicular line. connect these two marks. construct a 45° angle from the base and our perpendicular line. the cross section of the 45° line and the connection between the marks is the height we want. Then it is just dropping perpendiculars, marking intersections and drawing lines.
You could also do something based off the first method, the graph, using only a straight edge and compass, which also involves mirroring the 'graph.' Draw the triangle you want to draw a square in. Then draw horizontals extending from the middle vertex, and verticals extending from the outer two vertices, this will create two right triangles that form a circumscribed rectangle around the original triangle. Then connect opposite vertices of the rectangle to create diagonals, this will find two vertices of the square along the two non-horizontal edges of the triangle, both points of which are parallel to the triangle's horizontal line. Then drop verticals from these vertices and draw the horizontal line between them. Anyone familiar with drawing regular polygons with the straight edge and compass method will probably realize this is the simplest method.The second method in the video, the similar triangle method, gets way too large to be simplistic. While the third method involves so many construction lines just to find one vertex of the final square, which is way too much work when you're using the straight edge and compass method when you care about angles being accurate. These other methods are still just as valid, but work better as proofs, rather than being realistic uses. Hell, even my method can be optimized. Only do half the rectangle, find only one vertex of the square, a compass will give you two more vertices from that point, and you can draw the rest of the square from that.
You can link the first and last methods by drawing a square which has the triangle height as a side and adjacent to the base. Then you draw the line which connects a vertex of the triangle to the opposite vertex of the square and the job is done.
Passed over a really simple solution, if you draw a perpendicular from the left vertex to the height of the triangle, the you can connect the new point with the right vertex, and the line should cross the triangle at the height of the square. Same method as the first example using the ruler measurement, but only a straight edge and compass required.
@@marijnbliekendaal I don't get why not. The line from the right side is finding the width of the square that connects with the right side of the original triangle (2 units added to the width is one unit away from the center) Unless the left side needs to be a 45° angle, forcing a square from the right side, then you would extend to the point on the side of the original triangle. Still can be done without a ruler.
The way I though of initially is different: since the square is “in the middle” of the triangle, the midpoint of the bottom edge of the square will always be the midpoint of the bottom edge of the triangle. So, the measure of the angle from a vertex of the triangle to the midpoint to the vertex of the square corresponding to the corner on the side of the triangle whose ends are the initial vertex and the top vertex is always 60 degrees, since it corresponds to a right triangle with legs in 1:2 ratio. Hence, simply construct a 60 degree angle from the midpoint of the base of the triangle, then drop the perpendicular from the point of intersection. Edit: “midpoint of base of triangle” -> “point of intersection of base with perpendicular (with respect to base) through top vertex”
Unfortunately, the midpoint of the bottom edge of the square need not coincide with the midpoint of the bottom edge of the triangle.Maybe you are assuming that the triangle is isosceles?
The graph at 2:55 is labeled wrong, this had me confused for a while. With those labels the blue line doesn't make any sense. The x-axis represents rectangle height, while the y-axis represents height for the blue line and width for the green line.
I'm not sure quite what the graph does show, but the x-axis isn't always height - the x-intercept of the green line comes at a value equal to the maximum width of the rectangle, which is more than the maximum height.
I also struggled for a bit with the graph. Here's how I convinced myself that it is possible to find the square by intersecting the two lines: The green line is a set of points, where each point represents a possible rectangle that can be placed inside the triangle as described in the video. The x coordinate represents the rectangle's width and the y coordinate represents the rectangle's height. Now, we want a square, aka a rectangle where the width is equal to the height. If we plot a line representing all possible squares (where the x coordinate is the width and the y coordinate is the height) we get the blue line from the video. It then stands to reason that the intersection of the green and blue lines will give the dimensions of the square that can be placed inside the triangle (satisfying the criteria to lie both on the green and blue lines). I'm not sure if that's the way Dr. Lin intended the explanation to go, but it worked for me!
@@lukecoughlan5035 splendid explanation, made it click for me as well. But I do have one correction I would like to make, and it is that the green and blue lines also represent rectangles and squares "outside" the triangle, but they don't exceed the height of it. Thought I would put it out there, because I got hung on that for a little while.
he could totally have done that measuring construction within the original triangle with just a ruler and compass. draw a perpendicular to the base through one of its ends and a paralell to the base through the top vertex. connect the intersection of these two with the other end of the base. the intersection of this conection with one of the sides of the triangle is at the point where the square touches the side of the triangle.
Nice video, I like the third way best. It is worth saying that this video doesn't explain everything, only the concept. For example unless I missed it, it doesn't explain how to draw a perpendicular line from a point. It's basic geometry, but still probably many viewers don't know. Compass is used much more than what is showed in the video when you are doing it "for real".
Yeah I don’t think we can show and discuss compass straight edge construction EVERY TIME. This video shows a lot more, from memory. m.ruclips.net/video/6Lm9EHhbJAY/видео.html
The 1st method could have been done without the ruler: 1. Raise a perpendicular from the left end of the triangle base. 2. Draw a line parallel to the base, thru the apex. 3. Bisect the angle of the base and the perpendicular from 1. 4. Draw a line from the intersection of the perpendicular from 1 & the parallel from 2, down to the right end of the base. 5. The intersection of 3 & 4 is the height of the square (just as in his measured diagram). Draw a line parallel to the base thru that point, and where it intersects the left and right edges of the triangle are the upper corners of the square. 6. Drop perpendiculars from the corners from 5 down to the base. Square completed.
If there's one thing I learnt during this video it's that the Singaporean accent is very similar to the Belter dialect in The Expanse! XD Great video Dr Lin.
Very clever use of geometry. I can use this in my woodworking! But there is one point I see that makes it easier... a 45 degree triangle, to run the line from one point on the triangle to begin with...
I have a question: If you can construct something (with a ruler and compass) can you also construct it with an arbitrally small amount of ink? (not including the final thing!) For using this as an example it would be: can you make the square in the triangle so that the ink needed to figure out the sides of the square (without actually drawing them) would require -> 0 ink.
You can draw one rectangle too wide and one too tall, then use their intersection point as distances from the top corners to find a new point on the triangle's edge to draw rectangles from, giving you two rectangles that are slightly more square. Iterate until you have a perfect square. There may even be a mathematical expression for this method that leads to a way to calculate square roots!
Павел Менжинский - or, use a construction to trisect one of the sides of the triangle (easy using simple projection techniques). The square will touch the triangle two thirds of the way up, every time.
What is the relationship between the angle made by the line from the vertex through the corner of the squares and the vertex it came from? Does it bisect the angle, or is it a cosine or something? Can this property be used to draw a square with a compass and straight edge? I'm hoping he talks about it in the extras! Keep up the great work.
I think something might be wrong with the first method. Could you verify? I think the idea is to plot the height and width with respect to a single parameter t so that the intersection point (t,h(t))=(t,w(t)) will yield h(t)=w(t). Since the units used for the coordinates are the same as in the original triangle, putting the triangle and the plot side by side and then drawing a horizontal line through (t,h(t)) will give the desired point on the original triangle. Using geogebra (base length 5, acute angles both 30 degrees), I have verified this will work. However, in the video, there are two graphs. One is for the height with respect to the width and the other the width with respect to the height. In particular, the graph for the length (color blue in the video) will match the graph in the paragraph above but the one for the width (color green in the video) will not. The one in the video will have the point (0, height of triangle) but the one in the prior paragraph will have (0, base length of triangle). In fact, the construction in the video is equivalent to drawing a vertical segment (with length equal to the height of the triangle) at an endpoint of the triangle (with an acute angle) and drawing another segment from the other triangle endpoint to the end of the constructed segment. Testing with geogebra (same example as in prior paragraph) yields an incorrect height. I think the problem with the construction in the video is that if one axis is rectangle height and the other is rectangle width, the relationship between the two would just yield a single graph (color green in the video). We don't get new information if we look at width vs height after already knowing height vs width. The blue graph in the video is actually the height of the rectangle with respect to time (or how far along you are at the base of the triangle), not the width of the resulting rectangle. The reason the particular example in the video worked is that the acute angles are about 45 degrees. In that special case, one can work out algebraically that the point of intersection in my suggested method and the method in the video is the same. In fact, by working out the expressions, the method I suggested is equivalent to everything as in the video except the green graph is replaced by the segment going through (0, base length of triangle) to (base length of triangle/2, 0). The blue graph is the same: a copy of the left leg of the triangle. A friend gave the following. Ignore the blue graph in the video. The green graph in the video (line segment from (height of triangle, 0) to (0, base length of triangle) gives the relationship between height of rectangle vs width of rectangle. This can be verified to indeed be linear. Then intersecting this green graph with the graph of y=x (to get where the height of a rectangle would be the same as the length of a rectangle) would yield the desired height. This works, as tested in geogebra. TLDR: I think the correct method is to plot the height and the width with respect to the same parameter. A friend gave a method where one just looks at the green graph in the video and intersects it with the graph of y=x. The method in the video is to plot both height with respect to width and width with respect to height in the same coordinate plane, though there seems to be some errors.
Simpler ideas to draw square into triangle , pretty amazing , lovely and enjoying but more intriguing that never thought about that , so thanks for great video 🍰🍰🍰🍰🍰🍰🥪🥪🥪🧇🧇🧇🧇 putting wafers and sandwich into triangle slice
it would be actually more accurate if you draw a square which is bigger than the triangle at number 3. In this way, when you shrink down the square, the errors get smaller instead of getting bigger.
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I love how elegant the second one is. (Though the practicality of the third being contained within the original triangle is the most practical.) Thank you Dr Lin.
I don't think the third is that practical, it relies too much on the ability to draw perfect squares when the others take already existing lines to base on the construction of those. Not trying to argue, just saying.
The third one isn't practical at all. Have you ever actually constructed polygons using the straight edge and compass method? A modification of the first method is the most practical, and doesn't even need a ruled edge. The second method's scale becomes too large to be practical, and the third requires way too many construction lines to be practical. What's shown here are very approximated polygons that only use a straight edge, but when you consider accurate angles, you also need a compass to do this.
In the first one, you have to draw two rectangles and a whole graph with perfect orthogonal axis. In the third one, you just have to draw one square, of course it's the most practical.
@@xaytana Squares are pretty easy to construct though.
@@adrianoaraujo1692 All you need to do it in the first is already provided by the triangle and trace a 45° angle line. The guy is just trying to make a clean example drawing to the side. In the final, you have to construct your own squares from zero, which completely mocks on the objective of the exercise in the first place. That is more like a way to make sure that the squares are well constructed. If you already had a way to construct perfect squares, why would you need to use any of these techniques to put them into the tringle?
Keeping us educated and entertained in a time of confinement; social distancing doesn’t need to include intellectual distancing! Thank you!
Very well spoken
So true
??.
I love this channel because it covers math ideas rather than making you solve equations on my math lab and getting the whole equation wrong because you forgot to include the negative sign somewhere
More geometry stuffs, please. I really like this topic.
agreed. it's like the only thing I can follow lol
Try a game called Euclidea in Google store, it has tons of puzzles similar to this, just pure geometry puzzles, it starts pretty easy but ends up with super complicated stuff.
@@Yatapower yeah I downloaded that, pythagorean, x section for fun, I still stuck at beta levels in euclidea though lol.
I love how the third one blows away the complication of the first two.
Day 11 of quarantine really got me drawing squares inside triangles rn
We can only hope that _Drawing Squares Inside Triangles_ helps to _Flatten The Curve_ - j q t -
??.
Method #3 requires the least amount of paper.
And ink :]
Also reduces square waste
Save the trees!
??.
1:22 You may be cool, but you'll never be Dr. Lin dropping the perpendiculus cool.
There is a way to do the first way with compass and straight edge. Here is how
1) label the traingle ABC with A on top.
2) extend BC through C to D such that BC=CD
3) draw perpendicular to AD through C.
4) draw a line parallel to BD through A and mark its intersection with the perpendicular through C as E.
5) draw segment ED
6) draw a line parallel to AB through C and mark its intersection with ED as G.
7) draw a line parallel to BD through G mark its intersections with AB and AC as H and I respectively.
8) drop perpendiculars through H and I to BC.
You now have created the square inside the triangle
@@stuartryerse you are correct, the problem was I typed this up on my cell phone from memory of the construction I had done earlier but unfortunately got the point names messed up. I have fixed it
loved this one.....the topic, the ideas, the explanations...more of Mr. Lin please?
also, if he's up for it, let him thicken up the singaporean accent... just for the few of us in here to enjoy :D
If you wanna find the side length of the square that’s inside the triangle you can use
If angle is less or equal to 45
a = base(tan x)/2+(tan x)
If angle is greater than 45
a = base(tan x)/1+ (tan x)
Base being the entire base of the triangle
X being one of the two congruent angles of the triangle
And y being the side length of the square
That is what I love about maths. It is exact. You can chose different paths and always arrive at the same solution.
5:17 Parker line
i have understood this reference
@@nicokuhne3255 Me too :-)
Aleph Null
Love me some straight edge and compass construction, great job folks!
Then you might like the app "Euclidia" (definitely available for Android, probably for iPhone as well, but not sure about that).
It gives you tons of different tasks that can be constructed with only these two items, making it kind of a game where you have to look for the solution requiring the fewest steps (there are even bonus points for additional, symmetrical solutions in some tasks...).
There are a lot of solution videos here on youtube as well, for every single task, in case one gets stuck.
@@Shadow81989 You're right, I love euclidia, though I haven't played it in forever. Definitely worth going back to now that I've forgotten all the solutions!
Me: Studying hard for my Calculus exams.
RUclips recommended: 3 Ways to Draw Squares Inside Triangles - Numberphile.
Me: Exams never were that important anyways.
New numberphile video?
Sleep is over rated anyway..
I was thinking, isn't there a Calculus concept having to do with finding the minimum and maximum values? Wouldn't that be somewhat relevant here?
wait a sec doesn't due to covid ,exams all over the world are postponed(I mean even if it was a joke,u may have used some other topic)
Colleges are for the most part still rolling on a mostly regular schedule just with online learning instead of on campus.
How are you going to an exam? Aren't exams postponed because of Covid-19? I know online education is a thing but I can't imagine they would let anyone do their exam online as it is easy to cheat. Maybe exams are still going just with precautions but wouldn't think so
I really encourage these kind of construction videos. Simple enough to get kids hooked on maths. Had good fun for a few hours trying it out. The third method was too good. We need more Dr Lin. Very infectious enthusiasm.
This is easily one of my most favourite video from this channel. So simple and elegant and beautiful. Thank you for sharing it.
This is one of the challenges in the Euclidea app. My approach was to skew the original triangle until it's a right-angled triangle, and bisect the right angle, to find the correct height. It's similar to approach 1 but keeping everything drawn in-place.
More than similar, it *is* method 1, but in-place and straight edge instead of ruler. Excellent work.
I love Euclidea! But which level is this one?
@@Dalroc It's quite a long way in, "nu" 4.
This is why I always loved geometry in a school. Most of the time everything you need to solve the problem is right in front of you; you just need some cleverness to connect all the dots and make deductions to each step to reach the desired end. Plus it's visual, which I was always better at. You can see how everything interacts.
Everybody: Panicking about Corona
Numberphile: LET’s DRAW SQUARES IN TRIANGLES!!!
The emptyness of that office still reflects the situation.
Math will keep COVID19 at bay
You chose to play this video!
Please make a video on Neal conjecture
Times are tight at Numberphile that they couldn't afford to give this poor man another sheet of paper.
Great episode, your speaker explained this very well and seems to really enjoy his work. Thanks for sharing and hope to see this speaker again soon!
Watching this, for the first time, on the 2-year-anniversary of the upload date of this video: 22.11.2022 😎.
More geometry! The papers are always so cool at the end, and I know how much you like the final papers, Brady!
He makes it easy! That was a great explanation and visual guide :)
Thanks! Glad you liked it.
@@calvinlin5753 Compared to the other hosts, you are a really fine presenter.
This is the first numberphile video I've watched in a while that I've paused and tried it for myself! I got the third solution on my own!
4th Way: raise the perpendiculars of the base to form a square then draw lines from the top corners of that square down to the center of the base. Didn't actually try yet, just thought of it right after seeing the second method.
Edit: As john osullivan has pointed out, this only works for isosceles triangles. I originally thought this was a requirement, but it's not, and the other 3 ways work on ANY triangle that has acute angles on both sides of its base, not just isosceles triangles.
That won't work. Most triangles aren't isoceles.
Whoops, I thought that was a requirement but I just re-watched the beginning and indeed the ONLY requirement is that they are acute. Thanks :)
His Singaporean accent is so genuine lol
i was wondering where the "then hor" is
Kenji Gunawan i 💖 his accent
I thought he was german lol
lolol i can identify
i searching his name after listening to the accent
What a beautiful, lucid demonstration. Thank you Dr. Lin
Method 1: makes sense
Method 2: NANIIII?!?!?! WOAHHHHH
Method 3: WOAAAHHHH
Overall: MIND = BLOWN
7:40 "parallel line"
Parallel to the base of the triangle*
Being precise is hard.
This is actually really useful for me- I'm doing an artists book of geometric constructions and I'm doing the Magnum Opus symbol (square within a triangle within a circle). Hitherto I've been constructing the shape by starting with the square and the constructing an equilateral triangle from its top edge, but that doesn't give me control over the size of the final circle (important when laying out a book). Using the third method I should be able to construct the symbol in reverse order- big circle, then triangle, then square. Thanks Dr Lin! More geometry videos please!
If you do one on the side, you could just run any measure through a multiplier to get whatever size you want, starting anywhere
0:44 *Horror music starts to play*
Here before this comment blows up.
Slenderman entered the chat
It should be Jiorno's theme tho!
Man. This guy is amazing ....
Its amazing that human mind can understand language from speakers of different ethnicity
Thanks for showing us how to draw it, I can't draw a straight line whatsoever, so i'm probably going to need a ruler for this
Just put your pencil in the paper, then rotate your shoulder instead of your wrist. Works better if your arm is infinitely long.
I've always seen Calvin on Brilliant but have never heard his voice. Awesome! He's a genius.
Cool ! More Calvin Lin
Would love to :)
I love the first one. That kind of thinking is great.
I never used to like geometry. Even when I had a really good geometry teacher, I didn't like it. I had finally started to understand algebra, and then they put me in geometry. I have a much better appreciation for, and even love of, geometry now. Partially because of that teacher, partially because of physics (you really need all kinds of math in order to do the problems on physics tests), and partially because of videos like this.
One of the really neat things about geometry in general, is so much of it can be done without any numbers, kind of like what Calvin Lin does in this video. I will definitely be showing some of my students these methods (I tutor part-time).
Aaah. I truly feel the fun of childhood, when I was figuring out by myself how to build cubes and other 3D objects out of paper. Playful geometry is the best.
You can also do the first method contained almost in the same footprint as the original triangle: Construct a triangle t' with the same base as the original triangle t, but with the top vertex shifted horizontally to the left such that the bottom left corner is 90°. Then construct the angle bisector of the bottom left corner and the intersection with the right edge of t' gives you the height of the square.
Great suggestion! I'd use that the next time. I originally kept the approaches distinct, but decided to combine it as we started filming. The glee at the end was pure joy.
This video made me remember that mathematics is beautiful. Thank you.
It is!!!
Hey ! One of the first time I figure out the solution myself ! So proud !
Congrats!
You can indeed use method 1 with a compass and straight edge.
construct a perpendicular line to the base in the bottom left point of the triangle.
mark the length of the top left edge of the triangle on the base.
mark the length of the base on our perpendicular line.
connect these two marks.
construct a 45° angle from the base and our perpendicular line. the cross section of the 45° line and the connection between the marks is the height we want.
Then it is just dropping perpendiculars, marking intersections and drawing lines.
I like this guy! We gotta get more of him :)
Brady reads comments, so :cross_fingers:
1:47 he draws some lines, then in the next shot they've disappeared and he draws them again in a different place.
The triangle with the green and blue rectangles are also different, which means he completely redrew everything.
You could also do something based off the first method, the graph, using only a straight edge and compass, which also involves mirroring the 'graph.'
Draw the triangle you want to draw a square in. Then draw horizontals extending from the middle vertex, and verticals extending from the outer two vertices, this will create two right triangles that form a circumscribed rectangle around the original triangle. Then connect opposite vertices of the rectangle to create diagonals, this will find two vertices of the square along the two non-horizontal edges of the triangle, both points of which are parallel to the triangle's horizontal line. Then drop verticals from these vertices and draw the horizontal line between them.
Anyone familiar with drawing regular polygons with the straight edge and compass method will probably realize this is the simplest method.The second method in the video, the similar triangle method, gets way too large to be simplistic. While the third method involves so many construction lines just to find one vertex of the final square, which is way too much work when you're using the straight edge and compass method when you care about angles being accurate. These other methods are still just as valid, but work better as proofs, rather than being realistic uses.
Hell, even my method can be optimized. Only do half the rectangle, find only one vertex of the square, a compass will give you two more vertices from that point, and you can draw the rest of the square from that.
And it works for asymmetrical triangles as well! Great entertainment when you're in quarantine.
Highly enjoyable video!
Could watch him draw squares and be happy about it all day :-)
Please explain beal conjecture
Using string and a pen for his compass. Awesome!
You can link the first and last methods by drawing a square which has the triangle height as a side and adjacent to the base. Then you draw the line which connects a vertex of the triangle to the opposite vertex of the square and the job is done.
Compass and straightedge puzzles are always my favourite
Passed over a really simple solution, if you draw a perpendicular from the left vertex to the height of the triangle, the you can connect the new point with the right vertex, and the line should cross the triangle at the height of the square.
Same method as the first example using the ruler measurement, but only a straight edge and compass required.
I've tried it and it doesn't seem to work
@@marijnbliekendaal I don't get why not. The line from the right side is finding the width of the square that connects with the right side of the original triangle (2 units added to the width is one unit away from the center)
Unless the left side needs to be a 45° angle, forcing a square from the right side, then you would extend to the point on the side of the original triangle.
Still can be done without a ruler.
I love the third solution, it's very fast and easy to remember.
That was more interesting than I expected it to be :)
The way I though of initially is different: since the square is “in the middle” of the triangle, the midpoint of the bottom edge of the square will always be the midpoint of the bottom edge of the triangle. So, the measure of the angle from a vertex of the triangle to the midpoint to the vertex of the square corresponding to the corner on the side of the triangle whose ends are the initial vertex and the top vertex is always 60 degrees, since it corresponds to a right triangle with legs in 1:2 ratio. Hence, simply construct a 60 degree angle from the midpoint of the base of the triangle, then drop the perpendicular from the point of intersection.
Edit: “midpoint of base of triangle” -> “point of intersection of base with perpendicular (with respect to base) through top vertex”
Unfortunately, the midpoint of the bottom edge of the square need not coincide with the midpoint of the bottom edge of the triangle.Maybe you are assuming that the triangle is isosceles?
Calvin Lin you’re absolutely right. What I meant was the perpendicular through the top vertex.
Triangles are magic!
That was neat. Self-proving methods where one can easily use a second way to validate the first findinds.
Viewers: Why do they keep doing interviews when the disease is spreading?
Numberphile: Actually, we recorded this last year.
Viewers: TRIGGERED
The second one is the one you draw in High School geometry and scream, "I GET IT NOW" when it finally comes out
I love this guy! Does he have other videos?
Short and sweet!
5:34 we could do this infinitely, over and over and over again!
That was oddly satisfying.
The second method is so pretty, love it.
I loved the first method.
The graph at 2:55 is labeled wrong, this had me confused for a while. With those labels the blue line doesn't make any sense. The x-axis represents rectangle height, while the y-axis represents height for the blue line and width for the green line.
I'm not sure quite what the graph does show, but the x-axis isn't always height - the x-intercept of the green line comes at a value equal to the maximum width of the rectangle, which is more than the maximum height.
I also struggled for a bit with the graph. Here's how I convinced myself that it is possible to find the square by intersecting the two lines:
The green line is a set of points, where each point represents a possible rectangle that can be placed inside the triangle as described in the video. The x coordinate represents the rectangle's width and the y coordinate represents the rectangle's height.
Now, we want a square, aka a rectangle where the width is equal to the height. If we plot a line representing all possible squares (where the x coordinate is the width and the y coordinate is the height) we get the blue line from the video.
It then stands to reason that the intersection of the green and blue lines will give the dimensions of the square that can be placed inside the triangle (satisfying the criteria to lie both on the green and blue lines).
I'm not sure if that's the way Dr. Lin intended the explanation to go, but it worked for me!
@@lukecoughlan5035 splendid explanation, made it click for me as well. But I do have one correction I would like to make, and it is that the green and blue lines also represent rectangles and squares "outside" the triangle, but they don't exceed the height of it. Thought I would put it out there, because I got hung on that for a little while.
This was an interesting one
he could totally have done that measuring construction within the original triangle with just a ruler and compass. draw a perpendicular to the base through one of its ends and a paralell to the base through the top vertex. connect the intersection of these two with the other end of the base. the intersection of this conection with one of the sides of the triangle is at the point where the square touches the side of the triangle.
Nice video, I like the third way best. It is worth saying that this video doesn't explain everything, only the concept. For example unless I missed it, it doesn't explain how to draw a perpendicular line from a point. It's basic geometry, but still probably many viewers don't know. Compass is used much more than what is showed in the video when you are doing it "for real".
Indeed. I decided to skip that as it would have taken much longer otherwise, and might make it hard to see the beauty.
Yeah I don’t think we can show and discuss compass straight edge construction EVERY TIME. This video shows a lot more, from memory. m.ruclips.net/video/6Lm9EHhbJAY/видео.html
This is amazing🤩
The 1st method could have been done without the ruler:
1. Raise a perpendicular from the left end of the triangle base.
2. Draw a line parallel to the base, thru the apex.
3. Bisect the angle of the base and the perpendicular from 1.
4. Draw a line from the intersection of the perpendicular from 1 & the parallel from 2, down to the right end of the base.
5. The intersection of 3 & 4 is the height of the square (just as in his measured diagram). Draw a line parallel to the base thru that point, and where it intersects the left and right edges of the triangle are the upper corners of the square.
6. Drop perpendiculars from the corners from 5 down to the base. Square completed.
compass + straightedge = ruler anyway
Wow, just amazing. It is as nice as simple, thank you!!!
That's pretty amazing
beautiful methods!
I like this guy
3:33 Now that's thinking outside the triangle
Yes
I want to quarantine with these type of videos..
Is Dr Lin Singaporean/ from an ASEAN country? I know its not relevant but I feel this instant affinity with his accent 😂
He is from SG, recognised the accent immediately
Please some trigonometry and calculus stuff!!
If there's one thing I learnt during this video it's that the Singaporean accent is very similar to the Belter dialect in The Expanse! XD
Great video Dr Lin.
Very clever use of geometry. I can use this in my woodworking! But there is one point I see that makes it easier... a 45 degree triangle, to run the line from one point on the triangle to begin with...
I have a question:
If you can construct something (with a ruler and compass) can you also construct it with an arbitrally small amount of ink? (not including the final thing!)
For using this as an example it would be: can you make the square in the triangle so that the ink needed to figure out the sides of the square (without actually drawing them) would require -> 0 ink.
Theoretically yes, practically you need sufficient ink so that the constructed lines are visible
You can draw one rectangle too wide and one too tall, then use their intersection point as distances from the top corners to find a new point on the triangle's edge to draw rectangles from, giving you two rectangles that are slightly more square. Iterate until you have a perfect square. There may even be a mathematical expression for this method that leads to a way to calculate square roots!
the problem with this method is that you have to do it infinitely often until you get the correct result.
@@PeterAuto1 Same problem for finding e. Iterating infinite times to converge on an answer is not a problem for a mathematician!
Anyone else absolutely adore the way Calvin says “Perpendiculars”?
And recTANGle.
Fourth method: drop a height and move one side (parallel slide) at this point - where it intersect other side, there is a point for square build.
Павел Менжинский - or, use a construction to trisect one of the sides of the triangle (easy using simple projection techniques). The square will touch the triangle two thirds of the way up, every time.
That is so cool!
Thank you!!!!
last methed. wow. amazing
Easy, just draw a triangle around myself.
*bruh*
Very neat!
Any new numberphile podcasts coming?
What is the relationship between the angle made by the line from the vertex through the corner of the squares and the vertex it came from? Does it bisect the angle, or is it a cosine or something? Can this property be used to draw a square with a compass and straight edge? I'm hoping he talks about it in the extras! Keep up the great work.
I think something might be wrong with the first method. Could you verify?
I think the idea is to plot the height and width with respect to a single parameter t so that the intersection point (t,h(t))=(t,w(t)) will yield h(t)=w(t). Since the units used for the coordinates are the same as in the original triangle, putting the triangle and the plot side by side and then drawing a horizontal line through (t,h(t)) will give the desired point on the original triangle. Using geogebra (base length 5, acute angles both 30 degrees), I have verified this will work.
However, in the video, there are two graphs. One is for the height with respect to the width and the other the width with respect to the height. In particular, the graph for the length (color blue in the video) will match the graph in the paragraph above but the one for the width (color green in the video) will not. The one in the video will have the point (0, height of triangle) but the one in the prior paragraph will have (0, base length of triangle). In fact, the construction in the video is equivalent to drawing a vertical segment (with length equal to the height of the triangle) at an endpoint of the triangle (with an acute angle) and drawing another segment from the other triangle endpoint to the end of the constructed segment. Testing with geogebra (same example as in prior paragraph) yields an incorrect height.
I think the problem with the construction in the video is that if one axis is rectangle height and the other is rectangle width, the relationship between the two would just yield a single graph (color green in the video). We don't get new information if we look at width vs height after already knowing height vs width.
The blue graph in the video is actually the height of the rectangle with respect to time (or how far along you are at the base of the triangle), not the width of the resulting rectangle. The reason the particular example in the video worked is that the acute angles are about 45 degrees. In that special case, one can work out algebraically that the point of intersection in my suggested method and the method in the video is the same.
In fact, by working out the expressions, the method I suggested is equivalent to everything as in the video except the green graph is replaced by the segment going through (0, base length of triangle) to (base length of triangle/2, 0). The blue graph is the same: a copy of the left leg of the triangle.
A friend gave the following. Ignore the blue graph in the video. The green graph in the video (line segment from (height of triangle, 0) to (0, base length of triangle) gives the relationship between height of rectangle vs width of rectangle. This can be verified to indeed be linear. Then intersecting this green graph with the graph of y=x (to get where the height of a rectangle would be the same as the length of a rectangle) would yield the desired height. This works, as tested in geogebra.
TLDR:
I think the correct method is to plot the height and the width with respect to the same parameter.
A friend gave a method where one just looks at the green graph in the video and intersects it with the graph of y=x.
The method in the video is to plot both height with respect to width and width with respect to height in the same coordinate plane, though there seems to be some errors.
Simpler ideas to draw square into triangle , pretty amazing , lovely and enjoying but more intriguing that never thought about that , so thanks for great video 🍰🍰🍰🍰🍰🍰🥪🥪🥪🧇🧇🧇🧇 putting wafers and sandwich into triangle slice
it would be actually more accurate if you draw a square which is bigger than the triangle at number 3. In this way, when you shrink down the square, the errors get smaller instead of getting bigger.
Yyoooouuu've been had. That is pretty darned funny. Didn't see that coming.
No entiendo el ingles muy bien pero si las matematicas.... y claro la geometria mucho mas