Sunday Puzzle gang rise up!! the most splendid video of all time: ruclips.net/video/Qt4pdX-w2JA/видео.html Join Wrath of Math to get exclusive videos, lecture notes, and more: ruclips.net/channel/UCyEKvaxi8mt9FMc62MHcliwjoin
This is a really interesting puzzle. In a few of my previous classes, I've done geometric constructions, but doing it without a compass? Thats a whole new challenge.
@dominiquefortin5345 Perpendicular lines are two lines that intersect at a right angle. Through every pair of points is a straight line (axiom). Find those points with just a compass. According to the Mohr-Mascheroni theorem, it can be done.
@@Rack979 Yes you can find all the points necessary with the Mohr-Mascheroni theorem, but you still need a straight edge to draw the two perpendicular lines. Without lines you just have points and no perpendicular.
@@dominiquefortin5345 'It must be understood that "any geometric construction" refers to figures that contain no straight lines, as it is clearly impossible to draw a straight line without a straightedge. It is understood that a line is determined provided that two distinct points on that line are given or constructed, even though no visual representation of the line will be present.', straight from the wiki on the theorem.
@WrathofMath It would have been nice to also have the demonstration for when P is between A and B, because the construction is deferent (the 2 inscribed triangles are on the same side of the diameter), because, at first glance, it is not obvious that the same theorems could be used to find the perpendicular.
If P is on the line AB (whether inside or outside the circle) then it just becomes a case of constructing a perpendicular line through a point in general. I know how to do this with a compass... I'm not sure how you would do it without one.
@@codahighland "...Such a disclaimer suggests that it ISN'T possible." That and if PB is perpendicular to AB you also can't prove it, but I'm not 100% sure.
@@dominiquefortin5345 That's my point, yes. You can't use a compass, so from what I understand and from what I've seen trying to look it up, it can't be done, but I'm acknowledging that there might be some obscure trick I didn't know.
step 1. extend AB step 2. align the markings of the ruler (for transparent rulers) or the edge of the ruler with the line AB step 3. shift the ruler along AB until it touches P step 4. draw the line with the ruler
@@ik-now178 You might want to read my comment again. I said “listen” precisely because this video is about solving the problem with a straight edge 0:13 And not with a ruller like in the problem that this video is based on. I don’t care what’s written in the original problem’s statement, this video is about solving it with a straight edge, which is exactly what was said.
@@fullfungo I read carefully the title and it is ruler. Now I can put two random dots on a straight edge (or use its full length) and still solve it. (edit) You didn't say "listen"
After getting the hints, I came to the same construction, but with a diffrrent interpretation. I considered the triangle ABP, of which AQ and PR are two heights. The third height then of course has to go through the same point, and being a height, it's prrpendicular to the line AB.
Almost got there - When I took Geometry back in High School (1972/73) we had to PROVE EVERYTHING! Including constructions! I was sort of getting there but was missing how I could PROVE that the final intersection was indeed a right angle. Interesting!
That's a much nicer proof than I had. I solved it by intuition and then spend several minutes calculating angles of various triangles until I proved the result.
I miss interpreted the challenge and was surprised that you were worrying about properties of triangles when you could just draw a line extension and use the junction of two arcs to solve it in a few seconds
Very interesting problem. Reminds me of taking a Linkages class in college. I initially thought to bisect the circle, with using a ruler as a compass. By using an additional pencil or finger as a compass needle, would hold the ruler firm and allow the main pencil to draw another circle. Engineering mind at it again haha. I love the method used the in the video, very fun! :) Have a good day!
How would the construction change if P were still outside the circle but more “between” A and B, rather than being so far from one point and close to the other?
My bachelor thesis was actually on paper folding and how it's at least as powerful for geometric construction of points (and lines) as a ruler and compass. So I'd ditch the ruler too and just solve it with the paper. (Depending on what you allow as a construction, you can actually also trisect angles and take arbitrary cube roots. So you can solve arbitrary quartic equations over the complex numbers. Look for the "Beloch fold" and see if you'd allow it.)
I haven't read about geometric constructions via folding in a long time - it would be a fascinating topic to cover. Thanks for reminding me of this topic! Would love to check out your thesis if you have a link.
@ I did find it, but is indeed in Dutch. I did find the sources I used though. The part about the Beloch fold and cubic equations was based on this article called "Solving Cubics with Creases: The Work of Beloch and Lill" by Thomas C. Hull. I also used the book "Geometric Constructions" by George E. Martin, specifically chapter 10 it seems.
I will try to solve it before watching the video. -Extend the diameter -draw a circle with center p such that it cuts the diameter in 2 points -make the perpendicular bisector of those 2 points
I used a much simpler geometric construction. Extend AB. Measure the distance from A to P. Find a point on AB extended, Point C, so that the distance from PC is the same as PA. Measure distance AC. Find point D on the line AB extended which is half the distance AC. Line PD is perpendicular to AB. This method is much easier if you use a compass. Also I ignored the instructions and used a pencil AND a ruler.
You cannot measure the distance. You can only use the straight line of the ruler, but not to measure. You could measure if you had a compass, but you don't.
@@IvanToshkov Where did you pull that one from Ivan? A classroom ruler has marked scales so you can measure. The problem does not state that you cannot measure. If measuring was not permitted, then the problem would have stated that the student could only use a straight edge.
@johnspathonis1078 maybe you are rolling, but in case you are serious, that's what they mean by "the straight edge of the ruler." Otherwise, the problem becomes trivial.
It took me so long to realize the pitch bending and seemingly random noises were from the video and not from something 2 rooms away from me. I thought I was going crazy at first
What if P is inside the circle What if P is "above" the circle (more precisely: if it is outside circle, but between tangent from A and tangent from B)
Before watching the video, here's my solution: Draw line PA. Its intersection with the circle is B'. Draw line PB. Its intersection with the circle is A'. Draw lines AA' and BB'. Their intersection is H. Draw line PH. This is the line that was asked for. This works because BB' is constructed perpendicular to AP and passes through B, so it's the height of triangle ABP that passes through B. Similarly, AA' is the height of ABP that passes through A. Therefore their intersection, H, is the orthocentre of ABP, and PH is the height of ABP through B, which passes through P and is perpendicular to AB.
How would I continue the construction if BP is the tangent at point B? This would force B and R two be the same point and than the construction afterwards no longer works
holy crap I actually did it, kind of... I did it in GeoGebra, but I kind of cheated via software. I did half of it though if that means anything, even if I can't prove it
Please, don't. Just elongate AB, put center of compass on P and find B' on AB. Next, by drawings 2 circles from B and B', find P' and PP' is Your answer.
I'd probably cheat and fold the right angle corner of the paper to construct it instead of the ruler. If they told me no paper I'd just say "No paper, no problem. Therefore there's no problem which requires a solution" and then walk away mysteriously into the mist (smokebomb optional).
Can't you just continue the diameter to any arbitrary point, then use the compass to make 2 intersecting arcs (one from the point and the other from the point of intersection of the original arc with the diameter), and that will pass through the middle of your small segment while being perpendicular to it?
It seems the puzzle is either underspecified or your solution misses the case that the arbitrary point P happens to be located on the straight line through A and B.
"...mirror P on this line into P'..." You can't magically find P'. You have to find it's position with a construction of lines where you are guaranteed that the line PP' is perpendicular to the line AB.
@dominiquefortin5345 no magic required, just use the calipers and mark 2 random points on line AB with calipers centered on P. Center the calipers on each, and the 2 circles drawn would intersect in P and P'
The diameter AB (scalar here) does not depend on the direction and is the same at any angle. A and B are simply opposite points. You simply connect the point O with P in a straight line and this line will be perpendicular to the circle and therefore also perpendicular to the diameter. The problem does not say that you have to take the orientation of line AB as it is drawn. Diameter is also the wrong term here, you should have taken the term vector given by A and B, then the orientation of the diameter vector would make sense as a prerequisite and thus the solution shown to construct a perpendicular vector to the vector AB. And I don't see that the problem is solved. A perpendicular vector on the vector AB, this vector is defined from A to B and does not lie outside the circle.
@dominiquefortin5345 it would be called a straight edge if doing traditional geometry. Most rulers like 95+% are rectangular. It can easily be verified by drawing a square with it, and then showing the diagonals and midpoint intersect at the same point. Check and checkmate.
@Zacristanda wrong. This solution which uses a perpendicular connected straight edge as he used the word ruler, is equally as rigorous and requires far less steps. Making it superior. And inferior solutions are wrong as being "the solution" it is merely "a solution ". Flawless victory.
@@PriyanTekchandani-t6i I said the _line_ AB. That is, it can be outside of the _segment_ AB. But even assuming it's inside, how does this help us create the perpendicular?
Pausing video to think this through. Um. I'd use the dot product to project one vector onto the other. Do we even have standard vectors? Do we have angles to make a polar coordinate system? I'm not clear on what the rules are or what methods we can use to get data... Let me read the instructions again.
Ugh. "Difficulty level medium" Let's blame the hangover. I mean can you measure? I guess so. DB is a vector and I have the mag. OP is a vec and I can make an intersection OC for another unit vector if I make the unit the radius. But where's my basis? Don't have one... What about P to the tangent of the circle? Then I can intersect that with the origin. Is that a right angle? Nah. Ok, now I'm depressed. Might be the hangover.
If PB is perpendicular to AB, then you're already done by definition -- you have found the line perpendicular to AB that passes through P. If you need more evidence, consider that PB must be tangent to the circle, because otherwise it would have intersected the circle in one additional point. And all tangents to a circle are perpendicular to the radius at the tangent point.
@@dominiquefortin5345 The proof is that the line doesn't pass through another point on the circle. This is the definition of a tangent. In compass-and-straightedge constructions, you are assumed to be able to identify intersection points perfectly.
extend segment AB, make a circle from point P with a radius larger than the distance between point P and line AB, construct 2 equally sized circles from the points where the circle meets the line, create a line through point P and the point where the two equal circles meet. easy problem edit: aw fuck there’s no compass
@@renka-chan9213 we use a cheap trick in school. A ruler has 90° angles built-in. The question doesn't say "use a straight edge". It allows a ruler. Fault of question. Why use more brain when you don't have to?
@@adorp It only allowed a ruler, not because to measure the sides nor to measure the angle, but because there is no ideal straightedge in the real world (an ideal straightedge is one without markings, and has an infinite length)
@renka-chan9213 well, the question is theoretical anyway. In practice, you will not get a more accurate drawing using Thales and Orthocentre instead of simply using the ruler's corner. When doing genoetry in school, I noticed that the official perpendicular drawing methods actually introduce more room for error than anything else. Simply using a set square is way more accurate. Geometrical drawing is theoretical.
I used other method: use the ruler to measure length of BP then draw an isosceles triangle BPC which has BC is extended from AB. After that, extend BD from BP so that BD = BP, extend CE from PC so that CE = PC. BCED is an isosceles trapezoid. F is the intersection of 2 diagonals of the isosceles trapezoid. FP is perpendicular to AB
yes, because the green line we may define as the third altitude of the triangle, which comes from the vertex A. The other two altitudes of the triangle met at B, and it has to be that all altitudes of the triangle meet at a common point, so this third altitude (the green line) has to go through B as well.
@@WrathofMath ah so there's a law about all three lines passing through that point? Makes sense. I'm not very versed in all this stuff, great video though!
there is both a way easier and way faster method that creates less visual clutter. extend AB out some distance. place the end of the ruler on the line (make sure its line up) and move it around until you can draw a line from P to line AB.
![Seungmin "그렇게, 천천히, 우리(As we are)" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/kAzmhLHePqU/mqdefault.jpg)
Sunday Puzzle gang rise up!! the most splendid video of all time: ruclips.net/video/Qt4pdX-w2JA/видео.html
Join Wrath of Math to get exclusive videos, lecture notes, and more:
ruclips.net/channel/UCyEKvaxi8mt9FMc62MHcliwjoin
Ok.
Ok.
Me using the right angle of the ruler edge
Clean and simple
Same
Thought the exact same thing
good luck getting it exactly 90
@ I can use the edge of the paper to double check
This is a really interesting puzzle. In a few of my previous classes, I've done geometric constructions, but doing it without a compass? Thats a whole new challenge.
The same problem can be done with just a compass, right? No lines, of course but you can find points on the perpendicular to the diameter through P.
"...No lines, ..." What is the definition of a perpendicular ?
@dominiquefortin5345 Perpendicular lines are two lines that intersect at a right angle.
Through every pair of points is a straight line (axiom). Find those points with just a compass.
According to the Mohr-Mascheroni theorem, it can be done.
@@Rack979 "... every pair of points is a straight line ..." False. You can always make a straight line using 2 points but 2 points are not a line.
@@Rack979 Yes you can find all the points necessary with the Mohr-Mascheroni theorem, but you still need a straight edge to draw the two perpendicular lines. Without lines you just have points and no perpendicular.
@@dominiquefortin5345 'It must be understood that "any geometric construction" refers to figures that contain no straight lines, as it is clearly impossible to draw a straight line without a straightedge. It is understood that a line is determined provided that two distinct points on that line are given or constructed, even though no visual representation of the line will be present.', straight from the wiki on the theorem.
5:32 that is the reddest red I've ever seen
yes. when I saw it it was so bright and searing.
I read the title wrong. I thought it said "You can only use an EULER!"
My proof was:
Let's suppose the answer is e.
QED
Using an Euler in math is overpowered
0:33 impossible because you don’t have anything to draw
you're permitted a straightedge and a feather whose tip is coated in the blood of your enemies
@@WrathofMath Oops, used one coated in my own blood, sorry for failing the challenge
I just repeatedly rubbed the edge of the ruler to the paper to burn in a dark line
I was just gonna comment that the instructions didn't mention a pen or pencil 😂
@WrathofMath It would have been nice to also have the demonstration for when P is between A and B, because the construction is deferent (the 2 inscribed triangles are on the same side of the diameter), because, at first glance, it is not obvious that the same theorems could be used to find the perpendicular.
If P is on the line AB (whether inside or outside the circle) then it just becomes a case of constructing a perpendicular line through a point in general. I know how to do this with a compass... I'm not sure how you would do it without one.
I found another statement of this puzzle that specifically called out that P is not on AB. Such a disclaimer suggests that it ISN'T possible.
@@codahighland "...I know how to do this with a compass..." but can't use a compass per the rules.
@@codahighland "...Such a disclaimer suggests that it ISN'T possible." That and if PB is perpendicular to AB you also can't prove it, but I'm not 100% sure.
@@dominiquefortin5345 That's my point, yes. You can't use a compass, so from what I understand and from what I've seen trying to look it up, it can't be done, but I'm acknowledging that there might be some obscure trick I didn't know.
step 1. extend AB
step 2. align the markings of the ruler (for transparent rulers) or the edge of the ruler with the line AB
step 3. shift the ruler along AB until it touches P
step 4. draw the line with the ruler
You can't proved that your line is perpendicular !
You are not allowed to use a ruler. You are only given a straight edge.
Next time make sure you listen carefully to the assignment before solving.
@@fullfungo "you can use ONLY a simple ruler"
@@ik-now178 You might want to read my comment again. I said “listen” precisely because this video is about solving the problem with a straight edge 0:13
And not with a ruller like in the problem that this video is based on.
I don’t care what’s written in the original problem’s statement, this video is about solving it with a straight edge, which is exactly what was said.
@@fullfungo I read carefully the title and it is ruler. Now I can put two random dots on a straight edge (or use its full length) and still solve it.
(edit) You didn't say "listen"
After getting the hints, I came to the same construction, but with a diffrrent interpretation. I considered the triangle ABP, of which AQ and PR are two heights. The third height then of course has to go through the same point, and being a height, it's prrpendicular to the line AB.
Almost got there - When I took Geometry back in High School (1972/73) we had to PROVE EVERYTHING! Including constructions! I was sort of getting there but was missing how I could PROVE that the final intersection was indeed a right angle. Interesting!
That's a much nicer proof than I had. I solved it by intuition and then spend several minutes calculating angles of various triangles until I proved the result.
I miss interpreted the challenge and was surprised that you were worrying about properties of triangles when you could just draw a line extension and use the junction of two arcs to solve it in a few seconds
me: starts building a pythagorean triangle calculating the correct values for each side so i can make it exactly 90°
My first thought exactly
You can't do that though, since you're not allowed to use the ruler to measure when doing construction problems
I forgot about the orthocentre thing, as soon as you mentioned it, it was easy.
Very interesting problem. Reminds me of taking a Linkages class in college.
I initially thought to bisect the circle, with using a ruler as a compass. By using an additional pencil or finger as a compass needle, would hold the ruler firm and allow the main pencil to draw another circle. Engineering mind at it again haha.
I love the method used the in the video, very fun! :) Have a good day!
Circles are the result of a compass, though. This problem shouldnt even give you a circle if it really wanted you to only use a straight edge
How would the construction change if P were still outside the circle but more “between” A and B, rather than being so far from one point and close to the other?
a²+b²=c²
My bachelor thesis was actually on paper folding and how it's at least as powerful for geometric construction of points (and lines) as a ruler and compass.
So I'd ditch the ruler too and just solve it with the paper.
(Depending on what you allow as a construction, you can actually also trisect angles and take arbitrary cube roots. So you can solve arbitrary quartic equations over the complex numbers. Look for the "Beloch fold" and see if you'd allow it.)
I haven't read about geometric constructions via folding in a long time - it would be a fascinating topic to cover. Thanks for reminding me of this topic! Would love to check out your thesis if you have a link.
@WrathofMath It's been about 15 years, so I don't have it.
Also, if I remember correctly, it's in Dutch.
@ I did find it, but is indeed in Dutch. I did find the sources I used though.
The part about the Beloch fold and cubic equations was based on this article called "Solving Cubics with Creases: The Work of Beloch and Lill" by Thomas C. Hull.
I also used the book "Geometric Constructions" by George E. Martin, specifically chapter 10 it seems.
I will try to solve it before watching the video.
-Extend the diameter
-draw a circle with center p such that it cuts the diameter in 2 points
-make the perpendicular bisector of those 2 points
Ok sorry, i thought you had compas
I wanna try this now
I used a much simpler geometric construction. Extend AB. Measure the distance from A to P. Find a point on AB extended, Point C, so that the distance from PC is the same as PA. Measure distance AC. Find point D on the line AB extended which is half the distance AC. Line PD is perpendicular to AB. This method is much easier if you use a compass. Also I ignored the instructions and used a pencil AND a ruler.
You cannot measure the distance. You can only use the straight line of the ruler, but not to measure. You could measure if you had a compass, but you don't.
@@IvanToshkov Where did you pull that one from Ivan? A classroom ruler has marked scales so you can measure. The problem does not state that you cannot measure. If measuring was not permitted, then the problem would have stated that the student could only use a straight edge.
@johnspathonis1078 maybe you are rolling, but in case you are serious, that's what they mean by "the straight edge of the ruler." Otherwise, the problem becomes trivial.
This is fantastic
It took me so long to realize the pitch bending and seemingly random noises were from the video and not from something 2 rooms away from me. I thought I was going crazy at first
What if P is inside the circle
What if P is "above" the circle (more precisely: if it is outside circle, but between tangent from A and tangent from B)
Before watching the video, here's my solution:
Draw line PA. Its intersection with the circle is B'.
Draw line PB. Its intersection with the circle is A'.
Draw lines AA' and BB'. Their intersection is H.
Draw line PH. This is the line that was asked for.
This works because BB' is constructed perpendicular to AP and passes through B, so it's the height of triangle ABP that passes through B. Similarly, AA' is the height of ABP that passes through A. Therefore their intersection, H, is the orthocentre of ABP, and PH is the height of ABP through B, which passes through P and is perpendicular to AB.
That is only one possible case. It doesn't work for when P is inside the circle.
@@dominiquefortin5345Before watching the video they wouldn't know that it was supposed to be generalized.
Bro just fold the paper man
How would I continue the construction if BP is the tangent at point B? This would force B and R two be the same point and than the construction afterwards no longer works
This makes me scared of what’s gonna happen later in my geometry class…
You gotta give me more time to consider that next time
holy crap I actually did it, kind of... I did it in GeoGebra, but I kind of cheated via software. I did half of it though if that means anything, even if I can't prove it
With the ruler, can I draw a line from P that is tangent to the circle ?
Or is it against the rules ?
me just using the lines on my ruler to make it good enough perpendicular
yes, my teacher hated me
Doesn`t work for an ARBITRARY point P though. Because there`s no solution if P lies in AB (at least as far as i can tell)
Please, don't. Just elongate AB, put center of compass on P and find B' on AB. Next, by drawings 2 circles from B and B', find P' and PP' is Your answer.
compass is not permitted
@@WrathofMathdammit, You are right, the whole "construction" motive confused me
I'd probably cheat and fold the right angle corner of the paper to construct it instead of the ruler.
If they told me no paper I'd just say "No paper, no problem. Therefore there's no problem which requires a solution" and then walk away mysteriously into the mist (smokebomb optional).
Can't you just continue the diameter to any arbitrary point, then use the compass to make 2 intersecting arcs (one from the point and the other from the point of intersection of the original arc with the diameter), and that will pass through the middle of your small segment while being perpendicular to it?
What compass?
It seems the puzzle is either underspecified or your solution misses the case that the arbitrary point P happens to be located on the straight line through A and B.
Cant you just lengthen AB infinitely, mirror P on this line into P', and connect P with P'?
Clip would be 30s long and would not require using words like "orthocentre"...
"...mirror P on this line into P'..." You can't magically find P'. You have to find it's position with a construction of lines where you are guaranteed that the line PP' is perpendicular to the line AB.
@@Wakans orthocenter is a common geometry term tho
@dominiquefortin5345 no magic required, just use the calipers and mark 2 random points on line AB with calipers centered on P. Center the calipers on each, and the 2 circles drawn would intersect in P and P'
@@F1R3S74R73R "... use ONLY a simple ruler." Where does it say calipers ?
What if P is on extention of AB?
The is a solution when P is on the line AB but not between A and B. Look for the answer of @hughcaldwell1034 on one of my comments.
wonderful music choice
The diameter AB (scalar here) does not depend on the direction and is the same at any angle.
A and B are simply opposite points.
You simply connect the point O with P in a straight line and this line will be perpendicular to the circle and therefore also perpendicular to the diameter.
The problem does not say that you have to take the orientation of line AB as it is drawn.
Diameter is also the wrong term here, you should have taken the term vector given by A and B, then the orientation of the diameter vector would make sense as a prerequisite and thus the solution shown to construct a perpendicular vector to the vector AB.
And I don't see that the problem is solved.
A perpendicular vector on the vector AB, this vector is defined from A to B and does not lie outside the circle.
why can't you just extend the line AB, then draw a line down from P. then erase the excess lines
Wrong. Extend AB. Use the fact the ruler is rectangular to draw a perpendicular from that extended line towards P. Gg.
"...the fact the ruler is rectangular.." Is it really ... a perfect rectangle ? and how can you guarantee that you will align it perfectly ?
@dominiquefortin5345 it would be called a straight edge if doing traditional geometry. Most rulers like 95+% are rectangular. It can easily be verified by drawing a square with it, and then showing the diagonals and midpoint intersect at the same point. Check and checkmate.
You are still wrong in saying his solution is wrong. His solution is right, and much more rigorous than your supposed solution.
@Zacristanda wrong. This solution which uses a perpendicular connected straight edge as he used the word ruler, is equally as rigorous and requires far less steps. Making it superior. And inferior solutions are wrong as being "the solution" it is merely "a solution ". Flawless victory.
@@gregorymorse8423 "...Check and checkmate..." Wow! Are you still in middle school? You have never done classical geometry in your life.
2:06 isnt it perpendicular bisector instead of altitudes?
Both are true. The point of intersection between the perpendicular bisectors is called the circumcenter
@@prp1epr1ncezz ohhh. confused the orthocenter and circumcenter lol.
It's a good one! But what if P lies on the line AB?
If P lies of AB itself, then the diameter would be the sum of, say, AP + PO + OB = AB
@@PriyanTekchandani-t6i I said the _line_ AB. That is, it can be outside of the _segment_ AB. But even assuming it's inside, how does this help us create the perpendicular?
@@IvanToshkov I apologize but I am too inexperienced in math for this. Could you please explain what you are saying?
@PriyanTekchandani-t6i no need to apologize :)
What I'm saying is that the construction described in the video won't work if P lies on the AB line.
@IvanToshkov Look at the answer given by @hughcaldwell1034 in one of my comments below. The answer is for P being outside the circle on the AB line.
Pausing video to think this through. Um. I'd use the dot product to project one vector onto the other.
Do we even have standard vectors? Do we have angles to make a polar coordinate system?
I'm not clear on what the rules are or what methods we can use to get data... Let me read the instructions again.
Ugh. "Difficulty level medium"
Let's blame the hangover. I mean can you measure? I guess so. DB is a vector and I have the mag. OP is a vec and I can make an intersection OC for another unit vector if I make the unit the radius. But where's my basis? Don't have one... What about P to the tangent of the circle? Then I can intersect that with the origin. Is that a right angle? Nah.
Ok, now I'm depressed. Might be the hangover.
Ok, just unpaused for a few secs. I literally was discussing Thales yesterday with someone as an example of an early proof. Zero excuses. Hey ho.
How the fuck did you get magnitudes and dot products involved this is 2 dimensional euclidean geometry
@WrathofMath Also if PB happens to be perpendicular to AB, how do you construct the proof ?
If PB is perpendicular to AB, then you're already done by definition -- you have found the line perpendicular to AB that passes through P. If you need more evidence, consider that PB must be tangent to the circle, because otherwise it would have intersected the circle in one additional point. And all tangents to a circle are perpendicular to the radius at the tangent point.
@@codahighland This is Math. It is not enough to think it looks like a tangent, you need proof.
@@dominiquefortin5345 The proof is that the line doesn't pass through another point on the circle. This is the definition of a tangent. In compass-and-straightedge constructions, you are assumed to be able to identify intersection points perfectly.
@@codahighland Yes, you can create a tangent PB' but without proof, you can't say that B' = B.
@@dominiquefortin5345 There is literally no problem, just continue the construction from the video. You are imagining problems that don’t exist.
just use the short side of the ruler
(edit was because i said flat instead of short which makes no sense)
I have to use only two construction.
10 hoot and a half's out of 10
appreciate you!
extend segment AB, make a circle from point P with a radius larger than the distance between point P and line AB, construct 2 equally sized circles from the points where the circle meets the line, create a line through point P and the point where the two equal circles meet. easy problem
edit: aw fuck there’s no compass
Class 😎
Extend AB
Contruct the perpendicular of P and AB
Was that wrong?
only a little
I mean, you can do that, however, how do you show that the line you constructed is perpendicular to AB?
@@renka-chan9213 we use a cheap trick in school. A ruler has 90° angles built-in.
The question doesn't say "use a straight edge". It allows a ruler. Fault of question. Why use more brain when you don't have to?
@@adorp It only allowed a ruler, not because to measure the sides nor to measure the angle, but because there is no ideal straightedge in the real world (an ideal straightedge is one without markings, and has an infinite length)
@renka-chan9213 well, the question is theoretical anyway. In practice, you will not get a more accurate drawing using Thales and Orthocentre instead of simply using the ruler's corner.
When doing genoetry in school, I noticed that the official perpendicular drawing methods actually introduce more room for error than anything else. Simply using a set square is way more accurate. Geometrical drawing is theoretical.
Why on earth do you call a ruler a ‘straightedge’? What other edge would a ruler have?
straightedge is just what it's called in this context
Yesss, i love geometric construction!
I would really just use the numbers on the ruler as a right triangle ruler 📐😂😂😂😂
That was my thought as the wording of the puzzle said “simple ruler” rather than “straightedge”.
I used other method: use the ruler to measure length of BP then draw an isosceles triangle BPC which has BC is extended from AB. After that, extend BD from BP so that BD = BP, extend CE from PC so that CE = PC. BCED is an isosceles trapezoid. F is the intersection of 2 diagonals of the isosceles trapezoid. FP is perpendicular to AB
Ruler ≠ straight edge
Can I use two rulers. 😅
i don't get why the green line "must" go through B, just because the other two did...?
yes, because the green line we may define as the third altitude of the triangle, which comes from the vertex A. The other two altitudes of the triangle met at B, and it has to be that all altitudes of the triangle meet at a common point, so this third altitude (the green line) has to go through B as well.
@@WrathofMath ah so there's a law about all three lines passing through that point? Makes sense.
I'm not very versed in all this stuff, great video though!
there is both a way easier and way faster method that creates less visual clutter.
extend AB out some distance. place the end of the ruler on the line (make sure its line up) and move it around until you can draw a line from P to line AB.
The title says ruler, if you know the difference between a ruler and a straightedge, that's easy.
Sure, a ruler is a king or emperor. ;-)
A ruler has marking to measure lengths, as a straight edge just guarantees to be straight.
Why his he wearing the V O I D
"Nerds" BS. This was way too simple to think, didn't even bother watching the whole video. Or... I am just a nerd 🤓.
Zora River music ftw! BotW or ToTk?
5th I guess, anyways, cool math!
Please get a straight-edge that’s not also a ruler.