LOL now every time a math question with a diagram has "not drawn to scale" on it, I'll remember this video and be like :thinking:
@jimdecamp7204
6 жыл бұрын
The danger in life is not a math problem with a set solution. It's when you sketch a diagram to help yourself understand something. Examine your assumptions. But first be aware of your assumptions.
@brambl3014
6 жыл бұрын
First thing I notice is that line from E to CD is not perpendicular
@Mixa_Lv
6 жыл бұрын
🤔
@fakepistonnotrealuser9
2 жыл бұрын
KZitem really just recommended everyone this lol
@JLConawayII
6 жыл бұрын
You made it pretty obvious with that janky "perpendicular" bisector. You can just eyeball it and see that they meet below line AB.
@eran0004
Жыл бұрын
Meeting below AB doesn’t in itself invalidate the proof. The proof fails because they meet so far below AB that the line z on the right side falls outside of point B. Had z been inside of point B then the proof would still hold (although with some adjustments to the algebra, you’d have alfa-beta = 90 and 100 instead of alfa+beta = 90 and 100). It would have been much harder to spot the problem if the figure was drawn with the intersection already below AB, it could be an interesting variation of the problem :)
@hqppyfeet7513
Жыл бұрын
that perpendicular bisector actually really annoyed me
@jemandanderes7075
9 ай бұрын
Yes, but only for hte case where it specifically goes across B. @@eran0004
@user-vn2xq9wb1d
6 жыл бұрын
When you said "the picture is not drawn to scale " specifically , it kind of gave away the idea 😂
@aashsyed1277
3 жыл бұрын
Yeah
@wombat4191
2 жыл бұрын
It really didn't, or at least shouldn't, because not drawing a diagram to scale is a very common thing to do either to better demonstrate a problem, or because it doesn't really matter. And it should always be mentioned. The fact that in this case mentioning it was pretty imperative to the problem doesn't mean that it was anything out of ordinary.
@rudy3396
2 жыл бұрын
@@wombat4191 it did because he’s done things like these before with images not drawn to scale
@vratha9740
2 жыл бұрын
Yeah
@eomoran
2 жыл бұрын
@@rudy3396 right however no diagram is ever “to scale”
@Shonicheck
2 жыл бұрын
The first thing i check when someone says "lets draw something" is if your drawing is representing the real thing or if you just have eyeballed it. Learned it the hard way back in school when i was participating in math olimpiads, did too many wrong assumptions based on my incorrect drawings - non-existant intersections are one of them
@kishtarn555
2 жыл бұрын
Have you heard of the wide dot theorem? All three lines intersect in a single point if you draw the point wide enough.
@asr2009
Жыл бұрын
@@kishtarn555 lol.
@samuelbruyneel
6 жыл бұрын
0:43 It’s very obvious that line is not drawn perpendicular at all...
@olaf98
6 жыл бұрын
"Diagram is not to scale"
@jaagup
6 жыл бұрын
But what is drawn doesn't matter.
@mattgsm
6 жыл бұрын
But they should at least make the diagram similar to the scale. It may not be TO scale but it should be appropriately drawn first
@TheBigRedskull
5 жыл бұрын
Exactly
@TheBigRedskull
5 жыл бұрын
BossProGamer Should have had to 90 degrees symbols on either side of the line. Therefor not perpendicular
@rybiryj
2 жыл бұрын
This was pretty hard for me. My first thought was that E should be outside of ABCD, namely below the line AB. But that doesn't fix the issue! Only after noticing that E can be even further down - so far that it is to the down-right of the line BC, the problem was fixed. Also, I love when someone describes their diagram as "not to scale" when the phrase "totally botched" would be a much better description xD
@user-wi6uf3xy6n
5 жыл бұрын
90% accuracy is good enough for most things.
@PuppyPrincess
2 жыл бұрын
I like your username xD
@shayanazizi1907
2 жыл бұрын
90 = 100 😀
@archturusdeydas3989
2 жыл бұрын
@@PuppyPrincess 142 likes but the last comment was just two days ago! Wow indeed
@thereaction18
2 жыл бұрын
90 = 100 just like 0.999... = 1. Obviously. It's just a matter of significant digits.
@thereaction18
2 жыл бұрын
You're just jealous that Donald J Trump is hung better than you.
@aritradey8270
6 жыл бұрын
Can it be mathematically proven that the perpendicular bisectors have to meet exterior to the quadrilateral..?
@rohitharapanhalli5584
6 жыл бұрын
Yes. I did it by finding the equation of the perpendicular bisector to AB (x = a/2 for a being the length of AB) and for the line CD, and plugging in a/2 into the second equation. Then, I showed that it lies below AB by letting A be the origin (0,0) and B (a,0), and demonstrated that the second equation yields a value less than 0.
@Josh8far
6 жыл бұрын
I assume so given the information that two sides are of equal length.
@stevethecatcouch6532
6 жыл бұрын
The video itself is a proof by contradiction that the bisectors don't meet in the interior. Rohit Harapanhalli's idea of using linear algebra to locate the intersection is more direct.
@camembertdalembert6323
6 жыл бұрын
PeasantZilla it can't be a square because one angle is 100°.
@sukhvinderbura1572
6 жыл бұрын
Raphaël D yes
@Marnige
2 жыл бұрын
... I was literally like... But the diagram makes no sense! The top one doesn't look at all like a 90°! All the concepts seem right but I was convinced it was an impossible shape. My test method would've been to try and draw a shape that meets all the criteria. Im glad to know that the diagram itself WAS the issue.
@bluerizlagirl
6 жыл бұрын
I knew it was something to do with the "not drawn to scale" warning! It's an impossible drawing as you have it in your video. BC is twisted clockwise with respect to AD. And CD is longer than AB. The midpoint of CD must therefore lie somewhere to the _right_ of the midpoint of AB. So wherever E is, it must lie somewhere to the right of the midpoint of AB, and therefore outside the figure ABCD.
@gazzawhite
4 жыл бұрын
It's not just that the intersection point lies outside ABCD, it's also because it lies on the other side of line BC extended.
@quintopia
2 жыл бұрын
paused at 2:17 to solve. result: alpha=175, beta=85, mistake is not so much that E is outside the quadrilateral but rather that segment EC is outside the quadrilateral. Thus, the relevant equations are alpha=beta+90 and alpha+beta+100=360.
@jasondeng7677
2 жыл бұрын
even in the more accurate diagram 3:45 those "congruent" triangles don't look congruent. i guess it's just not a perfectly accurate diagram... also i wonder what would happen if the currently 100 degree angle increased or the length of the side on the bottom was shorter, because you could definitely make the intersection inside the quad
@midnightposting
6 жыл бұрын
In some cases, if the intersection point is below the line AB you can still obtain the same contradiction. What you actually want is the point of intersection to be below the line BC
@reggietheanarchistrat8805
2 жыл бұрын
I’m at 1:50, And I’m guessing now, this would only work on a parallelogram, not just in a quadrilateral
@jimdecamp7204
6 жыл бұрын
(I haven't looked, yet) The mistake is the assumption that the intersection of the perpendicular bisectors lie inside FIGURE ABCD. In fact, they intersect below line AB. The point "E" is below AB on the diagram. Presh blows some smoke in our faces early by saying that the figure is not to scale. He piles on by saying "we will mark these sides 'z'" to lead us to question whether or not the two z's are equal. They are. I do not know if that was intentional. And the con continues from there. Notice how "wonky" the "perpendicular" of bisector of CD looks at 1:14. The "perpendicular" containing E and D is clearly smaller than the one containing D and C. Let A be the origin of a Cartesian coordinate system. Let the AB be along the X-axis, AD along the Y-axis. To avoid confusion rename length x = 1 = |AD| = |BC|. Not knowing length |AB|, or its size relative to x, one cannot explicitly solve for the location of E in terms of AD and AB, but one can solve for it parametrically. It turns out that the value of the y coordinate for the intersection is about -5.715... times |AB|, expressed in units where x = 1. (After seeing the result: Don't bother to scale off the figure at 3:20, which is "more accurate", but still not nearly to the correct scale.)
@jimdecamp7204
6 жыл бұрын
I got the same result, but actually "proved" that E lies below AB, which Presh correctly asserts. At this point he has done his job, of pointing out the mistake, and in a popular exposition is just about right. I was gonna make the same point as Presh, take pains to identify and verify your assumptions. Identifying your assumptions is generally more difficult than verifying. I say this from the perspective of an engineer with years of experience in testing complex systems. Presh only demonstrated it for one value of |AB|. I verified it parametrically in |AB|. Result is the same for |AB| negative. For |AB| = 0, A = B = E = origin, which the formula correctly computes as y =0.
@MindYourDecisions
6 жыл бұрын
Fair point! The only inaccurate part not to scale, is the angle is more like 120 degrees. The general fallacy is to prove "any obtuse angle is equal to 90 degrees" but I felt a concrete and terse title like "90 = 100" would be better. It would be a fun extension to prove point E is "below" ABCD like is illustrated in the video.
@rmsgrey
6 жыл бұрын
Just lying below AB is not sufficient to wreck the proof. If you extend AD and BC to intersect at G, then if E lies within triangle ABG you can fix the proof by looking at the differences of the two angles rather than their sums. For the two angles to be unequal, E has to lie below BC.
@seanocansey2956
6 жыл бұрын
Smert boi
@jimdecamp7204
6 жыл бұрын
I agree. The slope -5.715.... is tantalizingly close to the -tan(80), but not quite. I just solved it numerically using about 10 rows of Excel and lost interest. If it were important, if the results of a million dollar test depended on it, I would dig deeper. Some tests do cost a million dollars, or much more, and are worth it.
@biggbarbarian224
6 жыл бұрын
I would like to mention that E lying below ABCD is not sufficient here. The crucial part is that E lies far enough below ABCD so that the triangle BCE flips otherwise you still get the wrong "proof".
@bobzarnke1706
3 жыл бұрын
This is similar to the "paradox" that all triangles are isosceles. This is "proven" with a diagram that shows the angle bisector of one vertex of a triangle meeting the right bisector of the opposite side inside the triangle. If A is the origin, B is (b,0) and angle ABC = θ then the right bisectors of AB and CD meet at (b/2, b cos(θ)/(2(1-sin(θ))), which is below AB since cos(θ) < 0 and sin(θ) < 1 for π/2 < θ < π.
@FireyDeath4
2 жыл бұрын
What really bothers me is this. Can you change the 100° angle so it's closer or further to 90°, so that the intersection point actually falls within the shape?
@JacksonBockus
2 жыл бұрын
The answer is no. I played around in Desmos, and for any angle other than 90 degrees, the intersection falls outside the shape. (And of course for an angle of 90 degrees, the two perpendicular bisectors are the same because the shape is a rectangle.) In fact, this video serves as a proof that the intersection falls outside of the shape, since assuming the intersection falls within the shape leads to a contradiction.
@vizender
2 жыл бұрын
@@JacksonBockus I guess it has to do with the definition of 4edge figures, whereas triangles for exemple all have the intersection in the middle.
@JacksonBockus
2 жыл бұрын
@@vizender not necessarily! On a triangle, the intersection of the sides’ perpendicular bisectors is called the circumcenter, and for an obtuse triangle the circumcenter lies outside the triangle.
@vizender
2 жыл бұрын
@@JacksonBockus ok thanks 8
@jbtechcon7434
6 жыл бұрын
To quote XKCD, "Communicating poorly and acting smug when you're misunderstood is not cleverness."
@onradioactivewaves
3 жыл бұрын
"Always use radians, never degrees" -me
@broes
4 жыл бұрын
1:25 is not true. BAE is ABE, but BAD is not ABC. So EAD is not EBC
@binhanh296
Жыл бұрын
Spotted the problem right at 0:44 where the question drew the perpendicular bisector of CD, that line does not look like a right angle at all, but the question said that it is a right angle.
@aarcaneorg
6 жыл бұрын
Hey Presh, a while back I posted a complaint that your talking speed was getting incredibly, condescendingly slow. I'd like to report that after a few months' time, I've noticed that your pacing has improved muchly and I no longer feel talked down to. Thank you! I love the content, and keep up the good work!
@cannot-handle-handles
6 жыл бұрын
Still feels comfortable to watch this at 125% speed, though. :-)
@StRanGerManY
6 жыл бұрын
Are you a native speaker by any chance? Or just fluent in english. In this case i might see why it was annoying; however, consider people with intermediate english skills. They would appropriate slow and accentuated pronounciation, simular to the one used in kindergardens. Maybe he cared for the international audience? I'm pretty sure that was the case
@jaysethi733
6 жыл бұрын
‘Condescending’? We’re just throwing in random adjectives without knowing what they mean right now, aren’t we?
@simplebutpowerful
6 жыл бұрын
Agreed. The pacing & tone are improved which makes these videos even more enjoyable. Good work, Presh!
@PafiTheOne
6 жыл бұрын
Did you have enough time to search the triangles he was talking about, and to confirm they were really congruent?
@cmilkau
2 жыл бұрын
E is actually outside the quadrilateral, but that only changes the sign of β.
@timothy8428
6 жыл бұрын
I got it but not before having to draw it in Inkscape to find the problem.
@ayaghsizian
6 жыл бұрын
I thought point E might lie on line AB and that was the con. Under line AB makes more sense in hindsight.
@willmunoz1638
6 жыл бұрын
Paused at 2:38, my guess is that the perpendicular bisectora dont intersect inside the quadrilateral.
@mioszantas4462
Жыл бұрын
Brings me back to high school where the teacher constantly reminded us that the drawing cannot suggest anything
@colinslant
2 жыл бұрын
"Diagram not to scale" = "Diagram is deliberately misleading".
@jessehammer123
6 жыл бұрын
We did this puzzle in New York Math Circle HSB2 this summer. Thanks Misha!
@guyonYTube
Жыл бұрын
the interesting thing here is that the "proof" still stands when E is outside the triangle. i drew the diagram such that part of the line CE is still in the quadrilateral, and the proof stood. The actualy diagram had the line CE not even touch the quadrilateral at all(except at C)
@michaelweiske702
2 жыл бұрын
Interesting problem, but my guess is that one of the premises must be incorrect. Perpendicular bisectors are equidistant from the parent points, so that's not it. Congruent triangles have congruent angles, so that's not it. One thing we did take for granted, though: that E exists as we see it. It could be possible that E lies outside of the quadrilateral, which would make the the triangles fall outside the quadrilateral, and thus they can't be fully contained by the corners.
@velosobruno
2 жыл бұрын
Yeah, right from the start… hey, that,s not perpendicular, whythe square angle sign???
@TheEulerID
3 жыл бұрын
Where "not to scale" means intentionally misleading. The intersection of those two perpendicular lines bisecting the sides will be outside the quadrilateral. Now to watch the rest of the video, but I think it's pretty obvious where this is going to go. Usually these sort of things involve making an unwarranted assumption.
@Zollaho
2 жыл бұрын
Yes. This is rather childish.
@QuasiELVIS
5 жыл бұрын
I immediately thought it was a rough looking right angle at the top in the original sketch.
@eris4734
2 жыл бұрын
interestingly, this proof would still work with E in a small range outside ABCD, but because it passed the continuation of line segment BC, it ended up flipping the right side triangle, invalidating the proof.
@brianneill4376
Жыл бұрын
The "Difference" is the same difference in these 2 Symbols "~" & "=", one means approaches and touches the next "Equivalent", the means that the bodies in measure are the same weight and or size or each other being duplicates or redundancies of the other 1~1=2, 1~1=0,-...>|0|
@cmilkau
2 жыл бұрын
E is on the wrong side of (BC). So angle ABC is NOT α+β.
@fred8780
6 жыл бұрын
despite the top perpendicular not being drawn perpendicular, the right base (100) angle is 10 degrees more than the left base (90) angle. So the left "a" angle does not equal the right a (that angle is a+10). If the right side and left side triangles do not have the same angle values which were both marked as "a", then they cannot be congruent. In addition, it is obvious, that at point E, there are no vertical angles.
@m10domedia
3 ай бұрын
I looked at the top and I was like... "I trust that right angle marker about as much as I trust a hungry polar bear to not kill me."
@Khalid.115
2 жыл бұрын
Good think I didn't forget that in a quadrilateral, opposite angles sum up to 180°; and that top right angle definitely does not look like a right-angle. It's an impossible shape, basically.
@markou2376
2 жыл бұрын
Can you find the mistake? Me: **looks at the mirror**
@tjseagrove
6 жыл бұрын
I saw it when you did it but then you removed it so it could be forgotten. As my math teacher said...”show all work”. 😬 Great as always!!
@MatheusFreitasOrangeMaths
2 жыл бұрын
This proof is actually a proof by contradiction that the intersection is outside of the figure.
@silverdragon2462
6 жыл бұрын
Had this been in a test setting or something, how could you know/prove that point E is in fact below AB? In the first, wrong, diagram, I noticed immediately that the perpendicular for DC was off, just by eyeballing the fact that it is was definitely more than 90 degrees. But is there a way we can prove that (possibly using the fact that the two sides are equal - x)?' Please reply, I would be interested to know :)
@rmsgrey
6 жыл бұрын
You don't need to prove that E is in fact below AB (and, more importantly, to the right of BC) - to answer the question posed, you only need to show that if E is not between AD and BC then the proof fails - the onus would then be on the person proposing the proof to show that E is in the region that gives a paradoxical result.
@silverdragon2462
6 жыл бұрын
that is true, but i am still interested in how you would know it is either below AB or the right of BC. I know that it is not necessary (and it is the person proposing the proof's job), but still interested
@stevethecatcouch6532
6 жыл бұрын
If your were asked to prove that E is not in the quadrilateral, assume that it is inside and proceed to show that implies 90 = 100. The contradiction means your assumption was incorrect and that the intersection is not inside the quadrilateral.
@indiaglowing
6 жыл бұрын
You could mark A as the origin (0,0), B as (a,0) and proceed to work with appropriate coordinates for other points and form equations of lines and end up solving those to show that the intersection of the two perpendicular bisectors lies below the line AB (the X axis)
@stevethecatcouch6532
6 жыл бұрын
Daas, yeah. The coordinates for C are (a + .17x, .98x). The y coordinate for E is ugly, but I think I convinced myself that it was negative.
@hirotrupp
Жыл бұрын
The perpendicular bisectors would never meet within the object
@evanbelcher
6 жыл бұрын
This was a tricky one, I liked it
@kenhaley4
6 жыл бұрын
Presh, I think the error might have been less obvious (and more challenging to find) if (1) you hadn't told us "the diagrams are not to scale" and (2) if you placed the point E much closer to the bottom of the rectangle (but still inside), so that the supposed perpendicular bisector of DC looked more correct. In your diagram the DC "perpendicular bisector" was clearly NOT perpendicular, so I could immediately see that by straightening it out in my mind, the point E would be outside the quadrilateral.
@Loganl1980
2 жыл бұрын
I figured it out as soon as that 90 degree symbol was put on the top line WAY off from square. I mean, it's a geometry problem. It was glaringly simple.
@daviddejong3894
6 жыл бұрын
@MindYourDecisions and @ Everyone that has mentioned that E must be below the BC line, or asking for a proof that E lies below AB that does not use the question's contradiction or an impression from drawing the diagram. I asked the same questions. I think there is a geometric proof that E lies below AB and also one that E lies below the extension of BC. Also, apologies ahead for my lack of adherence to conventional notation, and perhaps making some leaps along the way, not invoking the right theorems, etc.... Also I reposted this to be a comment on the OP instead of a reply to someone else's comment. 1) Because E is an intersection of bisectors, it is also the common centre of 2 concentric circles of radii y and z, respectively. Let us call them Circle Y and Circle Z respectively. A and B are points on Circle Y. C and D are points on Circle Z 2) For any point F on the Circle Y there are only two points (let us call them G1 and G2) on Circle Z that are exactly distance x away from F, given that both x > |z - y| and x < y+z. You can see this by drawing out a circle of radius x centred on any general point F of one of two non overlapping concentric circles. 3) Any EFG1 and EFG2 must be congruent to all other EFG1 or EFG2, as they have the same length sides. They can all be pictured as the same triangle rotating around its corner E, or as a reflection of such a triangle across a line passing through E. (I think you can prove the latter by showing that all EFG1's are a reflection of its corresponding EFG2 over the axis of the EF line, and then that all rotations to an EF'G2' will result in a triangle that is a reflection across an axis that has rotated half as much around E. This axis would also be the common bisector of G1 G2' and F F') 4)EAD and EBC must be of that set of triangles. 5) We can eliminate the case that a reflection is involved with the transformation from EAD to EBC. If there were a reflection involved, then the bisectors of AB and DC would be a common line. So, they must be only of the rotation case. (Probably need a proof for this one, but I think it follows from 3's proof sketch) 6) As Triangles EAD and EBC must be rotations of each other around E, then the difference of 10 degrees in orientation (angle? not sure the best term here) between the segments AD and BC must be equal to the rotation around E that can transform EBD to EAD. There is no other means to have changed that angle with only rotations around E. i.e. it is a 10 degree rotation, so angle AEB and Angle DEC are both 10 degrees. 7) the bisectors of DE and AB split these angles DEC and AEB respectively in half. i.e. Let P be a mid point on AB. The Angle PEB = (10 degrees)/2 = 5 degrees. Similarly: For Q as a mid point of DC, Angle QEC = 5 degrees 8) I suspect there are far more elegant proofs to show that E is "below" AB, but this is what I came up with: 8A) Consider the quadrilateral A'B'CD with the following properties: it shares CD with ABCD of the problem, |A'D| = |B'C| = x, but A'B' is parallel to DC. Note that the bisectors of DC and A'B' would be a common single line, so the P' of A'B' would be on the same line as QE. Note that interior angle of A' = interior angle of B' > 90. i.e. they are the same obtuse angle. 8C) Let us now create a distorted version of this quadrilateral A''B''CD, such that CD is again in common and all side lengths remain constant. We can control the distortion of A''B''CD by rotating side DA'' around D to move A'' farther away from C than A'. As we increase angle A''DC (i.e. the interior angle of D with C and A''), then the following will also happen: * interior angle A'' will shrink towards 90 and A'' will move on a path from A' towards A, * interior angle of B'' will get larger. * Both A'' and B'' will move to the "left", but, * in terms of perpendicular distance to DC, A'' will move away until interior angle A'' becomes 90 and B'' will move closer, * A'' and B'' will move over A and B when interior angle A'' = 90, and A''B''CD will be equal to the original ABCD. Thus A'B' will have rotated in a counterclockwise fashion, as well as translating in a "left" ish direction 8D) Let P' and P'' be the midpoints of A'B' and A'' and B''. With this distortion, P'' must move to the "left" as well away from P' and towards P. thus, the A''B'' bisector has also moved to the "left" of the QE bisector line, at least in a space close to P'. As the A''B'' bisector is perpendicular to A''B'' it must also rotate counterclockwise, the same rotation direction as the A''B'' segment. In other words, above the various AB lines, the A"B" bisector is getting perpendicularly farther from the CD bisector, while immediately below the various AB lines, the A''B'' bisector is getting closer The only candidate for an intersection between the two bisectors after the distortion (which must exist, as they are not parallel) must lie on the other side of the segment A''B'' from DC, or "below" it. As ABCD is a special case of A''B''CD, E must be "below" AB. Now, for proving E is below the BC line: 9) Note that if we extend BC to cross the bisector of AB, and define the intersection as point R, then Angle PRB = 10 degrees 10) As E, P and R are on the same line by construction, and E is also below AB by 8) the only further construction that allows Angle PRB > Angle PEB requires R to be between P and E. 11) therefore E must be below the line that is the extension of BC Several of my steps involve eliminating all other possibilities so that one must remain, which is pretty tenuous I think, but this still feels satisfying enough to me to answer the questions "how can we prove E is Below the extension of BC?" and "How can we prove that E lies below AB without using the question's contradiction or an impression from drawing the diagram?"
@stevenwilson5556
6 жыл бұрын
I had to draw this figure out myself to see the problem. The perpendicular bisectors are almost parallel and do not intersect inside the figure, so the "not drawn to scale" is indeed rather tricky since what is described in the video is simply impossible. If anything this proof indicates that the intersection cannot lie inside of quadralateral ABCD.
@user-wr8yg4eo9i
Жыл бұрын
This problem can also be simplified to a triangle with a bisector of an angle and perpendicular bisector of the opposite side. Follow the mistake and you’ll prove every triangle is isosceles
@bkucenski
3 ай бұрын
You really had to stretch to call that line on DC perpendicular. "Not to scale" was just a way to excuse an angle that was clearly not drawn at 90 degrees.
@robair67
Жыл бұрын
Looks just like a typical race car roll centre diagram, but on its side. One of the few problems on here that I actually managed to spot instantly!!!
@elSethro
Жыл бұрын
I figured it out. I realized that the perpendicular bisector for AB is parallel with line AD (since both are perpendicular to AB). Since the two lines are parallel, A and D are both the same distance from the perpendicular bisector of AB. We know that B is also the same distance too, so A, B, & D, are all the same distance from that line. But if you look at any point along the bisector of AB within the polygon, there is no point where C & D are equidistant to that line: C is always farther away since it doesn't line up with B the same way D lines up with A. Therefore, the intersection point E must be outside of the polygon.
@tcfs
2 жыл бұрын
At 0:45 the perpendicular for DC looks so distorted, that rang my bell...
@severindusell2468
2 жыл бұрын
I didn't exactly get that the point of this was to demonstrate that you should be careful of where to assume a constructed point is, but I knew something was wrong when you said that triangles ADE and BCE were congruent when they are constructed of entirely different angles.
@AhsimNreiziev
6 жыл бұрын
I'm not certain, but I *think* that the relatively small difference between 90° and 100°, resulting in the positively *minute* downwards slope of CD, makes it so the intersection of the perpendicular bisectors will always lie outside of the quadrilateral ABCD.
@Tiqerboy
5 жыл бұрын
The bisector of line CD meeting at point E is not a perpendicular bisector. It can't be because CD > AB.
@imflaq7145
2 жыл бұрын
It's 3 am, KZitem suggested me this and now I'm here watching this with my potato brain
@ecyranot
2 жыл бұрын
I must be missing something. The first one is the sum of the horizontal and vertical moves of the stairs. The 5 is a diagonal line so of course it's shorter. Right?
@marcandreservant8824
6 жыл бұрын
On the diagram it was obvious visually that the perpendicular bisector of CD was not actually perpendicular. But I immediately realized that if you take the limit of the length AB when it tends to zero, you have an isoceles triangle with angles ADC = ACD. The perpendicular bisector of CD therefore goes through AB at point A. If the length AB is increased, then the bisectors must cross *beyond* segment AB, because the angle between them is narrower. This is true for any angle ABC between 90° and 180°.
@hardikkumarsrivastava878
6 жыл бұрын
Can you please upload a proper proof as to why they intersect outside? BTW love your videos always
@Gooey
4 жыл бұрын
Proof by contradiction The initial argument was that they intersected inside, and that lead to a contradiction so the initial argument is invalid. :)
@mahdiramezani8536
2 жыл бұрын
The exact problem with the first diagram is that E is drawn on the left hand side of the BC vector. If you draw E outside of the quadrilateral but still on the left hand side of the BC vector, you can still prove that 90 = 100.
@Tofukatze
2 жыл бұрын
"Because E is on the perpendicular bisector of CD, it is equidistant to C and D" my brain immediately: uhh...no it's not?
@IoEstasCedonta
Жыл бұрын
"Not to scale / mark the intersection" - oop, got it! ...I've been watching too many of these videos I think.
@fiercethundr_
6 жыл бұрын
“a” measures are not the same as each other (After looking): Ok I saw the bisector was off but didn’t want to complicate my answer. But still, going more generally “a” measure 1 and “a” measure 2 could not be the same physically
@the1barbarian781
4 жыл бұрын
Fierce Thunder they are the same
@TGMResearch
2 жыл бұрын
Next up: MYD draws a circle, which is actually a triangle only not to scale, and the question is why the angles don't add up to 180.
@utopes
4 жыл бұрын
1:03 aaaaaa E is obviously not equidistant to C and D yet the proof against this is far more complex than the simple answer I came to
@ryanlangman4266
2 жыл бұрын
I think there is another much more important point that you skipped over. This does show that CBE+FBE != ABC and FAE+EAD != BAD, so a+b != 90 and a+b != 90, but using the chain of logic described in the video you should still be able to show that CBE-FBE = ABC and EAD-FAE = BAD so this should show that 90 = a-b = 100. The problem actually comes from earlier in the proof where you assume that EAD = EBC. While this is true, the a that he wrote down was a reflex angle, so couldn’t have formed a triangle. The actual a that does form a triangle was 180 minus the a that he wrote down. This is where the real mistake came from.
@divyanshtulsian8194
2 жыл бұрын
the line is perpendicular but doesn't divide the line into 2 equal parts. If it does, it won't meet at point E @MindYourDecisions
@Lorenzo_der_Ritter
Жыл бұрын
oh yeah I knew that one. It's always suspicious if a supposedly 90° angle isn't 90° in a diagram, even if it's not to scale
@rupasarkar8276
4 жыл бұрын
Excellent way to clarify mistake we often make.
@OskarCzechowicz-OmniMusician
2 жыл бұрын
Ok, it enraged me a bit. I wanted to solve this only on paper, without use of a computer or geometric tools, I wanted it to stay on theory level. I had the idea of the bisectors meeting outside the figure, but didn’t know how to prove it without drawing it, which I considered a lazy solution. I was only able to find out that for the triangles to be similar, they’d need to be inside a rectangle and any other figure wouldn’t work, but this didn’t clash with any of the claims made in the question.
@WesternPuchuu
4 жыл бұрын
Is there any way to prove that, regardless the distance between the x lines, this E point is always outside of the quadrangle?
@orivandenhoven5283
2 жыл бұрын
When seeing the title I was like "oh just a case of dividinh by 0 or a-b where a=b" I couldn't be more wrong
@Melpheos1er
2 жыл бұрын
Well the diagram around 1:00 clearly shows the perpendicular bisector to DC is not perpendicular at all. It's not that it's not to scale, it's completely wrong
@lionbryce10101
6 жыл бұрын
Going to stop at 1:31. angle b was 100, a was 90. The triangle AEB has 2 sides that are the same length so the angles b - alpha and a - alpha have to be the same and 100 - x is not equal to 90 - x.
@SleepMastR
6 жыл бұрын
There's also the fallacy of the isosceles triangle: en.wikipedia.org/wiki/Mathematical_fallacy#Geometry
@s888r
2 жыл бұрын
I realised the mistake was in the perpendicular bisectors but didn't completely know the solution.
@Foro5Cibercafe
6 жыл бұрын
Statement at 1:18 is wrong. So i did stop the video from that point forward
@_maul_001
2 жыл бұрын
Is it necessary that every time the perpendicular bisectors do not intersect inside square , even at increasing 100 degrees further , if yes . Whyy
@idkwtvr4844
2 жыл бұрын
I think this problem was on the PROMYS application last year
@alexandreman8601
4 жыл бұрын
1:18 EAD and EBC don't have three equal sides.
@krj.
4 жыл бұрын
They do
@imshiruba
9 ай бұрын
90=100 9=10 1=10/9 1≈10/9, well I guess you should've put an approximate sign instead of an equal sign
@apollosun6268
2 жыл бұрын
The perpendicular Bisector of CD is not drawn correctly. I don't think the two perpendicular bisectors would intersect within the original quadrilateral.
@gdmathguy
Жыл бұрын
So basically: If you see the phrase "Diagram not to scale", immediatly, if possible, redraw it in true version
@robinlindgren6429
5 жыл бұрын
I would like to draw a picture to explain this, but in short, the intersection point E always falls outside (specifically below) the quadrilateral if the angle B is greater 90 degrees, in fact it always ends up so far outside that the CBE triangle does not overlap the quadrilateral and the CEB angle ends up not having any interesting relation to the internal angles of this shape at all, therefore trying to relate it to them in the way the video does is a fallacy.
@CDubs720
6 жыл бұрын
lol, 45 seconds in. The bisector of CD is not drawn at a right angle, even if it's not to scale....
@illesfleischman2814
4 жыл бұрын
Im from hungary too. It's nice to see that im not the only hungarian who likes math.
@kurzackd
2 жыл бұрын
yeah, but how are you supposed to know what the "real scale" is without the dimensions being given explicitly to you?
@raydorepp
2 жыл бұрын
I think you're supposed to use some math laws.
@AndreySvyat
5 жыл бұрын
Let mark middle point of DC as M and middle point of AB as N. So the perpendicular ME has no ways to be IN the figure
@hikariwuff
2 жыл бұрын
90 = 100 Divide both sides by 0. undefined = undefined Profit.
@gmutubeacct
3 жыл бұрын
Nice problem. This reminds me of something my classmate and I did back in the mid 80's. Might have been the same problem drawn differently. I feel your answer in incomplete. It is not adequate that E be exterior to ABCD. The key requirement is that the entirety of CE is exterior to ABCD. You can prove that if any portion of CE is (incorrectly) drawn interior to ABCD, you can show that ADC = BCD and hence DC || AB (because AD=BC), and hence their perpendicular bisectors are either on the same line or will never intersect. Hence CE has to be purely exterior to ABCD. That is the important part that is missed in your explanation. Thanks for sharing these.
@sriram3710
2 жыл бұрын
Yes. Drawing CE incorrectly to the interior of the quadrilateral would lead to proving plenty of wrong statements. But since it's given that DAB is right angle, there cannot be a scenario where the perpendicular bisectors never intersect. Either they are same line or they intersect.
@boopersindoopers
2 жыл бұрын
if you take away a zero from 90 and 100, you get 9 and 10, so if you multiply 9 x 10 you get 90, so 90 equals 100 because 90 and 100 are the same
@notwildcard377
6 жыл бұрын
1:05 the bisector of CD doesn't seem to be perpendicular. Looks like it was forced to be perpendicular.
@janeluooo
2 жыл бұрын
I think the two perpendicular bisectors actually meets outside of this quadrailateral
@kmarasin
2 жыл бұрын
The answer is literally outside the box
@matthewsaulsbury3011
2 жыл бұрын
I noticed the mistake not too long after he drew the bisector from the top side. Great video by the way! 👍🏼😀
@HotelPapa100
2 жыл бұрын
It became pretty obvious when you drew the second bisector clearly not at 90°. I immediately thought: "E is outside the quadrilateral!"
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