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12 Aug Should I study Mathematics at university? Dr. Andrew Lobb



The Maths Faculty – University lectures for secondary schools
For more videos, go to www.TheMathsFaculty.org

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MOST RECENT COMMENTS
21 Comments
  • Lorenzo Panetta
    Posted at 07:47h, 12 August

    If the angle bisector and the perpendicular bisector are not the same lines. Then the top left triangle would have a smaller/larger base then the right triangle. Also the sine rule only applies to the same triangle I think and used the sine rule assuming both triangles were the same which is what he is trying to prove in the first place.

  • Hristina Nikolic
    Posted at 07:47h, 12 August

    When you look the last pair of triangles, they are not the same and the sides that are part of the big triangle are not equal (sorry for bad english explanation)

  • Giorgos Zafiris
    Posted at 07:47h, 12 August

    He assumed that the angle bisector and the perpendicular bisector of the opposite side meet inside the triangle.
    In reality the two lines will meet in the midpoint of the arc "BC" ,of the circumcircle of the triangle (the arc which does not include A that is).
    I know this as the "South Pole theorem"
    When the triangle is isosceles the angle bisector and the perpendicular bisector are the same line, thus every point on them is a "common point".
    I was bothered from the start when he drew the intercection inside the triangle, i kept thinking, "that's not supposed to be there!" xD

  • Kevin Smyth
    Posted at 07:47h, 12 August

    This was posted three years ago but I have just found it.
    An ingenious idea but I am not sure if it is really true. I have yet to see a convincing construction of the Irrationals (either in a book or on a you tube vidoe). This worries me but does not seem to worry many mathematicians. I have to admit, though, I passed my maths degree almost 50 years ago in 1969 and can honestly say I did not understand a word of elementary real analysis at the time. I have recently looked at the subject again and would say I understand most of it but just don't believe it.
    I think most people contemplating taking maths would see the flaw in your proof quite quickly so the persistence would not have been tested. But I 100% agree this is a basic requirement, also later in the working world. For example in my own career as a software engineering. If you are not prepared to sit in front of a computer screen for long hours trying to find that difficult bug you are in the wrong job. I am assuming, of course, the same test would apply for those contemplating taking a computer science course.

  • Ster Chess
    Posted at 07:47h, 12 August

    My left ear enjoyed this

  • Mahmood42978
    Posted at 07:47h, 12 August

    I think, and I maybe wrong , is that In the beginning, he said that the triangle is a "random" triangle, meaning we don't know if the successive properties would hold for any random triangle. the argument he gave would only work if he assumes that it has to be isosceles first.

  • Samia Zaman
    Posted at 07:47h, 12 August

    Funny. I was speaking in my mind the words 'but it bothers me that it's like that', and first rewatched part of the video, before hearing the professor read my mind and actions from before. Wow.

  • Ilkinond
    Posted at 07:47h, 12 August

    Well he's drawn a triangle that looks almost isosceles… I don't think he could get away with this if his starting triangle was very obviously not isosceles…. It must be right at the beginning, and it's about if and where the two lines (top angle bisector and bottom side bisector) meet.

  • Raisie Wong
    Posted at 07:47h, 12 August

    The angle bisector and the perpendicular bisector cannot meet inside the non-isosceles triangle. Even if there is a meeting point, it is outside of the triangle. This can be proved by two methods: by drawing and by trigonometry. For drawing, if you accurately draw a non-isosceles triangle, you will realise that it is impossible to find a meeting point. As the triangle becomes very close to an isosceles triangle, the angle bisector that meets the base is only slightly behind the midpoint of the base. Hence, in the diagram drew in the video, it is hard to notice by eye that the it is slightly off and they should not meet in reality. To prove by trigonometry, simply draw a non-isosceles triangle, and connect the top vertex to the midpoint of the base. Then,by applying the sine rule and some comparison, it can be proved that the angle bisector is always behind the midpoint of the base unless it is a isosceles triangle. It is hard to describe here, but the problem in the video it is simply impossible to have a non-isosceles triangle where their angle bisector and perpendicular bisector meets INSIDE the triangle. As the first step is wrong, the following steps are of course, invalid.

  • Ominöse Omimöse
    Posted at 07:47h, 12 August

    If the Triangle is isosceles, the angle bisector and the perpendicular from the Point in the middle of the lower side would be the same line, which means that no intersection exists

  • Kaif Bhana
    Posted at 07:47h, 12 August

    the perpendicler bisector to the bottem and the line from the top u dont know when they intersect so u cant say tht the sides are equall

  • Striped Zebra
    Posted at 07:47h, 12 August

    The triangle is assumed to be isosceles from the outset, just by considering position of its orthocentre (i.e. the point at which the three perpendiculars to the sides of the triangle coincide) lying on the perpendicular bisector of its base. This is followed by a circular argument in which this assumption is then purportedly proved.

  • Calcul8er
    Posted at 07:47h, 12 August

    The first congruency proof is invalid, side-angle-side congruency is only applicable for when the angle is located between the two sides in question

  • Always Anonymous
    Posted at 07:47h, 12 August

    just because the triangle can be made up of isocilies doesnt mean all triangles are isoseles obviously. the biggest triangle is not an isoseles, if it were the perpendicular bisector would split directly down the centre and form the right angles but in this example it does not… or is that the stupidest thing ive ever said in my life? 😂 either way the trick is that hes given you a triangle and done about a thosand things to focus your attention on the inner triangles but looking at the bigger picture theres no way that the statement can be true because its based on a non isoseles triangle to begin with…

  • Scarly Kyn
    Posted at 07:47h, 12 August

    I love how initially I was no I'm not bothered then just calmingly re wrote the triangle and goes so I'm guessing it's your assumption of the meeting points because if it wasn't an isosceles forming a bisector between the 3 points of the triangle they would be at angles and would not meet the bisector from the opposite side. >_< then i read the comments and go ahh….still don't know whether to do Maths at Uni or not I'm stuck :/

  • Roofus Onna
    Posted at 07:47h, 12 August

    The right angles that make up the larger triangle are not isosceles.

  • D. Net
    Posted at 07:47h, 12 August

    The drawing is at fault here, a random (non isoscele) triangle would not give us a meeting point between an angle bisector and the opposing perpendicular bisector. If the angle bisector and perpendicular bisector would meet, then the triangle would be isoscele (or equilateral). In all other cases, the would not meet. We could use an inner circle and an outer circle to that an angle bisector and they opposite perpendicular bisector are not touching each other… Not sure how to demonstrate it fully though 😉

    Your conclusion should rather be that all triangle that have opposing angle bisector and perpendicular bisector that are touching themselves are isosceles. Or something like that.
    Nice video

  • lilyRANDOM1
    Posted at 07:47h, 12 August

    If we have an isosceles triangle with a bisector running straight thru, we would have two isosceles triangles separated by the bisector. Knowin tht one angle is 90, the other two can either be 45, 45 or 60,30. I dnt know if im right, but wat was done in the video looked as though he was operating under the assumption that a 60 30 triangle was impossible. Because following the 60 30 90 triangle would literally disprove this entire (false) theorem that all triangles are isosceles -_-

  • Anteater23
    Posted at 07:47h, 12 August

    Where did you 'pull the wool over our eyes'

  • Jatt Prime
    Posted at 07:47h, 12 August

    The angle bisector and the perpendicular bisector at the beginning must be the same line?

  • BearCubster
    Posted at 07:47h, 12 August

    The construction where the stepped thru reasons lead to conclusion doesn't hold all triangles

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