Aggvent Calendar Day 17

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  • Опубліковано 23 гру 2024

КОМЕНТАРІ • 101

  • @Memetist
    @Memetist День тому +117

    We WILL catch up!

  • @KMills
    @KMills День тому +206

    The 30/60/90 triangle rule is getting some serious mileage this month

    • @LucasVieli
      @LucasVieli День тому +3

      I kinda surprised that in the last two ones he didnt directly used that the the height of the equilateral triangle is equal to 3 times r

    • @metsrule2000
      @metsrule2000 День тому +1

      I swear if I cant remember n, 2n, and root3n by the end of the month what even was the point!

    • @siddharathmanosenthil8394
      @siddharathmanosenthil8394 День тому

      Ikr

  • @rigaudio
    @rigaudio День тому +20

    Three "how excitings", truly a Christmas miracle.

  • @expobeau
    @expobeau День тому +45

    This may have been the hardest one yet! Not necessarily in terms of actual math work, but definitely in terms of creativity for how to get all of the sides you need. I could sit here for weeks and not come up with the ways you thought of to find the side lengths. This is why this channel is so cool to watch.

  • @Anab_Khan
    @Anab_Khan День тому +10

    Among many things, I love your articulate way of speaking and math-solving.

  • @PiAddiction
    @PiAddiction День тому +1

    This series have been the best holiday gifts.

  • @shaylevinzon540
    @shaylevinzon540 День тому +29

    Cool resolution with the right triangle to find c

  • @kenhaley4
    @kenhaley4 День тому +8

    Here's a shortcut for the second part of the solution starting at 4:01. Looking at the green triangle, we know that the bottom angle is 60°, and has side lengths a, b and c with c opposite the 60° angle. So, using the law of cosines we have
    c² = a² + b² -2ab(cos 60°).
    cos 60° is 1/2, so
    c² = a² + b² - ab.
    This can be substitued for c² in the 2nd equation resulting in
    (√3/4)( a² + b² - ab + ab) = ?
    or simply
    (√3/4)( a² + b²) = ?
    which is the same as the top equation left-hand side. Hence ? = 100.

    • @JennyBlaze253
      @JennyBlaze253 День тому

      Good catch there! :D

    • @Z-eng0
      @Z-eng0 День тому

      True, but I personally prefer a pure geometric method whenever possible trig-free, since for me, I'm here mainly to improve my pure geometry skills (which Andy is really helping me with here).
      Besides, there was no point in his going through drawing the height of the scalene triangle as b and a base of √3 * a / 2, since the right triangle with hypotenuse c was gonna be drawn anyway, we can just use it to find the area of the scalene triangle and add that to the purple triangle

    • @haitianGK
      @haitianGK День тому +1

      How did you prove that the green triangle was a right triangle in order to use this?

    • @JennyBlaze253
      @JennyBlaze253 День тому

      @@haitianGK You don't. The law of cosines can work for any triangle. Technically the Pythagorean Theorem is just the Law of Cosines for the hypotenuse but because cos 90 is 0, the only part of the Law of Cosines we need to take into account is the a^2+b^2 part.

    • @kenhaley4
      @kenhaley4 День тому

      @@haitianGK It's definitely not a right triangle. See reply from JennyBlaze for a nice answer.

  • @cube-san7351
    @cube-san7351 День тому +10

    I immediately thought it was a 100, but I can't come up with a neat solution, nice work as always! How exciting
    I just imagined a scenario where all triangles would have equal area and since it will still satisfy the conditions then if one area is 50 then the sum of two triangles is 100. I reckoned it'd just be the same for all of them

  • @TimMaddux
    @TimMaddux День тому +3

    I couldn't solve this the hard way, but the easy way is to notice either a or b are zero then the green triangle collapses and c=a (or b) giving ?=100. If a=b then you have 4 congruent equilateral triangles and ?=100.

  • @keex0003
    @keex0003 День тому

    How exciting indeed. I solved it in a similar way and it put a smile on my face. Thank you and have a wonderful Christmas!😀

  • @MrKrille93
    @MrKrille93 День тому +4

    This challenge was tougher than most of the other ones. How exciting.

  • @pedllz
    @pedllz День тому +3

    Since the only cue is that triangles A B & C are equilateral you can assume them being equal, and in that scenario D becomes a same size equilateral...so A+B = C+D
    It's great how you worked out the math !!

    • @bachmee5665
      @bachmee5665 День тому

      That's exactly how I knew the solution is 100 or yellow + blue = purple + green.
      But knowing and proving are two different things. Andy did a wonderful job at proving it!

  • @m.h.6470
    @m.h.6470 День тому +2

    Solution:
    So we know, that the area of any equilateral triangle is a²√3/4 or a²√(3/16).
    Let's call the base lengths of the equilateral triangls y(yellow), b(blue) and p(pink)
    We therefore have the equation:
    100 = y²√3/4 + b²√3/4 = √3/4 * (y² + b²) |*4/√3
    400/√3 = y² + b²
    We also have the law of cosines. With it, we can say on the green triangle, that:
    p² = y² + b² + 2yb*cos(gamma)
    Since gamma HAS to be 60° (because yellow and blue are on the same line and both have 60° angles, so 180° - 2 * 60° = 60°) and cos(60°) = 1/2, we get:
    p² = 400/√3 - yb |*√3/4
    p²√3/4 = 100 - √3/4 * yb
    So the area of the pink triangle (p²√3/4) is 100 - √3/4 * yb
    Now let's focus on the area of the green triangle.
    Since it shares the same angle at the bottom, it shares it's height with both yellow and blue, depending which baseline we use.
    Lets use the larger b baseline and therefore have a height of √3y/2 (typical for equilateral triangles).
    It's area is therefore:
    1/2 * √3y/2 * b = √3/4 * yb
    This is the unknown term from form earlier, so we have everything we need.
    The area of the pink and green triangle together is:
    A = 100 - √3/4 * yb + √3/4 * yp
    A = 100
    Or in other words, the yellow and blue triangles together have the same area as the green and pink triangle together.

  • @Z-eng0
    @Z-eng0 День тому +1

    This is the first time my solution was faster, and somewhat more elegant than Andy's.
    For reference I started with the right triangle of hypotenuse c that Andy drew, instead of drawing the base b and height √3 * a / 2

  • @1115991
    @1115991 День тому +1

    when you declare the height of triangle b is identical to the height of triangle a, I think you need to prove that the two sides are parallel to each other by stating angle a and angle b are both 60 degree on that straight line.

  • @feyh
    @feyh День тому +2

    Everytime we barely have any number, I just try the edge cases. If the green and yellow triangle had area zero, the pink would be equal do the blue, so it would be 100. If the blue and yellow would be equal, the four would be equal equilateral triangles, each with area 50, and the answer would be 100. I know this is no proof, but at least I would know ahead the answer I was looking for.

  • @charimonfanboy
    @charimonfanboy День тому +6

    fun one, immediately got 32 because if both squares are the same height, each square is 4 squared and the question implies it's the same all around
    Then I sat and stared at it for ages until it finally clicked
    side of yellow square is a and square red square is b
    the area of the two squares are a^2+b^2
    the diagonals of the squares are aroot2 and broot2
    if you connect all three corners that touch the circle together you get two diagonals, the diagonal of a square has 45 degrees, so the bottom corner of the triangle has two of them, so 90 degrees, it's a right angled triangle
    if you have a two chords meeting at a right angle, the line between the other points is the diameter, which in this case we know is 8
    so we have a right angled triangle, using pythag, 8^2=(aroot2)^2+(broot2)^2
    which simplifies to a^2+b^2=32 and this is what we were trying to find

    • @hashirwaqar8228
      @hashirwaqar8228 День тому +3

      oh my god brother , I also immediately got 32 after scaling both squares as equal and then took time to solve the problem traditionally by making a right angled triangle . How fugging exciting😂

  • @saurabhbunty123
    @saurabhbunty123 День тому +5

    How exciting!

  • @AzouzNacir
    @AzouzNacir День тому +1

    The diagonals of the two squares form a right triangle whose hypotenuse is the diameter of the circle. If we assume that a and b are the lengths of the sides of the two squares, then (a√2)²+(b√2)²=8², and from this a²+b²=32, so the sum of the areas of the two squares equals 32.

    • @Viktor_Johansson
      @Viktor_Johansson День тому

      Can we be sure the diagonal is the diameter of the circle and not just a chord-line?

    • @saiyendluri5776
      @saiyendluri5776 День тому +2

      The diameter always subtends an angle of 90 degrees anywhere on the circle. From the diagram, the two angles in between the diagonals of the squares add upto 90 degrees (45 deg + 45 deg) so the chord indeed is the diameter of the circle.

    • @AzouzNacir
      @AzouzNacir День тому

      ​@@Viktor_JohanssonThe diameter of a square divides the angle into two angles measuring 45°, and from there the two diagonals form an angle measuring 45°+45°=90°. The right triangle inscribed by a circle has a hypotenuse that is a diameter of the circle.

    • @AzouzNacir
      @AzouzNacir День тому

      ​@@Viktor_JohanssonThe diameter of a square divides the angle into two angles measuring 45°, and from there the two diagonals form an angle measuring 45°+45°=90°. The right triangle inscribed by a circle has a hypotenuse that is a diameter of the circle.

    • @AzouzNacir
      @AzouzNacir День тому

      ​@@Viktor_JohanssonThe diameter of a square divides the angle into two equal parts, each measuring 45°. From this, the diagonals of the two squares form an angle measuring 45° + 45° = 90°. The right triangle inscribed by a circle has a hypotenuse as a diameter of the circle.

  • @ianbrooks6816
    @ianbrooks6816 День тому +1

    How colourful!

  • @Skyhighjettalone
    @Skyhighjettalone День тому

    That almost feels like an axiom to me:
    If two equilateral triangles that share only one vertex have parallel sides, then a third equilateral triangle with a base that is perpendicular to a set of parallel sides of the other two triangles will have half the combined area of those triangles.
    Im sure if I went digging, I'd find a rule or proof that says pretty much that, if not a simpler form of it.

  • @cyruschang1904
    @cyruschang1904 День тому

    Answer to the next question:
    If you connect the upper-right and the lower-left corner of the large square, then the upper-left and the lower-right corner of the small square, you get two diagonals (d & D) that are perpendicular to each other. This means that the distance between the upper-right corner of the large square and the upper-left corner of the small square = the diameter = 2 x 4 = 8
    So (d)^2 + (D)^2 = (8)^2 = 64
    But the diagonal of a square = (√2)(side length)
    (diagonal of a square)^2 = (2)(side length)^2 = twice the area.
    So the combined area = 64/2 = 32

  • @ultimaurice
    @ultimaurice День тому +2

    Visual proof and how I solved it.
    It doesn't specify the size of yellow or blue which means that it doesn't matter for the solution.
    If both yellow and blue have the same area then all the triangles will be congruent equilateral triangles. Therefore they will have the same area.

    • @MartB1979
      @MartB1979 День тому

      Yeah, I thought that as immediate answer without needing to calculate anything. If the size difference between blue are yellow isn't needed in the question because red plus green is proportional to blue plus yellow, then if yellow and blue are same size, green must be an identical equilateral triangle too to fit the space inbetween. And as it shares a side with red, and red is equilateral, then red is identical too. It's not a proof, but it makes the calculations feel a bit meaningless.

  • @pedroamaral7407
    @pedroamaral7407 День тому

    I did it another way:
    - Using the information in the problem and applying the formula for the area of ​​an equilateral triangle, we have: a² + b² = 400/sqrt(3).
    - Applying the law of cosines to the middle triangle: c² = a² + b² -2ab*cos(60º) => c² +ab = 400/sqrt(3).
    - Applying the area formula to the middle triangle: A = (1/2).a.b.sen(60º) => A = ab*sqrt(3)/4
    - The desired area is: ab*sqrt(3)/4 + c².sqrt(3)/4 = [sqrt(3)/4]*[c²+ab]= [sqrt(3)/4]*[400/sqrt(3)] = 100.

  • @henrygoogle4949
    @henrygoogle4949 День тому

    Y'all now know "how exciting" isn't just a catch-phrase.

  • @Viktor_Johansson
    @Viktor_Johansson День тому +1

    Next day is 32 square units, if you can draw a diagonal from the corners of the squares that goes through the center of the circle so that; 8^2 = (a+b)^2 + (b-a)^2. Otherwise I dont think you can solve it?

    • @joeydifranco0422
      @joeydifranco0422 День тому

      I did it like that too.
      Also, another right triangle you could use is the diagonal of a and b with the hypotenuse still 2r.
      Then (sqrt2a)^2 + (sqrt2b)^2 = (2r)^2
      2a^2 + 2b^2 = 4r^2
      a^2 + b^2 = 2r^2 = 32

  • @TheDdnlL
    @TheDdnlL День тому

    To solve this problem, you can assume:
    a=0, or b=0, or a=b
    Which you will get c=b, or c=a, or c=a=b
    Then you will instantly get green + red =100
    What Andy did was to prove in all cases of a and b, green + red will always be the same with blue + yellow.
    How exciting!

  • @josephj9828
    @josephj9828 День тому +1

    Again with the obvious answer that takes some impressive math to prove. HOW EXCITING.

  • @montyyoutube
    @montyyoutube День тому +2

    At 2:00 how do you know that the height of the green triangle is equal to the height of the yellow triangle? Just by the fact that they share the same side?

    • @charimonfanboy
      @charimonfanboy День тому +1

      the base of the green triangle, as he is calculating it, is the side of the blue triangle and that intersects the horizontal base at 60 degrees
      the top of the green triangle touches the side of the yellow triangle which also intersects the horizontal base at 60 degrees
      this makes the two lines parallel
      and in an equilateral triangle the height is the same whichever side you call the base

    • @Epyxoid
      @Epyxoid День тому

      The blue triangle and the yellow has the same orientation and are similar. Since two of the green triangle's sides share a side with them, we know that perpendicular to those sides are the heights of those triangles, and the heights are parallel because of the orientation.

    • @nathanc6516
      @nathanc6516 День тому

      After he rotates the triangles you can see that the height of the green triangle equals the height of the yellow because the yellow is equilateral.

  • @brandongraham3509
    @brandongraham3509 День тому

    These are great and we'll catch up. What takes longer? Solving the initial problem or animating the living daylight out of it?

  • @bradramsay8299
    @bradramsay8299 День тому

    Since the heights of blue and yellow are not provided, we cannot assume that the drawing is to scale, so I made an assumption that they were equal. It was immediately obvious after this assumption that the answer was 100. The solution will always be the same, regardless of what the heights are. Easy peasy!

  • @malcolmpelletier2107
    @malcolmpelletier2107 День тому

    Here you can use the posing of the question to solve it quickly. Since the question was posed, presumably it has an answer. Since no information was given about the relative sizes of the yellow and blue squares, suppose they are the same (it shouldn’t matter, since the problem doesn’t specify). The pink and green triangles are now forced to be the same size as the yellow and blue, so of course they add to 100.

    • @malcolmpelletier2107
      @malcolmpelletier2107 День тому

      Similarly, suppose the area of the yellow triangle is zero (it is a single point). Then the pink triangle has the same side length as the blue, and the green triangle is also of area zero (it is a line segment). The pink and blue triangles now must each have an area of 100.

    • @malcolmpelletier2107
      @malcolmpelletier2107 День тому

      I should emphasize that this is not a proof - it relies on the fact that the information given is simultaneously enough to determine a numerical answer, and not enough to determine the relative sizes of the yellow and blue triangles - in other words, the relative sizes cannot impact the answer, so make them something simple and you can solve from there

  • @bando404
    @bando404 День тому

    Intuition immediately told me the answer but I can’t prove it.. there are 3 equilateral triangles connected like this, forming a fourth upside down triangle in the middle. If you wiggle them around without disconnecting them or changing the equilateral, wouldn’t the sun area for the two bottom ones always be equal to the area of the top and center one? It is that way if you move them all into one large equilateral triangle.

  • @thoperSought
    @thoperSought День тому

    isn't there some corollary on the Pythagorean theorem that you can do it with other shapes-incl. triangles? does that shorten this problem at all, I wonder?

    • @trelligan42
      @trelligan42 День тому

      Law of Cosines; c^2 = a^2 + b^2 - 2ab * (Cos C)

  • @SaintUpthrust
    @SaintUpthrust День тому +5

    2:13 "Mm" 🙂

  • @jelejacques
    @jelejacques День тому

    If I assume that the combined area of the two squares are always the same and I would say yellow is zero unitsquare than we have only a red square.
    Which would have a hypotenuse = 2r. So its 32unit squares?

    • @Syfes
      @Syfes День тому +1

      or you can assume two squares of equal size, they would also add up to 32

  • @txikitofandango
    @txikitofandango День тому +1

    I was worried that I overthought it, but my work was the same as yours! So maybe not!

  • @JayEff-c3x
    @JayEff-c3x День тому

    How does we KNOW the height of the green triangle was the same as a side of yellow? Looked to me the side of yellow was slightly longer than the height of the green.

  • @2BadgersBlue
    @2BadgersBlue День тому

    I struggled with this one, think I made a simple error somewhere as was getting some horrible equations 😂
    Glad it wasn't really straightforward though and it wasn't just that I was missing something really obvious.
    Loving this series!

  • @stiger23
    @stiger23 День тому

    🔥🔥🔥🔥

  • @balazsgyekiczki1140
    @balazsgyekiczki1140 День тому

    This was so interesting, I've learnt something new.

  • @Applebapple-qm2pk
    @Applebapple-qm2pk День тому +1

    It was my bday on the 17th!

  • @Stranglygreen
    @Stranglygreen День тому +2

    mind blown.

  • @devyadav6261
    @devyadav6261 День тому +1

    Lion goes back to go ahead

  • @mozzdog
    @mozzdog День тому

    Bro... that was exhausting, but;
    I'm excited

  • @radfue
    @radfue День тому

    Making a right triangle instead of using the rule of cosines was really cool

  • @sbusisomanqele4318
    @sbusisomanqele4318 День тому

    Quick question: how is it a hundred as the radical fraction is still there? I don't know how to explain my question but like
    A plus b part isn't multiplied by the radical 3 over four so it can't be 100?
    Can it?
    I'm missing something 😭

  • @hashirwaqar8228
    @hashirwaqar8228 День тому +1

    area of both square combined is 32

  • @mantasr
    @mantasr День тому

    How do we know the height of green is the same as yellow?

  • @TurtleGod2
    @TurtleGod2 День тому

    All that work just to end up Yellow + Blue = Pink + Green

  • @picknikbasket
    @picknikbasket День тому

    Wow, what a trip!

  • @pault726
    @pault726 День тому

    You need to settle down a little there. You had a premature exclamation: "How exciting!"

  • @A7m7y
    @A7m7y День тому

    Amazing ❤🎉

  • @gregboi183
    @gregboi183 День тому

    I was expecting some cool pythagorean proof given it's a triangle with similar shapes on all its sides. This was still cool though

  • @umbra0rpng173
    @umbra0rpng173 День тому +4

    In Andy math we trust

  • @radfue
    @radfue День тому

    Hint 1 for tomorrow's problem: Use the diagonals of the squares
    Hint 2 : Whats the angle between the diagonals?

  • @justinmuskivitch5545
    @justinmuskivitch5545 День тому

    “Oh no. We need to do more work…” 😎

  • @tkThePigeon
    @tkThePigeon День тому

    Wasn't this part of a proof you showed us in another video?

  • @I_Am_UnJaded
    @I_Am_UnJaded День тому

    4:02 this is where I got stuck😢

  • @5gearz
    @5gearz День тому

    I tried this and thank god i didn't try any further because golly!

  • @tellerhwang364
    @tellerhwang364 День тому +1

    day18
    (a·sqrt2)^2+(b·sqrt2)^2
    =(2R)^2=8^2=64
    2(a^2+b^2)=64→a^2+b^2=32😊

    • @tellerhwang364
      @tellerhwang364 День тому

      other method
      let a=0→b=4sqrt2
      →a^2+b^2=32😊

  • @chrishelbling3879
    @chrishelbling3879 День тому +3

    Who comes up with these clever problems? All these problems help explain why students learn Algebra first, then Geometry.

    • @cyruschang1904
      @cyruschang1904 День тому +1

      I have no idea who created these questions, but he came up with the word aggvent calendar because Agg is the last name of the person who forwarded these questions to him 😊

    • @thoperSought
      @thoperSought День тому +1

      catriona agg

  • @hustledude
    @hustledude День тому

    Very clever

  • @gabelordjumong
    @gabelordjumong День тому

    Anyone else using Desmos and excel to solve these problems?😢

  • @Mricy1
    @Mricy1 День тому

    Nice video

  • @gayathrikumar5643
    @gayathrikumar5643 День тому

    Day 18: 32 sq units

  • @Cyco-Dude
    @Cyco-Dude День тому

    ...and enhance. Lol, you're silly. Don't kill yourself getting all these done, man. It's a mistake to try to solve and upload every day; the correct way would be to have these all done in advance, then just upload them later. Content creators typically have weeks, sometimes months, of videos ready to go. That way they don't have to work so hard to get them out on time, as well as have stuff ready to go in case something happens (illness, equipment failure, etc).

  • @rld1982
    @rld1982 День тому

    I always find it weird how Americans say "one fourth" when their sports are broken into 4 QUARTERS. What do you guys think a quarter is, exactly?

    • @jimlocke9320
      @jimlocke9320 День тому

      Football has four quarters. Baseball divides the game into innings and there are 9 innings except in certain special circumstances.

  • @pedroamaral7407
    @pedroamaral7407 День тому

    Next problem: 32

  • @Name_Naminson
    @Name_Naminson День тому +2

    First comment

  • @thynedewaal1823
    @thynedewaal1823 День тому +1

    hi

  • @Farah-x5v1n
    @Farah-x5v1n День тому

    There was an easier way