Sweden Math Olympiad | A Very Nice Geometry Problem

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  • Опубліковано 22 жов 2024

КОМЕНТАРІ • 20

  • @harikatragadda
    @harikatragadda День тому +2

    Chord AB is inclined at 70° at B to its tangent, hence it also subtends the same angle at C. Since Arc BC subtends θ at A, Arc AC subtends 3θ/2 at B.
    θ+3θ/2=110
    θ = 44°

  • @psykomarvin
    @psykomarvin 23 години тому +5

    It's not said at the begining of the video that AB and BP are tangent lines to the circle and the drawing is misleading about it. Could be nice if you can edit it.

    • @oguzhanbenli
      @oguzhanbenli 12 годин тому

      Yes, the tangency is not declared

  • @ناصريناصر-س4ب
    @ناصريناصر-س4ب День тому +1

    The measure of the inscribed angle is proportional to the length of the arc that it encloses. From this we assume that

    • @Just0Me359
      @Just0Me359 21 годину тому

      ازيك يا عمنا عامل ايه؟بتعمل ايه هنا.انت منين؟

    • @soli9mana-soli4953
      @soli9mana-soli4953 21 годину тому +1

      Nice and simple! 👌

    • @ناصريناصر-س4ب
      @ناصريناصر-س4ب 20 годин тому

      @@Just0Me359 أنا من الجزائر واحب قنوات الرياضيات وانت من الظاهر مصري اليس كذالك؟

  • @PS-mh8ts
    @PS-mh8ts 10 годин тому

    △PAB is an isosceles triangle (because tangents PA and PB are of the same length). Hence ∠ABP=∠BAP => ∠ABP=(180°-40°)/2=70°.
    By alternate segment theorem (also known as tangent-chord theorem), ∠ACB=∠ABP
    Hence ∠ACB=70°
    Thus, in triangle △ABC, ∠C=70°. Angles ∠A and ∠B must then add up to (180°-70°=1110°)
    It's known that the length of an arc is proportional to the angle subtended by the arc. Thus, ∠A and ∠B must be in the ratio 2:3. We've already established that ∠A+∠B=110°. Thus, to find the values of ∠A and ∠B, we divide 110° in the ratio 2:3
    Hence ∠A=110°*2/(2+3)=44°

  • @marioalb9726
    @marioalb9726 15 годин тому +1

    Internal angles of circle:
    α₁ = 360° - 2*90° - 40° = 140°
    α₂+α₃ = 360° - α₁ = 220°
    Simple rule of three:
    α₂ = 2/5 (α₂+α₃) = 88°
    Requested angle:
    θ = ½α₂ = 44° ( Solved √ )
    Extremely complicated video solution, there no need to introduce variable "x" neither "r" radius of circle

  • @SGuerra
    @SGuerra 16 годин тому +1

    O desenho da questão não é claro inicialmente. Se A e B não são pontos de tangência a questão é difícil conclusão. Mesmo assim a questão (com pontos de tangência) é muito boa. Parabéns pela escolha. Brasil - Outubro de 2024.

  • @befactfull525
    @befactfull525 23 години тому +1

    done in just 2 minutes but please tell me from where you find these questions

  • @michaeldoerr5810
    @michaeldoerr5810 День тому

    The answer is 44°. Looks like I shall use this for practice!!!

  • @giuseppemalaguti435
    @giuseppemalaguti435 День тому

    140=2(180-θ-3/2θ)...5/2θ=110...θ=220/5=44

  • @와우-m1y
    @와우-m1y День тому

    asnwer=55 isit

  • @RealQinnMalloryu4
    @RealQinnMalloryu4 День тому

    (2)^2(3)^2={4+9}=13 {60°A+60°B+60°C}=180°ABC/13=10.50 10.5^10 2^5.5^2^5 1^1.1^2^1 2^1(ABC ➖ 2ABC+1).