Solution 87: Chasing the Angle HGI

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  • Published on Apr 14, 2019
  • We partake in an 11-minute-long angle chasing, throughout which we employ many tools like cyclic quadrilateral, trigonometry, and triangle similarity to facilitate the journey.
    Congratulations to Serengeti Ghasa, santosh tripathy, Minh Cong Nguyen, dev gupta, Dan Bollows, Evyatar Baranga, and Viraj Agrawal for successfully solving this math challenge question! Serengeti Ghasa was the first person to solve the question.
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Comments • 14

  • LetsSolveMathProblems
    LetsSolveMathProblems  2 months ago +3

    Here are some extra challenges if you want to explore the problem further:
    1) Prove that FGIC and BHGE are isosceles trapezoids.
    2) Prove that HGD is isosceles.
    3) Prove that FGD is similar to ACB.

    • Serengeti Ghasa
      Serengeti Ghasa 2 months ago +1

      Lots of love Sir ,
      ∆FGA~∆FDE
      =>FG/AF=FD/FE (1)
      Between
      ∆FGD and ∆FAE
      Angle GFD=angle AFE=60°
      Also consider (1) above
      ∆FGD~∆AFE~∆ABC prove of (3)
      =>∠GDF=∠GDH=75°
      again∠GHD=∠G+∠F=15°+60°=75°
      =>ΔGHD is isosceles prove of (2)
      With high regards
      Prabhat Ku Sahu
      Singhijuba

    • LetsSolveMathProblems
      LetsSolveMathProblems  2 months ago +2

      @Nicolas Nauli It can be proven that altitudes of any triangle intersect at one point, called the orthocenter of the triangle. There are multiple ways of proving this (Ceva's Theorem certainly does the trick), but perhaps the most elegant way is to derive the result from the fact that perpendicular bisectors of any triangle meet at the same point. (If you have not seen the proof that perpendicular bisectors concur, I encourage you to first look up its proof online.) As for the existence of orthocenter, I believe the aforementioned method is illustrated in the Khan Academy video linked below:
      www.khanacademy.org/math/geometry-home/triangle-properties/altitudes/v/proof-triangle-altitudes-are-concurrent-orthocenter

    • Nicolas Nauli
      Nicolas Nauli 2 months ago +1

      @LetsSolveMathProblems, at 3:50 how did you know that all the feets of the triangles intersect at one point?

  • Thanikachalam Mannagatti

    Most of the time voice is not enough...
    We need clear written description for the solving problem...
    For example Kindly check "mind your decisions" videos

  • dean jenny dean jenny
    dean jenny dean jenny 2 months ago +1

    Please find this number:
    1/(1*4)+1/(4*15)+1/(15*56)+...+1/(a*b)+1/(b*(4b-a))+...=?

  • Harshit Khandelwal
    Harshit Khandelwal 2 months ago +1

    This channel is very underrated.

  • XaXuser
    XaXuser 2 months ago

    Can u upload a video about the legendary question 6 in the mathematical olympiade 1988 with a comprehensible solution plzz

  • Risu0chan
    Risu0chan 2 months ago +1

    Watched 20 times. I still have no idea how you magically prove AFE is 60°.

    • LetsSolveMathProblems
      LetsSolveMathProblems  2 months ago +2

      @Risu0chan I apologize if the explanation was unclear. As you remarked, AFE = AOE follows directly from the fact that AFOE is cyclic. The fact that EODC is cyclic is used only to derive that AOE = 60, not to prove the equality of angles.

    • Risu0chan
      Risu0chan 2 months ago +1

      Oh, now I see it, AFOE are cocircular, and the subtended angles are equal because of that. I was confused because it has nothing to do with EODC you were talking about before, the quadrilaterals are clearly not similar.

  • Saro Harutyunyan
    Saro Harutyunyan 2 months ago

    Only the condition that angle A is 45 degree yields that angle HGI is 135 degree. For the proof one may see IMO 2013 problem 6 www.imo-official.org/problems.aspx or artofproblemsolving.com/community/c6h1181536p5720184

  • Btd Pro
    Btd Pro 2 months ago

    Hi

  • Keshav Ramamurthy
    Keshav Ramamurthy 2 months ago +2

    great video!