Japanese Math Olympiad | A Very Nice Geometry Problem

Рет қаралды 4,232

Күн бұрын

Пікірлер: 15

@TheAlavini 3 ай бұрын

Nice solution. The solution development help to understand the the relationship aming quarter circle and an inside square. Congrats.

@MathBooster 3 ай бұрын

Thank you 🙂

@santiagoarosam430 3 ай бұрын

Radio del cuarto de círculo = OQ=OB=OP=3+2=5. E es la proyección ortogonal de C sobre OQ y F lo es de B sobre OP → Los triángulos AFB, DOA y CED son congruentes y CE=OD=AF=a. Triángulo BFO: 5²-3²=OF²→ OF=4→ a=4-3=1 → EQ=5-a-3=1=OE→ CQ=1*√2=√2 =X. Gracias y saludos.

@sumanbasak3507 3 ай бұрын

Nice solution ❤

@Latronibus 3 ай бұрын

Pretty much what you did: draw a perpendicular from B to PO, label that point M. Draw a perpendicular from C to OQ, label that point N. Use the fact that you have a square to conclude that MAB, ODA, and NCD are all congruent. Now CQ=sqrt(CN^2+NQ^2). If the short side of all these congruent right triangles is x then CQ=sqrt((2-x)^2+x^2) so it suffices to find x. Draw OB and get (3+x)^2+9=25, so x=1 and CQ=sqrt(2). I wonder if there was any quicker way to conclude that M is the midpoint of PA?

@michaeldoerr5810 3 ай бұрын

The answer is sqrt(2). I just want to comment that at the 4:50 mark, the fact that THREE congruent triangles result in both line segments summing up the radius as well and showing that the radius is also the hypotenuse is a really unexpected result. I better practice this and combine that with a similar problem from the PreMath channel. I am starting that not only are there circle theorems, but also semicircle theorems. And I better use that as something to visualize and intuit correctly.

@aksiiska9470 3 ай бұрын

there is a similar problem with pb=bq instead of known distances pa and ao

@vcvartak7111 3 ай бұрын

How will you construct a square ABCD from given data . Now that you know it's side is √10 you may justify. But from given data it's difficult to construct such a square

@Nnbvcxfz 3 ай бұрын

which app are u using

@RealQinnMalloryu4 3 ай бұрын

(3)^2=9 (2)^2=4 {9+4}= 13 90°ABCDNOQX/13 =6.12ABCDNOQX 6.3^4 3^2.3^2^2 1^1.3^1^2 3^2 (ABCDNOQX ➖ 3ABCDNOQX+2).

@NTA_1 2 ай бұрын

Beautiful problem 🎉

@Emerson_Brasil 3 ай бұрын

Linda solução! 🎉🎉🎉

@giuseppemalaguti435 3 ай бұрын

ADO=α il lato AD=3/sinα.risultano 2 equazioni...1)legge del coseno..25=(3/tgα)^2+(√2*3/sinα)^2-2*(3/tgα)*(√2*3/sinα)cos(α+45)..si ottiene cos2α=-4/5..α=71,565...2) x^2=((3/sinα)*cosα)^2+(5-3-3/tgα)^2=18(ctgα)^2+4-12ctgα=2...ah ah sicuramente non è la via più breve.. comunque X^2=2