Hungary Math Olympiad | A Very Nice Geometry Problem | 2 Methods

[@saronohandoyo4958]

since all right triangle has 3:4:5 ratio, we can use that ratio. triangle abc: ab=3, bc=4 so ac=5 by phytagoras square mnop let say 60x as the side triangle bmp, hyp mp =60x so bm = 3/5 * 60x = 36x and bp=48x triangle amn, mn=60x so an = 3/4 * 60x = 45x and hyp am = 75x bm + am = 3 36x + 75x = 3 111x=3 x=3/111=1/37 side=60x=60/37 and area of square mnop will be 60²/37²

[@oscarcastaneda5310]

Nice problem for a Sunday morning, Gracias : )

[@Xyz-sd7ub]

3rd method....as sides of triangle ABC are at the ratio of 3:4:5, hence hypotenuse (5) will also be subdivided in the same ratio from top to bottom because all mini triangles are similar .....so side of square will be 5/(3+4+5) *4 = 5/12 * 4 = 20/12 = 1.667 and Area of square = 1.667 x 1.667 = 2.78 sq. units approx

[@quigonkenny]

As AB = 3 and BC = 4, ∆ABC is a 3:4:5 Pythagorean triple right triangle and CA = 5. As ∆ABC and ∆MNA share angle ∠A, ∆ABC and ∆COP share angle ∠C, and ∠ABC = ∠MNA = ∠COP = 90°, all three triangles are similar. Let s be the side length of MNOP. NA/MN = AB/BC NA/s = 3/4 NA = 3s/4 CO/OP = BC/AB CO/s = 4/3 CO = 4s/3 CA = CO + ON + NA 5 = 4s/3 + s + 3s/4 5 = (16+12+9)s/12 = 37s/12 s = 5(12/37) = 60/37 Square MNOP: Aꜱ = s² = (60/37)² = 3600/1369 ≈ 2.63 sq units

[@prossvay8744]

Area=(60/37)^2=3600/1369

[@devondevon4366]

1369/3600 or 2.629 Answer the hypotenuse = 5 ( Pythagorean) Let the side of the square = n, then line AM = 5/4 n and line MB = 3/5 n since the triangles are similar Since line AB= 3, then 3/5 n + 5/4n =3 12/20 n + 25/20 n = 3 37/ 20 n = 3 n = 20 * 3 /37 n = 60/37 n^2 = 3600/1369

[@imetroangola4943]

Segundo método é praticamente similiar ao primeiro. Um segundo método interessante é usar áreas de triângulos semelhantes!

[@giuseppemalaguti435]

l*tg(arctg3/4)+l+l*tg(arctg4/3)=5...3l/4+l+4l/3=5...l=60/37

[@mykhaylodorozhovets6120]

Дуже складно. Простіше: 1) AC=sqrt(3^2+4^2)=5; 2) 3/5*x+5/4*x=3; 3) x=60/37 4) s=(60/37)^2.

[@santiagoarosam430]

El triángulo ABC es rectángulo : Si AB=3 y BC=4→ AC=5 → Los triángulos ANM; PBM y POC son semejantes a ABC→ Si NM=a→ AM=5a/4 y MB=3a/5→ AB=3=(5a/4)+(3a/5)=37a/20→ a=60/37→ a²=60²/37²=2,6296...ud² =Área cuadrado MNOP. Gracias y saludos.

[@jmlfa]

ALPHA = 37, BETA = 53 - x . cos beta + x/cos alpha = 3 > x=1.618

[@RealQinnMalloryu4]

(3)^2 (4)^2={9+16}= 25 180°ABC/25 =7.5ABC (ABC ➖ 7ABC+5).

[@imetroangola4943]

The problem is that only you understand this! Your math is from Mars! 😂😂😂😂

[@penserjenser]

1/x = 1/c + 1/h für alle Dreiecke.

[@michaeldoerr5810]

The area is (60/27)^2. I have noticed that the AA similarity postilate seems to be way more useful than expected. I have noticed that in yesterday's problem and the day before yesterday. I have noticed that in today's PreMath video. And I have noticed that since this is a Pythagorean Triple, apparently both methods start off with the Pythagorean Theorem and involve the direct usage of AA similarity- with a few differences and always with the same last step. This makes it similar to the last Pythagorean Triple problem in this channel. Furthermore, the answer to yesterday'a problem is 1/17[32*sqrt(17)] and the answer to the problem the day before yesterday is is 5/8(sqrt(65)). Once again I apologize for not doing a sanity check the day before yesterday!!!

[@fm01it75]

(60/37)^2🙂

[@Xyz-sd7ub]

2.78 sq. units

[@fm01it75]

2.63 sq. units🙂

[@ionvasilescu2380]

If you resolve geometry problems with algebra, something is really wrong. Did Euclid use algebra? I don't think so. Bad teaching. 😢

[@devondevon4366]

I don't think this is 'teaching.' He solves the problems the way he knows how to and welcomes posters to share their methods.