Hungary Math Olympiad | A Very Nice Geometry Problem

[@soli9mana-soli4953]

Considering that ODP is a 30,60,90 degree right triangle, the radius of the semicircle can be found with the equation: 6 - r = 2r

[@quigonkenny]

Let P be the center of the semicircle. Draw radius PD. As OA is tangent to semicircle P at D, ∠PDO = 90. As ∠DOP = 30°, ∠OPD = 60° and ∆PDO is a 30-60-90 special right triangle. As PD = r, OP = 2r and OD = √3r. As OP = 2r and PB = r, OB = 3r. As OB = 6, r = 6/3 = 2. The shaded area is equal to the area of sector AOB minus the area of semicircle P. Shaded area: A = (θ/360°)πR² - πr²/2 A = (30°/360°)π{6²) - π(2²)/2 A = 36π/12 - 4π/2 A = 3π - 2π = π ≈ 3.14 sq units

[@marioalb9726]

Angular sector: A = ½αR² = ½.30°(π/180°)6² A = 3π cm² Radius of semicircle: r = R/3 = 2cm Area of semicircle A = ½πr² = 2π cm² Shaded Area A = A₁ -A₂ = 3π -2π A = π cm² ( Solved √ )

[@germanalexandru]

PDC equilateral, OCD isoscel, so a=2

[@jimlocke9320]

At 6:00, recognize that ΔODP is a special 30°-60°-90° right triangle, for which the hypotenuse is twice as long as the short side. The short side is opposite the 30° angle, so is PD and PD has length a. The hypotenuse OP is twice as long, or 2a. OB = OP + PB = 2a + a = 3a, and OB = 6, so a = 2. Skip ahead to 7:30.

[@saronohandoyo4958]

Let P is the center of the semicircle so BP=CP=DP = R which is radius of semicircle sin 30 = 1/2 = R/OP so OP = 2R OB = OP+PB 6 = 2R+R 3R=6 so R=2 area shaded area = areasector AOB - area semicircle =30/360*π*r*r - 1/2*π*R*R =π*6*6/12 - 2*2*π/2 =3π-2π = π

[@Great-Britain2]

Bro I swear educational things like this shoudl get way more views

[@SGuerra]

Uau! Que questão legal. Parabéns pela escolha. Brasil - Novembro de 2024.

[@ناصريناصر-س4ب]

Let I be the center of the semicircle and let H be the perpendicular projection of point A on OB. According to the triangles OAH and OID, we conclude (6-r)/6=r/3, from which r=2, and from which the required area is π*6²/12-π*2²/2=π.

[@michaeldoerr5810]

The shaded area is pi units square. Also I think this problme would still be applicable even if the triangles are NOT special right triangles. No lie. I could wrong.

[@himo3485]

r+2r=6 3r=6 r=2 Shaded area = 6*6*π*30/360 - 2*2*π*1/2 = 3π - 2π = π

[@giuseppemalaguti435]

r/sin30+r=6..r(1+2)=6...r=2....Ablue=πR^2/12-πr^2/2..[R=3r]..=π(3/4)r^2-πr^2/2=(1/4)πr^2=π

[@santiagoarosam430]

Si S es el centro del semicírculo→ SD=r→ SO=2r y SB=r→ OB=6=3r→ r=6/3=2→ Área sombreada =(π6²/12) - (π2²/2) =π. Gracias y un saludo.

[@cyruschang1904]

(6)(6)π x (30°/360°) - (2)(2)π/2 = 3π - 2π = π

[@RealQinnMalloryu4]

(6)^2=36{60°A+60°B+60°C}=180°ABC/36=5ABC (ABC ➖ 5PiABC+5Pi).

[@AmirgabYT2185]

S=π≈3,14

[@yakupbuyankara5903]

3.14

[@mordechaimalka]

Excellent exercise !! Greeting from the Eternal Jewish people With a 3500 years history !!! From his Jerusalem Capital from the king David . Israel will stay for ether !!

[@sakshikausal]

= [(pi*6²/6 )- pi*2² ]/2 = pi

[@wasimahmad-t6c]

3×3=9×3.14159268=28.274334)(39.901122-28.274334=11.6267886÷2=5.813394