China Math Olympiad Geometry Problem | 3 Different Methods To Solve

China Math Olympiad Geometry Problem | 3 Different Methods To Solve
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[@daddykhalil909]

Very interesting solutions
Thank you

[@quigonkenny]

Extend BE and FD so that they meet at G. As AE = ED (given), ∠EAB = ∠EDG = 90°, ∠BEA = ∠GED (vertical angles), and ∠ABE = ∠DGE = θ (as AB and CG are parallel), then triangles ∆EAB and ∆EDG are congruent, and GE = 7.
As ∠ABE = ∠DGE = θ, ∆BFG is isosceles, and thus FG = BF = 9. Draw FE. As E is the midpoint of the base of an isosceles triangle and F is the opposite vertex, then FE is perpendicular to BG and is the axis of symmetry for the isosceles triangle, meaning ∆FEB and ∆GEF are congruent triangles. As ∆GEF contains the area of ∆FDE and a triangle congruent to ∆EAB (∆EDG), then ∆FEB does as well, and thus takes up exactly one half of the shaded area.
Triangle ∆FEB:
FE² + EB² = BF²
FE² = 9² - 7²
FE² = 81 - 49 = 32
FE = √32 = 4√2
A(red) = 2•∆FEB = 2(bh/2)
A(red) = 2(7)(4√2)/2
A(red) = 28√2 ≈ 39.598

[@ДмитрийИвашкевич-я8т]

Треугольник AEB равен треугольнику BEP. Треугольник PEF равен треугольнику EDF. Треугольник BEF прямоугольный (right-angled triangle).
[ADFB]=2[BEF]=BE*EF=7*√(81-49)=7*√32=28√2

[@harikatragadda]

Reflect ∆BAE about BE to form ∆BGE. Since EG = ED, ∆EGF is Congruent to ∆EDF.
Since ∠BEG = 90°-θ and ∠GEF =θ ,
∠BEF = 90° and EF = 4√2
Red area = 2*(Area∆BEF)
= 2*½*7*4√2 = 28√2

[@Sam10099]

Very nice. Clever solutions.

[@MathBooster]

Thank you 🙂

[@md.abdurrahmantarafder2256]

In first method when we got isosceles triangle then area of triangle = 14/4 whole rootover 4×9×9 - 14×14 = 28root2

[@OrenLikes]

Nice!
Very long, a lot of repetition, but not skipping steps(!).
Interesting accent... :)

[@vkr122]

Area BED=BAE=7 , area ADF=BDF=2 , area AFD=1/2ABCD=14 , ABCD=28.

[@olivier9125]

What is P 12:49 / 31:14

[@sonam7choi]

점 E에서 선분 BF에 수선을 내려 계산하면 합동인 두삼각형이 나와서 쉽게 계산됨

[@arnavvankudre2166]

In triangle BPF the two angles are equal then PF=BF ?

[@MathBooster]

Yes, they will be equal

[@txt.myhome7979]

I went by the 3rd method

[@juanalfaro7522]

I used the 2nd method to solve the problem. I obtained the same answer.

[@User-jr7vf]

I did the 3rd method, the most painful one :) because I didn't thought of the first or second method. In my opinion the first method is the most beautiful one.

[@murdock5537]

∎ABCD → AB = CD = DE + CE = a + a = 2a → BC = BF + CF = AD = b → sin⁡(ABC) = sin⁡(BCD) = 1
AE = 7; AF = AS + FS = b + (9 - b) = 9; DAE = EAF = ϑ; sin⁡(ASE) = 1 → ES = DE = a; FS = CF = 9 - b →
BF = 2b - 9 → DAE = EAS = CEF = FES = ϑ → sin⁡(FEA) = 1 → EF = √(81 - 49) = 4√2 →
a = 7(4√2)/9 = 28√2/9 → b = √(49 - a^2 ) = 49/9 → 2b - 9 = 17/9 →
area AFCD = 2ab - a(2b - 9) = (28√2/81)(98 - 17) = 28√2

[@annashakaia9376]

very bad numbers i had a right solution but in the end was very very big numbers and i had to use photomath but at the end i still got the answer