You need to know this trick! | Math Olympiad Geometry Problem

You need to know this trick! | Math Olympiad Geometry Problem
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[@tomfalvey569]

Assign AB = 15/x and BC=x. ABF and BCE are 30-60-90 triangles so respectively BF = 10*sqrt(3)/x and BE=2x*sqrt(3)/3. Now calculate the area directly using Area = 1/2*BF*BE*sin30. Voila! 5. (This method requires that you the generalized formula for area of a triangle = 1/2*side*side*sine of included angle.)

[@ludmilaivanova1603]

Another way of solving this problem: Let us rename all sides- AB=a. BC=b, EC=x, AF=y. The area of a triangle under the question is Area of BEF = Area of ABCD - Area of BAF - Area of BCE - Area of EDF. EC/BC=sqrt 3b/3 ( tan 30 degrees) or x = sqrt 3 b/3 AF/AB=sqr 3a/3 DE = a-x and AF = b-y . Substitue x and y and write expressions of all three triangles areas using only a and b letters. Then subtract the Sum of three triangles areas from the product ab and you will get 1/3 of ab.

[@Mediterranean81]

Let AB= a BC= b EC = c and AF= d 3@=90° @=30° By soh cah toa 2a/sqrt 3=FB BE = 2b/sqrt 3 Area of red triangle 1/2 absinc 1/2*2a/sqrt 3*2b/sqrt 3*1/2=ab/3=15/3=5

[@Latronibus]

If you're gonna get into the weeds with some algebra, you can just chase lengths and sum areas, without the need to draw this extra horizontal line. [ABF]=1/2 x^2/sqrt(3), [BEC]=1/2 y^2/sqrt(3), [FED]=1/2 (x-y/sqrt(3))(y-x/sqrt(3)). Adding those and the desired area up, you get 15, but then any appearances of xy in the expansion are also 15, and the other terms cancel out in the expansion.

[@Lead-lx8qx]

BF=x,BE=y AB=xcos30 BC=ycos30 xy(cos30)^2=15 so xy = 20 Shaded area = xysin30/2 = 5

[@xualain3129]

Bravo! I got exactly the same idea as yours.

[@marioalb9726]

Area of rectangle Ar = a. b = 15 cm² However, ratio between 'a' and "b' IS NOT GIVEN. This means we can choose this ratio, and even so we are meeting the original conditions. I chose to have a rectangle with ratio a/b = tan 30° Now, points D and E are unified, and area of right triangle BCE is half of rectangle area, is 15/2=7,5 cm² Area of shaded triangle is 2/3 of area of right triangle At = ⅔ 7,5 At = 5 cm² ( Solved √ ) Note that Ar = 3 . At Everytime we have this configuration, At =⅓.Ar Consequently, the easiest and shortest solution is: A = A / 3 = 15 / 3 A = 5 cm² ( Solved √ )

[@marioalb9726]

Vertex angle of shaded triangle is 30° = ⅓ 90° Consequently: Shaded triangle area = ⅓ Rectangle area At = ⅓ Ar At = ⅓ 15 At = 5 cm² ( Solved √ )

[@marioalb9726]

Area of rectangle Ar = a. b = 15 cm² However, ratio between 'a' and "b' IS NOT GIVEN. This means we can choose this ratio, and even so we are meeting the original conditions. I chose to have a square instead a rectangle Now a = b = √15 Shaded triangle became an isosceles triangle, with side s = a / cos 30° = a . 2/√3 s = √15 .2/√3 = 2√5 cm At = ½ s² sin30° At = 5 cm² ( Solved √ ) Note that Ar = 3 . At Everytime we have this configuration, At =⅓.Ar Consequently, the easiest and shortest solution is: A = A / 3 = 15 / 3 A = 5 cm² ( Solved √ )

[@zedopicadilho60]

You're doing a great work here, man, never stop doing this

[@imetroangola4943]

*Solução Elegante:* No ∆ABF: *AB= cos 30°× BF (1)* No ∆BEC: *BC= cos 30°× BE (2)* multiplicando as equações (1) e (2), obtemos: AB×BC=(cos 30°)^2 × BF×BE Ora, AB×BC=15 e cos 30°=√3/2. Daí, 15= 3/4 × BF×BE, isto é, *BF×BE= 20 (3)* [BEF]= (BF×BE× sin 30°)/2 Substituindo a equação (3), temos: [BEF]= (20 × 1/2)/2 *[BEF]=5.*

[@jimlocke9320]

Solution by adding the areas of the unshaded triangles and deducting from the area of the rectangle. As in the video, let AB = CD = a and AD = BC = b. Having found that Θ = 30°, ΔABF and ΔBCE can be found to be 30°-60°-90° special right triangles, AF = AB/(√3) = a/(√3), and CE = BC/(√3) = b/(√3). Area ΔABF = (1/2)(AB)(AF) = (1/2)(a)(a/√3) = a²/(2√3). Area ΔBCE = (1/2)(BC)(CE) = (1/2)(b)(b/√3) = b²/(2√3). DE = a - b/√3 and DF = b - a/√3. Area ΔDEF = (1/2)(DE)(DF) = (1/2)(a - b/√3)(b - a/√3) = (1/2)(ab - b²/√3 - a²/√3 +ab/((√3)(√3)) = ab/2 - b²/(2√3) - a²/(2√3) +ab/6. So, sum of areas ΔABF, ΔBCE and ΔDEF = a²/(2√3) + b²/(2√3) + ab/2 - b²/(2√3) - a²/(2√3) +ab/6 = ab/2 + ab/6 = 2ab/3. However, ab = 15 (area of rectangle), so 2ab/3 = (2)(15)/3 = 10. Deduct the sum of areas of unshaded triangles, 10, from the area of the rectangle, 15, leaving area of ΔBEF = 5, as Math Booster also found.

[@Irtsak]

Let EC=a , AF=b In right triangle BEC θ=30° => BE=2a . Apply Pythagoras theorem in BEC => ……. BC=a√3 Do the same in right triangle ABF and you have BF=2b and AB=b√3 Area (ABCD)=15 => BC⋅AB=15 => a⋅√3⋅b√3=15⇒ *ab=5* Let EP= height in triangle BFE (construction) . Notice that EP=BE/2 =2a/2=a => *EP=a* Now we have almost finish . Shaded area = l/2 BF⋅PE=1/2( 2ba)=ab=5 square units

[@oscarcastaneda5310]

Can y'all prove that the desired area is Always (1/3) of the rectangle's area ? OOOPS ! That's exactly what the proposer proved ! I hadn't seen the solution before getting "My Math On !"

[@sergeyvinns931]

AB=a, BC=b, BF=c, BE=d, c=2a/\/3, d=2b\/3, area triangle FBE = 2a*2b*sin30/3*2= 4*15/3*2*2=60/12=5.

[@murdock5537]

φ = 30°; ∎ABCD → AB = CD = CE + DE = b = 15/a; BC = AF + DF = a; CBE = EBF = FBA = φ → sin⁡(φ) = 1/2 → AF = 5√3/a → BF = 10√3/a; CE = a√3/3 → BE = 2a√3/3 → area ∆ BFE = (1/2)sin⁡(φ)(BE)(BF) = (1/4)(2a√3/3)(10√3/a) = 5

[@imetroangola4943]

Seus vídeos são ótimos! Parabéns pelo trabalho! 🎉🎉🎉

[@kikilolo6771]

Or we can use heron's formula and all the things magically cancel lol

[@johnbrennan3372]

|BE|= 2b oot3 (using angles 30,60 and 90) and |BF|= 2a oot 3(using angles 30,60 and 90). Area of triangle BFE = 1/2 by sine 30 degrees by |BE| by |BF|= 1\4 by 2a/root3 by 2b/ root 3= ab/3=5.

[@slavinojunepri7648]

Excellent work

[@quigonkenny]

Let DC = AB = x and AD = CB = y. The area of the rectangle is thus xy = 15. As ∠CBA = 90° and θ+θ+θ = ∠CBA, then θ = 90°/3 = 30°. cos(θ) = CB/BE √3/2 = y/BE √3BE = 2y BE = 2y/√3 cos(θ) = BA/FB √3/2 = x/FB √3FB = 2x FB = 2x/√3 Triangle ∆EBF: Aᴛ = EB(BF)sin(θ)/2 Aᴛ = (2y/√3)(2x/√3)(1/4) Aᴛ = xy/3 = 15/3 = 5 sq units

[@michaeldoerr5810]

If I understand this video, the trick is to establish both the relevant chords and the right angles so that the corresponding and congruent angles are established FIRST. After that, then you conclude that the shaded triangle is the summation of TWO triangles bc of the bisected side of the square. Looks like I will have to practice on that problem!!!

[@johnbrennan3372]

Very nice solution