Find the blue shaded area | Math Olympiad Geometry Problem | Important Geometry and Algebra Skills

[@sumithpeiris8440]

Simple Geometry Solution Obviously, BDCE is a cyclic quadrilateral with diameter BC = a and center F Hence < BDC = 90 and so < ACD = 30, hence < DFE = 2 X 30 = 60 So Triangle DFE is equilateral with DE = a/2, AD = b/2 and AE = c/2 since Triangles ADE and ABC are similar Let S = S(ABC) = S so that S(ADE) = S/4 S(BCD) = 2 X 8 = 16, S(BCE) = 30 We can write 3 equations S = (V3/4)bc ............................(1) S - 16 = (V3/8)b^2....................(2) S - 30 = (V3/8)c^2....................(3) (2) X (3) (S-16)(S-30) = (3/64)b^2.c^2 = (3/64)(16S^2 /3) from (1) Simplfying 3S^2 - 184S + 1920 = 0 (3S - 40)(S-48) = 0 S = 40/3 is not possible since S > 8+15 So S = 48 and the Blue Shaded Area = 48 - 8 -15 = 25 Sumith Peiris Moratuwa Sri Lanka

[@User-jr7vf]

I proceeded in a similar fashion. From Brazil.

[@isaacrichter9167]

Great work. Your way of explaining the root you take is so clear and charming to watch. Well done!! and all the best to you!!

[@User-jr7vf]

I love when she says "squuare"

[@manuelgabrielrojasleon5393]

There is a mistake: (min 18:39) the BDF area is [BDF]=(Xcos(P))(Xsin(P))=8, because Xcos(P) is the altitude and 2Xsin(P) is the base

[@hans7831]

Ich hab mit den Mittelpunkt F einen Thaleskreis um die Punkte B, D, E und C gezeichnet und fand mich dabei genial. Leider kam ich damit aber kein bisschen weiter.

[@suchaisuteparuk2897]

Geometry only Follow the video thru 8:15 min triangle ABC is similar to triangle AED AB/AE = AC/AD = BC/ED; but BC = 2ED then AB = 2AE, AC = 2AD let AE = x, AD = y Draw CD We get AB = 2x, DB = 2x - y, and area of DBC = 16 Since area of ABC : area of DBC = AB : DB (Both triangles have the same height) area of ABC:16 = 2x:(2x-y) area of ABC = 32x/(2x-y) **1 Draw BE We get AC = 2y, EC = 2y-x, and area of ECB = 30 Since area of ABC : area of EBC = AC : EC area of ABC = 60y/(2y-x) **2 **1 = **2; 32x/(2x-y) = 60y/(2y-x) 8x/(2x-y) = 15y/(2y-x) re-arrange 16xy - 8(x^2) = 30xy - 15(y^2) 8(x^2) + 14xy - 15(y^2) = 0 (4x - 3y)(2x + 5y) = 0 x = 3y/4, -5y/2 --> neglect Negative value x = 3y/4 Substitute x in **1 area of ABC = 32*(3y/4)/(3y/2 -y) = 24y/(y/2) = 48 **** Area of ADFE (blue) = area of ABC - area of BDF - area of CEF = 48 - 8 - 15 = 25 Geometry only no Trigonometry Even you are in grade 9, you can do it.

[@ashoke7467]

Why area of BCD is 16 and area of BCE is 30 ?

[@BRUBRUETNONO]

Thanks for this journey !

[@amanuellebassi7294]

Thank you so much! You are fabulous

[@kranthi143B]

Nice explanation sir

[@eduardohenrique2242]

Awesome

[@Istaphobic]

I'm curious, because triangles ABC and ADE are similar, why is the ratio of their areas equal to the division of the squares of their bases?

[@MathBooster]

If 2 triangles are similar then ratio of their areas is equal to the ratio of squares of the corresponding side. For example : If triangle ABC is similar to triangle DEF then ratio of their areas will be equal to AB²/DE² that will be equal to BC²/EF² that will be equal to CA²/FD²

[@Istaphobic]

@@MathBooster Thank you for responding. I understand that is the principle, but what is the proof? Is there a name for this theorem?

[@valentinaivanova4806]

​@@IstaphobicIts almost obvious. If in such triangles the two sides are related as a1 = k * a2 and b1 = k * b2, and the angle (A) between them is the same ( ! ), then the areas of the triangles calculated by the formula S = 1/2 * a * b * sinA will be related as k^2

[@hanswust6972]

Beautiful!

[@user00000x]

soo cool i like your solution ❤

[@hijodebakunin]

Maravillosa solución.

[@luigipirandello5919]

Amazing.

[@MathBooster]

Thank you 🙂

[@johnbrennan3372]

Excellent solution

[@arulbiswas1260]

But how to do it in short way? I can't solve any of your problems , is it normal? Am a class 10th student from cbse in India.

[@MathBooster]

It is a math olympiad problem for grade 12 students. That's why it it hard.

[@maherkh24]

I think it is difficult solution

[@markkusalmelainen8529]

I fell off to close my eyes..boring. There is far easier way to go.

[@ADIBKHAN-x1u]

are you Indian???????????????????????????

[@reeturani610]

1st commentor😂😂

[@tarunmajumdar6166]

Excellent solution