There is no one like you, you are the best teacher in the world🥰

@PreMath

14 күн бұрын

Thanks dear for your continued love and support!❤️ You are the best!

The barrier students must overcome is that there are not enough equations to solve for a and b. PreMath was able to narrow it down to one equation and 2 unknowns (a and b). However, we only need to know the product of a and b to find the area of the white triangle, not the individual values of a and b. One solution approach would be to find the area of the white triangle for the special case of ABCD being a square. Then, letting s be the side of the square, a = b = s and you have one equation with one unknown. The problem statement does not rule out a square. So, you solve, get a value for s² and subtract the combined area of the red, green and blue triangles to get the area of the white triangle. The problem statement implies that the same solution applies to the general case of the rectangle, so you present that as your solution. If the problem statement is modified to require students to solve ABCD being the general case of a rectangle, students are effectively given a clue that the solution is valid for a range of values of a and b, in this case, whenever the product ab equals 18 + 2√(51).

@PreMath

14 күн бұрын

Super! Thanks for the nice feedback ❤️

@shahdmohammed4597

2 күн бұрын

نشكر حضرتك للمجهود والمعلومة انا اوجدت الناتج بنظرية ايجاد المساحة الداخلية المظلله فى الشكل =مساحة الشكل الخارجى _مجموع مساحة الشكل الداخلى والناتج كان عدد صحيح وليس عدد عشرى. اسفة للاطالة وارجو التصحيح لو هناك خطا فى الحل

Let AD=BC=a ; AB=CD=b Area of rectangle ABCD=ab CF=12/a ; AF=10/b DE=AD-AE=a-10/b DF=CD-BF=b-12/a Area of triangle DEF=1/2(DE)(DF)=1/2(a-10/b)(b-12/a)=1/2ab(ab-10)(ab-12)=7 (ab-10)(ab-12)=14ab ab=18+2√51=32.28cm^2 White triangle area=32.28-(5+6+7)=14.28 cm^2.❤❤❤ Thanks sir Best regards.

@PreMath

14 күн бұрын

Super! You are the best🌹 Glad to hear that! Thanks for sharing ❤️

Nice! I finsihed this one very quickly ⏰! I think what maybe made a lot of people fail to do this puzzle is labelling all the sides of the triangles with different variables instead of writing the height out in terms of the base, since you are given the area. Its a common pattern i have seen in other problems on your channel!

@PreMath

14 күн бұрын

Nice work! Thanks for the feedback ❤️

Very well explained

Assumes that at angles at vertices of abcd are 90degrees, not given, anyone can make assumptions.

@waheisel

6 күн бұрын

He does say it is a rectangle.

@Deribus575

6 күн бұрын

Which is an assumption not given in the problem

Thank you Sir. I always like your explanations!

Thank you! I tried to do this on my own before watching the video, and I got the correct answer 🙂

Nice one👍

@PreMath

14 күн бұрын

Thank you! Cheers!🌹❤️

Dividimos ABCD en cuatro celdas trazando una horizontal por E y una vertical por F→ Área de las celdas: (2*5-a); a; 14; (2*6-a)→ (10-a)/14=a/(12-a)→ a=18-2√51→ Área ABCD =10+12+14-a=36-18+2√51=18+2√51→ Área BEF =(18+2√51)-5-6-7=2V51 =14,2828.... Gracias y saludos.

@PreMath

14 күн бұрын

Excellent!🌹 You are very welcome! Thanks for sharing ❤️

@think_logically_

13 күн бұрын

Un error pequeño: en lugar de (10-a)/14=a=(12-a) debe ser (10-a)/14=a/(12-a). But it's clear anyway (sorry my Spanish is far from perfect). Excellent solution. I was looking for something like that, but failed to complete.

@santiagoarosam430

12 күн бұрын

@@think_logically_ Gracias por advertirme de la errata en la ecuación. Corrijo y quedo muy agradecido. Hasta siempre.

@think_logically_

12 күн бұрын

@@santiagoarosam430 Por nada

2sqrt51 calling B the base of rectangle, H its height, FC = x, AE = y we can write: x*H = 12 y*B = 10 (B - x)*(H - y) = 14 => BH - xH - yB + xy = 14 BH = 36 - xy Now we can divide the rectangle in 4 smaller rectangles in which the upper right rectangle has area = xy, then tracing perpendicular to base and height. Calling xy=a we can calculate the areas of each rectangles as: 10 - a | a _____________ 14 | 12 - a doing the crossed product we have: 14 a = (10-a)*(12-a) a² - 36a + 120 = 0 a = 18 - 2sqrt51 = xy BH = 36 - (18 - 2sqrt51) = 18 + 2sqrt51 White area = 18 + 2sqrt51 - 5 - 7 - 6 = 2sqrt51

@PreMath

14 күн бұрын

Excellent! Thanks for sharing ❤️

Ok got it correct. At first it seemed to become too complex with sqrt(816) and that there may be a simpler solution. But just continuing I got 2*sqrt(51)

@PreMath

14 күн бұрын

Excellent! Thanks for the feedback ❤️

Thank you!

@PreMath

14 күн бұрын

You are very welcome!🌹 Thanks ❤️

by formula, A^2=(5+6+7)^2-4×5×6=18^2-120=204, A=2sqrt(51).😊

@PreMath

14 күн бұрын

Excellent! Thanks for sharing ❤️

@manojmiya9141

14 күн бұрын

Proof of (5+6+7)²-4*5*6

AE=bp => DE = b(p-1). Also DF=aq => CF=a(1-q). So we have abp=10; ab(1-q)=12; ab(1-p)q=14. But ab=s - the area of rectangle. So p=10/s => 1-p=1-10s=(s-10)/s and 1-q=12/s => q = (s-12)/s. Thus from third equation (s-10)(s-12)/s² = 14/s with leads to quadratic equation s²-36s+120=0. The rest is clear.

very good

The solution may be complex at first, but I was able to get it.

@PreMath

14 күн бұрын

Bravo🌹

Fine.

@PreMath

14 күн бұрын

Glad to hear that! Thanks for the feedback ❤️

❤❤❤

@PreMath

14 күн бұрын

Thanks dear 🌹❤️

14,283 cm2

Why the negative root is not possible?

Great problem and solution. Afraid I am one of the 95%.

نشكرك

خیلی زیبا و جالب

@PreMath

14 күн бұрын

ممنون فاطمه عزیزم 🌹❤️

my answer is 6 cm square

Why is 18-2√51 scenario rejected?

@DeathZebra

14 күн бұрын

Too small. Area>18 (the combined area of the coloured triangles)

@bfelten1

14 күн бұрын

@@DeathZebra@DeathZebra: Well, that motivation must be in the solution for it to pass the test. But you have DF = a - 12/b, which means that a > 12/b and therefore ab > 12 (likewise for ED, but only ab > 10). So 18 - 2 * sqrt(51) must be greater than 12 -- it is slightly less than 4, so dismisst. QED.

1) Let's baptize things! 2) x = 5 sq cm 3) y = 6 sq cm 4) z = 7 sq cm 5) Fortunately we have a General Formula for these cases : White Area = "sqrt((x + y + z)^2 - 4xy)" 6) WA = sqrt((18)^2 - 4*(30)) ; WA = sqrt(324 - 120) ; WA = sqrt(204)sq cm ; WA = (2*sqrt(51)) sq cm ; WA ~ 14,3 sq cm 7) Final Answer : The Area of the White Square is equal to approx. 14,3 Square Cm. NOTE: I reach to the conclusion that the Domain of the Solution (White Area) should be somewhere between : 8 sq cm The Total of Possible Ordered Pairs (X ; Y) of Integer Solutions is : S {(18 ; 0) ; (20 ; 2) ; (22 ; 4) ; (24 ; 6) ; (26 ; 8) ; (28 ; 10) ; (30 ; 12) ; (32 ; 14) ; (34 ; 16) ; (36 ; 18)} X = Area of Rectangle [ABCD] and Y = Area of White Triangle [BEF]. And X - Y = X - (5 + 6 + 7) sq cm ; X - Y = 18 sq cm. The difference must always be 18 sq cm.

@PreMath

14 күн бұрын

Excellent! Thanks for sharing ❤️

bad number selected. furhter ,when calculating delta, yoiu'd better keep facters(in this case, 4^2 rather than calculate it out.

√(5+6+7)^2---4×5×6

@PreMath

14 күн бұрын

Thanks for sharing ❤️

The top 5% that did solve this this International Mathematical Olympiad puzzle all went on too be Substitute Teachers! 🙂

@PreMath

14 күн бұрын

😀 Thanks ❤️

The area R of the whole rectangle is: R = AB*AE+ DE*DF + BC*CF - AE*FC = 10 + 14 + 12 - AE*FC (1) Since AE = 10/AB and FC = 12/BC, we get AE*FC = 120/(AB * BC) = 120/R Plugging this into (1) we get: R = 36 - 120/R R^2 - 36R + 120 = 0 This quadratic equation yields R = 18 + 2*sqrt(51) Now our answer = 18 + 2*sqrt(51) - 5 - 7 - 6 = 2*sqrt(51)