Find Blue Area | Math Olympiad Geometry Question

[@اممدنحمظ]

تمرين جميل جيد . رسم واضح مرتب . شرح واضح مرتب . شكرا جزيلا لكم والله يحفظكم ويرعاكم ويحميكم جميعا. تحياتنا لكم من غزة فلسطين .

[@honestadministrator]

angle DCB = π/2 = angle CBA ∆ APD is equilateral. If angle BPA measures x° one gets AB / AP = sin (x) , Hereby angle CPD = 120° - x CD / PD = sin (120° - x°) b /a= sin ( 120° - x°) / sin(x°) Herein. a = AB and b = DC sin (120°) cot ( x°) - cos(120°) =b/a cot (x°) = (b/a - 1/2 ) / (√3 /2) = (2b -a) /(√3 a) BP = AB cot (x°) = (2b -a) /√3 Well symmetry denands CP =CD cot(120° - x°) = (2a -b) /√3 Now tan (120° - x°) = ( tan(120°) - tan(x°)) / ( 1 + tan(120°) tan(x°)) = ( -√3 - tan(x°)) /( 1 - √3 tan(x°)) = (√3 cot (x°) +1) /(√3 - cot(x°)) = (2b/a) /(√3) (1- (2b -a) /(3 a)) = 2b√3 / ( 4a - 2b) = b√3 /(2a - b) Hereby CP = CD cot (120° - x°) =(2a -b) /√3 This implies BC = BP + CP = ( a + b) /√3 Hereby desired area = (AB + CD) ( BP + PC) /2 = (a + b) ^2 /(2 √3)

[@XDXD-om9rc]

Very good, great job! I have a lot to learn from you, since i am young!🙏

[@Ankitsingh-y4j8m]

Ar(trapezium )= 1/2 (5+7)BC . BC= BP+Pc . Angle APB and DPC is 60 ° . Using trigonometry we get BP= 5/√3 and PC= 7/√3. BC= 12/√3. So Ar( trapezium) = 1/2(12)(12/√3)=24√3.

[@janolsen7915]

There is something vrong in your calculation, Angle APB and DPC are not 60 degree.

[@Ankitsingh-y4j8m]

@@janolsen7915 APD is equilateral triangle hence Angle APD is 60 °

[@honestadministrator]

angle DCB = π/2 = angle CBA Hereby AB || CD ∆ APD is equilateral. If angle BPA measures x° one gets AB / AP = sin (x) , Hereby angle CPD = 120° - x CD / PD = sin (120° - x°) 7 /5 = sin ( 120° - x°) / sin(x°) sin (120°) cot ( x°) - cos(120°) = 7/5 cot (x°) = (7/5 - 1/2 ) / (√3 /2) = 3 √3 / 5 BP = AB cot (x°) = 3 √3 CP = CD / tan (120° - x°) Now tan (120° - x°) = ( tan(120°) - tan(x°)) / ( 1 + tan(120°) tan(x°)) = ( -√3 - tan(x°)) /( 1 - √3 tan(x°)) = (√3 cot (x°) +1) /(√3 - cot(x°)) = (9/5 +1) /(√3) (1- 3/5) = 7/√3 Hereby CP = CD √3 / 7 = √3 Hereby desired area = (AB + CD) ( BP + PC) /2 = (5 +7) (3 √3 + √3) /2 = 24 √3

[@MarieAnne.]

Here was my approach: First, we establish that △APD is equilateral (and therefore its angles = 60°) and that ABCQ is a rectangle, therefore DQ = 2 Let ∠APB = θ. Then: ∠CPD = 180 − θ − 60 = 120 − θ ∠PDC = 180 − (120 − θ) − 90 = θ − 30 ∠ADC = 60 + (θ − 30) = θ + 30 In △ABP, sin θ = 5/a, cos θ = BP/a In △ADQ, cos (θ+30) = 2/a Using angle sum formula, we get: cos(θ+30) = cos θ cos 30 − sin θ sin 30 2/a = (BP/a)(√ 3/2) − (5/a)(1/2) → Multiply both sides by 2a 4 = BP√3 − 5 BP√3 = 9 BP = 3√3 Using Pythagorean Theorem in △ABP we get: AP² = AB² + BP² a² = 5² + (3√3)² = 25+27 = 52 Using Pythagorean Theorem in △DPC we get: DC² + CP² = DP² 7² + CP² = a² = 52 CP² = 52 − 49 = 3 CP = √3 In trapezoid ABCD, we have bases AB = 5 and CD = 7 and height = BC = BP + CP = 3√3 + √3 = 4√3 Area(ABCD) = 1/2 × (5+7) × 4√3 = 24√3

[@ВалерийАртемьев-й5я]

Let DC = a ; AB = b ; BC = h; AD = AP = DP = c; BP = h1; CP = h2. From equations: h = h1 + h2 h1² = c² - b²; h2² = c² - a²; h² = c² - (a - b)² We get c² = (a² - a*b + b²)*4/3; h² = (a + b)²/3; h = (a + b)*sqrt(3)/3. Area of the shape is S = (a + b)/2*h = = (a + b)² * sqrt(3)/6. In our case S = (7 + 5)²/6 *sqrt(3)= = 24*sqrt(3).

[@vipinmasih9740]

Cut them short when b,c are already established that they are right angle and you draw a perpendicular from a, It's obvious that it's going to be a rectangle

[@vijayanmaster-x3r]

Fine

[@miguelgnievesl6882]

I solved it in the following way: BP=x and PC=y, therefore height=x+yy>0. We proceed by applying the Pythagorean Theorem in the 2 right triangles; a²=x²+25 (i) and a²=y²+49 (ii). We subtract (ii)-(i) and get x²-y²=24 or (x+y)(x-y)=0. At this point, as x+y

[@TrebolManger]

En español: revise mi planteamiento, a vuelo de pajaro, no hay 2 ternas pitagorica enteras diferentes que tengan la misma hipotenusa entera, y q exige el tercer triangulo que es equilatero! Espero sus comentarios!

[@MarieAnne.]

Mistake: you forgot to take into account that the hypotenuses of △ABP and △DCP are two of the sides of an equilateral triangle, and therefore the angle between them must be 60°. With x = BP = 5 and y = PC = 1, then ∠APB = arctan(5/5) = 45° and ∠DPC = arctan(7/1) ≈ 81.87° Therefore, ∠APD = 180° - 45° - 81.87° = 53.13°, and △APD is not equilateral

[@pwmiles56]

Well, x+y isn't an integer. There is a third side of the triangle for which we have a²=(x+y)²+4 (iii) You can subtract (ii) from (iii) for an equation linear in y, solve for y, get it to a quadratic in x^2 and go from there. Sorry details a bit fiddly to type.

[@bienvenidos9360]

​@@TrebolManger Just because 2 triangles have the same length of hypoteneuse, it doesn't mean the triangles are proportional to a Pythagorean triplet. Angle APD = 60°, therefore angles BPA and CPD add up to 120° which they do. Angle BPA = 43.9° + angle CPD = 76.1° = 120°

[@TrebolManger]

@@bienvenidos9360 the second solution its isosceles triangle

[@alinayfeh4961]

Or a=2√(13)

[@yakupbuyankara5903]

24×(3^(1/2)).

[@alinayfeh4961]

The Area Bluesqure is √(1727)