Math Olympiad Geometry Problem | Find the blue shaded area | Important Geometry skills Explained

[@اممدنحمظ]

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

[@krishnamoyghosh6047]

Very complicated solution. We can do easily by working out value of AD and sine of angle ADC. After that apply sin area of trigonometry.

[@harikatragadda]

Since ∠A + ∠C = 180°, ABCD is a Cyclic quadrilateral.
Since Chord AD subtends 45° at B, it also subtends 45° at C.
If AD = AB= a, then BD = a√2 and by Ptolemy's theorem,
2*a + 3*a = √2a*AC
AC = 5/√2
Blue Area = ½(AC*Sin45)*DC = 15/4

[@ΓΕΩΡΓΙΟΣΛΕΚΚΑΣ-μ9μ]

Πολύ ωραία και η λύση του χρήστη harikatragadda. Μπράβο που χρησιμοποίησε Πτολεμαίο.

[@miguelgnievesl6882]

If BCD is a right triangle, then BD = √13. ABD is an isosceles right triangle, so AB=AD=(√26)/2. We focus on triangles ABC and ADC which have side AC in common.

[@Diph64]

Great solution

[@ΓΕΩΡΓΙΟΣΛΕΚΚΑΣ-μ9μ]

Ακριβώς! Ακόμα, εναλλακτικά μπορούμε να εφαρμόσουμε τύπο του Ήρωνα στο ACD και με αρκετές πράξεις είναι η αλήθεια φτάνουμε στο ίδιο αποτέλεσμα. Έτσι χρησιμοποιούμε ΜΟΝΟ νόμο συνημιτόνων στη λύση. Πολύ ωραία και η λύση του χρήστη harikatragadda. Μπράβο που χρησιμοποίησε Πτολεμαίο.

[@dileepmv7438]

What was the need of considering it as a cyclic quadrilateral, calculating the distances to the center etc. we can do this from basics

[@marioalb9726]

Upper right triangle BDC:
d² = 2²+3²
d = √13 cm
Angle BDC :
tan α₂ = 2/3 -----> α₂ = 33,69°
Angle ADC :
α = α₁ + α₂ = 45° + 33,69°
α = 78,69°
Isosceles right triangle
b = d cos 45° = √13 / √2
b = √6,5 cm
Blue triangle area:
A = ½.b.h
A = ½ . √6,5 . 3 cos (90°-78,69°)
A = 3,75 cm² ( Solved √ )
Much easier than this complicated video !!!

[@marioalb9726]

This quadrilateral is cyclic, has 2 opposite right angles.
Can be applied Ptolemy theorem
[ a.c + b.d = e.f ]
Being 'a' the side of isosceles right triangle:
2.a + 3.a = (√2.a) . d
d = 5/√2 = 3,5355 cm
Area of blue shaded triangle:
A = ½ b.h
A = ½ b . d cos45°
A = ½ . 3 . 5/√2 cos 45°
A = 3,75 cm² ( Solved √ )

[@mohammadazadi4535]

Hi and thanks a lot. I solved this via formola that relate primeter and area of triangel.

[@misterenter-iz7rz]

Computation will be quite long for we have count the sum of areas of three sectors, as it is a cycilquadrilateral with radius sqrt(13)/2 then we locate the position of the center and count the length of AB=AD=sqrt(2)/2×sqrt(13)/2=sqrt(26)/4......

[@charlesbromberick4247]

nice job

[@raghvendrasingh1289]

BD = √13 and AD = AB = √26/2 because triangle ABD is right angled isosceles triangle
semi perimeter of this cyclic quadrilateral is (5+√26)/2
Area is √{ (s-a)( s-b)(s-c)(s-d)}
=√ { (5/2)(5/2)(√26+1)/2 (√26-1)/2 }
= 25/4
ratio of areas of triangles ADC and ABC is 3 : 2 because AD=AB and angles ADC and ABC are supplementary
hence required area = (25/4)× (3/5) = 15/4

[@jimlocke9320]

We really don't need to know that ABCD are concentric and where the circle's center is. Furthermore, the formula for the tangent double angle avoids the radicals. Construct BD and let

[@santiagoarosam430]

2²+3²=13=BD²=(a√2)²=2a² → a²=13/2 → a=√26 /2 →→ La horizontal por C y las verticales por D y B definen dos triángulos rectángulos con razón de semejanza s=2/3 y lados respectivos [b/h/3)] y [(2b/3)/(2h/3)/(2)] → h=a+(2b/3) =a+(2/3)(a - 2h/3) =(5√26 /6 )-4h/9 → 13h/9=5√26 /6 → h=15√26 /6 →→→ Área azul = ah/2 =(1/2)(√26 /2)(15√26 /6) =15/4
Gracias y un saludo cordial.

[@manojitmaity7893]

Sir please solve this sum :
If a,b,c are the three distinct side lengths of a triangle so that the area of the triangle is 1 unit². Then prove that b>_2½ ( a>_b>_c)

[@sandipanbanerjee5010]

For a given triangle, the area is (1/2)bcsinθ, where b, c are two of its sides and θ the angle between them. Here,
(1/2)bcsinθ = 1 => bcsinθ = 2 => (b^2)sinθ >= 2 (by assumption) => b^2 >= 2 (sinθ >= 1) => b >= √2. Ekhane distinction between sides ta kono kaje laglo na, thakle kono problem nei jodio.

[@martinpletikosic3053]

Hey math friends! Let's get playful and creative! Rotate triangle ADC around point A by +90°. I'll let you figure out why that works.

[@זאבגלברד]

There is the theorm: AC*BD = AD*BC + AB*CD ... Find BD then find AD .....

[@楊俊賢-f7y]

(1/2)×3×(5/2)=15/2

[@NaceerAlassady-x2l]

بمجرد النظر إلى معطيات السؤال يتضح أن الشكل مستطيل لأن الزاويتان المعلومات ...متقابلتان وقيمتها تسعين درجة فلا يمكن لأي ضلع من الاضلاع المتقابلة أن يكون أكبر من الاخر.
وعليه يكن مساحة الجزء المضلل هو مساحة نصف مستطيل اي يكون الناتج هو ٣
تحياتي....وارجو التصحيح ان كنت على خطأ

[@sandipanbanerjee5010]

You are not getting it right, opposite angles might be ninety degrees, that doesn't mean other two should also equal ninety degrees.

[@Mou7_Mou7an]

أنشئ المثلث ABD قائم الزاوية . أرسم الدائرة المحيطة بالمثلث. سيكون قطرها هو BD . إختر أي نقطة M من نصف الدائرة التي لا تضم النقطة A. المثلث BMD سيكون دائما قائم الزاوية في M .
أي ان كل النقط في نصف الدائرة تحقق شرط المعطيات دون ان يكون ضروريا ان يكون المضلع الرباعي ABMD مستطيلا.
تنويه : هناك نقطة وحيدة C تحقق شرط المسافتين BC=2 و CD=3.
و هناك نقطة وحيدة (مخالفة للنقطة C ) تحقق شرط ان رباعي الأضلاع هو مستطيل.

[@NaceerAlassady-x2l]

@@sandipanbanerjee5010 thats right...Iam ronge

[@luiscalpo3813]

15/4 is not 15 by 4. it is 15 over 4