A Nice Geometry Problem | Olympiad Mathematics | Important Geometry and Algebra Skills Explained

A Nice Geometry Problem | Olympiad Mathematics | Important Geometry and Algebra Skills Explained

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[اممدنحمظ]

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

[santiagoarosam]

Si tomamos CA como base del triángulo azul, su máxima superficie se alcanza cuando el vértice B está lo más alejado posible de CA; su posición se puede hallar trazando una paralela a CA que sea tangente a la circunferencia → Con el vértice B en el punto de tangencia, el radio OB debe ser perpendicular a la tangente trazada y a la base CA → Si OB corta a AC en D, la máxima altura del triángulo será OB-OD → AC²=OC²+OA² → AC=5√5 → La razón de semejanza entre los triángulos AOC y ADO es s=5/5√5=√5/5 → OD=10√5/5=2√5 → Máxima altura azul =6√5-2√5 =4√5 → Área máxima azul =(1/2)(5√5)(4√5)=50
Interesante problema. Gracias y un saludo cordial.

[SAHIRVLOGCLP]

You made Math interesting, thank you..

[quigonkenny]

The area of a triangle is the product of its base and its height, divided by 2. The base of this triangle, AC, is set, so maximum value of the blue triangle area is going to come when the height is at its maxiumum, when the vertex of the triangle, along the circumference, is furthest perpendicularly from AC. If D is the point where OB and AC intersect, then DB will be its longest when OD is its shortest, as OB is the radius of the circle and its value is set.

Draw OD, where D is the point on AC where OD is perpendicular to AC, and extend it to B. Let OD = x, so DB will be 6√5-x.

Triangle ∆COA:
OA² + OC² = AC²
5² + 10² = AC²
AC² = 25 + 100 = 125
AC = √125 = 5√5

If ∠OAC = α and ∠ACO = β, where β is the complementary angle to α (α+β = 90°), then ∠COD = α and ∠DOA = β, as ∆ODC and ∆ADO are right triangles. Thus ∆ODC and ∆ADO are similar to ∆COA.

OD/OA = OC/AC
OD/5 = 10/5√5 = 2/√5
OD = (5)2/√5 = 2√5

DB = OB - OD = 6√5 - 2√5 = 4√5

A = bh/2 = (5√5)(4√5)/2
A = 20(5)/2 = 50 sq units

[じーちゃんねる-vn]

The base of △ABC is AC=5√5, the height is maximum when AC⊥OB, so h≦6√5-2√5=4√5 ∴S≦(1/2)(5√5)(4√ 5)=50

[uttiyamajumdar]

If coordinates of point B be (x, y) then,
=1/2 (10x-50+5y)
=5/2 (2x+y-10)=A (say)
Therefore, A=5/2 (2x+y-10)
But the point B(x, y) lies on the quarter circle whose equation is given as, x^2+y^2=180.
So, A=5/2 (2x+√(180-x^2 )-10)
⇒dA/dx=5/2 (2-2x/(2√(180-x^2 )))=5/2 (2-x/√(180-x^2 ))
Setting dA/dx=0 we get,
x/√(180-x^2 )=2
⇒x^2=4(180-x^2 )
⇒5x^2=180×4
⇒x^2=36×4
⇒x=6×2=12
Now, (d^2 A)/(dx^2 )=-1/√(180-x^2 )-x/((180-x^2 ) √(180-x^2 )) is negative when x=12
Therefore, A is maximum when x=12, When x=12, y=6
Therefore, A_max=5/2 (2×12+6-10)=5×20/2=5×10=50 sq.units

[harikatragadda]

Lovely problem!
Extend AC to form a chord of the circle. The perpendicular distance of this chord from the circumference is the height of the triangle. The maximum height is the radial perpendicular bisector. The lower part of this radial line divides the right triangle AOC into Similar two right triangles, at a distance 2√5 from the center. Thus the max height= 6√5 - 2√5 = 4√5
Max Area = ½*(5√5)*(4√5)

[BRUBRUETNONO]

Thanks of your interesting problem.
Here is the way I solved it. Of course, I didn't look at your solution before trying to solve it.
Greetings !

First is to recall that the triangle area is the base multiplied by the opposite height divided by two.
In this case, the triangle ABC has got its side AC with a constant length value of
So to maximise, triangle ABC area we have to find the position of point B maximizing the heigth opposite to side AC.
As B is on the quarter circle, heigth length is maximized when (OB) is perpendicular to (AC).
So angle (AOB) equals (ACO).
Let's call H perpendicular projected point of B on side (AC), triangles (OHA) and (COA) are similar .
Then so OH=10/sqrt(5)=2sqrt(5)
So height length max of triangle (ABC) is then

Then max area of (ABC) is then
Conclusion: Max area of ABC is 50.

[Clock_Tune]

Sir through these problems you made mathematics intersting

[marioalb]

Base of white triangle is the same that base of blue triangle

Base of white and blue triangles:
b² = 10²+5²
b = √125 = 11, 18 cm

Height of white triangle:
b.h = b' h'
h = b'h'/ b
h = 5 . 10 / 11, 18
h = 4, 4721 cm

Maximum area of blue triangje is when both heights of triangles are aligned

Height of blue triangle:
H = R - h
H = 6√5 - 4, 4721
H = 8, 944 cm

Area of blue triangle
A = ½. B. H
A = ½ 11, 18 . 8, 944
A = 50 cm² ( Solved √ )

[spacer]

The tangent line to the circle at point B must be parallel to line AC, as otherwise you could move point B along the circle to get a larger b. Since any circle tangent line is perpendicular to OB, line AC must also be perpendicular to OB. With that, it's simple to work out a and b and then area of ABC.

[CapnbloodBeard]

Thank you for this beutiful videos its not just equations its art thank you