Math Olympiad | Can you find area of the circle? | (Step-by-Step Math Explanation) | #math #maths

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Math Olympiad | Can you find area of the circle? | (Step-by-Step Math Explanation) | #math #maths

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[jimlocke]

As ybodoN noted, the location of D on AC is not given, so the area of the circle should be the same for all valid locations of point D. As I noted in my comment on ybodoN's post, if D is the midpoint of AC, the center of the circle lies on BC and is outside ΔABC when AD > CD. If we take the case where A and D are the same point, ΔABC is inscribed and has side length 13. Point O is equally distance from the 3 vertices of equilateral ΔABC. From properties of an equilateral triangle, this distance is side length divided by √3, or 13/√3 in this case. This is the radius, so area = πr² = π(169/3)

[hcgreier]

Draw a perpendicular on the 13-chord through point O, intersecting with circle. This chord is of length 2r, but the 13-chord cuts it in pieces r/2 and 3r/2. Chord theorem gives (r/2)(3r/2) = (13/2)², thus 3r²/4 = 6.5² → r² = (4·6.5²)/3 → r = 2·6.5/√3 = 13/√3 = 13√3/3. Area of circle is πr² = π(13/√3)² = 169·π/3.

[prossvay]

units
Area of the circle=π(13√3/3)^2=169π/3=177 square units.❤❤❤ Thanks.

[vaggelissmyrniotis]

DOB angle =2*DCB angle=120
Using law of cosines in DOB triangle we find r^2=169/3, therefore area=π*169/3

[Copernicusfreud]

After determining that triangle BOD is an isosceles triangle, Angles ODB and OBD must be 30 degrees if angle BOD = 120 degrees. Using the laws of sines was easiest for me to find the value of r. r/(sin 30) = 13/(sin 120). sin 30 = 1/2. sin of 120 = √3/2. Plug-and-chug gives r = 13/√3. r^2 = 169/3. The area of the circle = 169 * pi/3.

[solimana-soli]

ACB = 60° then DOB = 120° (angle in the center)
called OH the projection of O on chord DB we get two 30°, 60°, 90° triangles DOH and HOB (because the height of equilateral triangle DOB) whose sides are x, 2x and x√ 3
being the major catetus 13/2 => r = 13/3√ 3
Area = 169/3 pi

[tombufford]

At a quick glance: Using Central angle theorem, Angle BOD = 2 * BCA = 120. Triangle ABC is equilateral with 60 degree angles. Angle ODB = 2* 60 = 120. Drawing a perpendicular from midpoint DB, E to O. Then angle ODE = 180 - 90 -60 = 30. Then EO = tan(30) * 6.5 = 3.753 and the radius ^2 = 6.5^2 + 3.753^2. Area = PI * R^2 = PI * 56.33 = 177 area units.

[ybodoN]

D could be anywhere on AC, so let's put it in the middle of AC.
In this special case, DC = BC/2 = radius of the circle = 13/√3.
Therefore, the area of the circle is π 169/3 ≈ 177 square units.

[LuisdeBritoCamacho]

The inscribed angle [DOB] = 120 degrees; because angle [BCD] = 60 degrees. Angles of an Equilateral Triangle.
The Triangle [BDO] is isosceles; OD = OB; and angle [ODB] is equal to angle [OBD] = 30 degrees
So:
OD (Radius) = (13/2) * cos (30) = 6, 5 / (sqrt(3) / 2) = 13*sqrt(3)/3 ~ 7, 5055 lu
Area of the Circle:
A = r^2*Pi
A = (13*sqrt(3)/3)^2 * Pi
A = ((169*3)/9) * Pi
A = 507/9 * Pi
A = 169/3 * Pi
A = 169*Pi / 3 su (exact measure)
A ~ 176, 976 su
Answer: The Area of the Circle is ~ 176, 976 square units.

[bothalex]

Another solution:
Draw diameter from D-O-F; in triangle DBF angle FBD=90 degrees, angle BFD=60 degrees (equal with angle DCB)=> angle BDF=30 degrees; DB=13
cos (BDF)=cos =>R=13/(SQRT3) => A=PI*169/3

[jan-willemreens]

... (1) Equilateral triangle ABC means all angles are 60 deg. .... (2) Drawing line segments OB & OD ... triangle ODB is an isosceles triangle, where Angle(DOB) = 2 * ANGLE(ACB) = 120 deg., ANGLES (BDO)& (DBO) are equal and measure 30 deg. ... (3) Applying SINE RULE on triangle ODB to find I OD I, and is equal to the radius (R) of the circle ... SIN(120)/13 = SIN(30)/ I OD I ... I OD I = SIN(30)/SIN(120) * 13 = SQRT(3)/3 * 13 = 13/3 * SQRT(3) = R ... finally AREA(circle) = pi * R^2 = 169/3 * pi .... thank you for your always clear presentation .... best regards, Jan-W

[genelowry]

Or using inscribed angle Theorem and law of sines, we get the same answer.

[MegaSuperEnrique]

Thank you for pointing out that we didn't need to rationalize the denominator, since we were squaring it at the end.

[nehronghamil]

Alternate Solution:
Extend line form O to intersect BC perpendicular @ D. Let s be side of equilateral triangle
Let theta = angle BDC
from central angle theorem: theta = 1/2 angle BOC =angle BOD

s * sin (60) = 13 sin (theta)
sin(theta) = [s * sin (60)] / 13 (1)

r * sin(theta) = s/2 (2)

(1) & (2) :
r * [s * sin (60)] / 13 = s/2
r = 13 / [s * sin (60)] = 13/(3^.5)

A = pi * r ^2 = pi * 169/3

[Tanduy]

30°tan = .57735
OE = .57735 x 6.5 = 3.75278
OB = Radical of 3:75278² + 6.5² = 7.50555 = radius
Area of the circle = 7.50555² x 3.1416 = 176.9768 or 176.98

[misterenter-izrz]

Good explanation, I is not smart enough to solve it in justifiable way, I solve in deluded way, if r is uniquely determined, but the side of the equilateral triangle can be changed, the extreme case is that the triangle is just inscribed in the circle, then the radius can be counted immediately. 😅

[marcelowanderleycorreia]

13 divided by sin of 60º = 2.R (Law of sines), so R^2 = 169/3, therefore, the area is (169.pi)/3

[rajendraameta]

By using Pythagoras theorm, r^2-(r/2)^2=(13/2)^2 gives3r^2 /4=169/4 & so r^2=169/3 & so A=pi.169/3

[marcelowanderleycorreia]

Very good sir. But I think is much more quick if we use the Law of sines.

[misterenter-izrz]

Just think the triangle inscribed in the circle, so r=13/2 /cos 30=13/2 2/sqrt(3)=13/sqrt(3), therefore the answer is 169/3 pi=177 approximately. 😊