Olympiad Mathematics | A Very Nice Geometry Problem

[raghvendrasingh]

Diameter= 5×2/√3 hence radius is 5/√3
if we join centre to upper vertex of triangle we have to find sum of areas of a triangle and a sector
area of triangle = (1/2)(1/2)(25/√3) = 25√3/12
area of sector = π× ( 25/3) × (1/6) = 25π/18
Total area 25 (2π+3√3 )/36

[jimlocke]

<CAB is an inscribed angle and is given as 30°, so it intercepts a 60° arc. After constructing OC, we find that central <COB intercepts the same arc, so is 60°. <AOB is a straight angle, thus 180°, leaving <AOC = 120°. Sum of angles of a Δ = 180°, so <ACO = 30°. Drop a perpendicular from O to AC and label the intersection D. AC is bisected. ΔADO and ΔCDO are congruent 30°-60°-90° special triangles. From ratio of sides of 30°-60°-90° special triangles, OD = 2.5/√3 and AO = CO = 5/√3. Area of each triangle = (1/2)(2.5)(2.5/√3) = 6.25/(2√3) = 25√3/24. The purple area = sum of areas of ΔADO and ΔCDO and sector COB. Sector COB = (60°/360°)π(5/√3)² = 25π/18. Total area = (2)(25√3/24) + 25π/18 = (25√3/12) + 25π/18 = (25)(3√3)/36 + (25)(2π/36) = 25(3√3 + 2π)/36, as Math Booster also found.

[IllllllllllIIlIIIIlIlllI]

I was able to solve the problem before seeing your solution! I'm really happy. What a great feeling

[lasalleman]

I just set up a 30, 90 60 degree triangle. Going Pythagorean, line AB comes out to 5.78. Total area of the triangle is 7.22. The line from C to B Is the sine of angle 30 and has a distance of 2.89. Then simply find the segment area between the arc CB and line CB. Use centroid of circle at O (60 degrees) as bases for angle. Use formula-- area of segment = R^2/2*(pi/360 *Angle at O - Sine of O)
Area of segment turns out to be .7566. Total purple shaded area is 7.97. Right on the money!

[solimana-soli]

Being angle CAB = 30°, we can also draw an equilateral triangle with side=5 inscribed in the circle, the area over the chord AC (segment) is 1/3 the difference between tha area of the circle and the area of the equilateral triangle.
Area equilateral triangle = s²/4*√ 3 = 25/4* √ 3
radius circle*2 = side/sin60°= 5/√ 3/2 = 5/3√ 3
Area circle = 25/3pi
yellow area = 1/2*area circle - (area circle - area triangle)*1/3
yellow area = 1/6*area circle + 1/3*area triangle
yellow area = 1/6*25/3pi + 1/3*25/4√ 3
yellow area = 25/18pi + 25/12√ 3 = 25/36*(2pi +3√ 3)

[ritwikgupta]

Nice method. Nice problem. So many methods. Another:
AO=OC, ∆AOC is isosceles, ACO=30°. So the external angle COB=30°+30°=60°
Cosine formula gives R=5*5/2*5/cos30°
So we get area of segment COB
Area of ∆AOC=Ht of ∆AOC*R/2
Ht of ∆AOC=5*sin30°
All methods are connected.

[saronohandoyo]

I try to find the diameter first since AB is diameter, so if point C is touching the circle then ABC will be right triangle.
AC=5. Right triangle with 30 degree will be x:x√3:2x for base:height:hypotenuse
from this we get height = x√3 = 5 so we get base of triangle x=5/√3
hypotenuse = 2x =10/√3 which also this is a diameter.

[Latronibus]

This is a relatively easy one by this channel's standards. Draw BC, ABC is 30 60 90, the longer leg is 5, so the shorter leg is 5/sqrt(3), so that area is 25/(2sqrt(3))=25sqrt(3)/6. Also, AB is the hypotenuse of ABC, so the diameter is 10/sqrt(3). Now draw OC. COB is 60 degrees, and OC=OB, so COB is equilateral by SAS, and its side length is the radius, which is 5/sqrt(3). Therefore the area of COB is sqrt(3)/4 25/3=25 sqrt(3)/12, and the area of that sector is 25 pi/18. So now we have ABC + sector - COB, to avoid double counting COB. So that adds up to 25 sqrt(3)/12 + 25 pi/18.

[michaeldoerr]

The shaded area is 25[(3sqrt(3)+2pi)/36] units square. Looks like I need to know that if a circle theorem is used to show why there does need to be a right angle constructed in a sector ever, this is the reason why. And I shall use this for practice!!!

[devondevon]

7.97333
Easy
Let's label the shaded region ABC
draw a line from B to C to form a right triangle (Thales Theorem)
This triangle is 30-60-90
Hence, the sides are 5, 2.88675, and 5.7735.'
Hence, the radius also = 2.88675 since it is one-half the diameter.

Draw a line from the circle's center to C to form an isosceles triangle,
(30, 120, 30 degrees) . Its sides are 5, 2.88675, and 2.88675
Use Heron's formula to calculate the area = 3.61
Since the angle to the left of the triangle is 60 degrees ( 180 - 120 degrees),
and since the radius of the circle is 2.88675, then 1/6 (60/360) of the circle is
area = remaining sector.
The circle area = 26.16 and hence, the area of the sector = 26.16/6 = 4.36
Hence, the area of the shaded region = 3.61 + 4.36 = 7.9733 answer

[anikdolui]

Let O be the centre and AB = 2R
Join O and C & B and C.
Hence, AO = BO = CO = R.
Angle ACB = 90 deg bcz AB is diameter of the circle.
Hence, applying Pythagoras theorem in right angled triangle ABC we get R = 5/✓3.
Area of ABC = 1/2.AC.BC
= 1/2.5.5/✓3
= 25/2✓3

Please comment if i have committed any mistake.

[alexniklas]

5^2+r^2=(2r)^2; 25+r^2=4r^2;
r= 5/√3;
S=πr^2×30°/360°= π×25/36
Thanks sir!

[pojuellavid]

АСB=90°, BC=x, AB=2*x.
x^2+5^2=(2*x)^2. --> x=5/√3
S=((5/√3)*5)/2=25/√12

[marioalb]

2R = 5 / cos 30° --> R=5/√3 cm
A = A₁ +A₂= ½cRsin30° + ½.⅓πR²
A = 3, 6084 + 4, 3634
A = 7, 97176 cm² ( Solved √ )

[YardenVokerol]

O=½[AB], H=½[AC],
R²=|AO|²=|AH|²+|HO|²
|AH|=½|AC|=½•5=5/2
∠A=30°⇒|HO|=|AH|/√3

R=5/√3
∠A=∠ACO⇒∠CON=30°+30°=60°=π/3
S(OC͡B)=R²•(π/3)/2

S=25/(4√3)+25π/18
S=25√3/12+25π/18