Reciprocal Sum in a Circle

In order to obtain the wanted ratios, we first calculate the fractions $\frac{AO}{AD}$ , $\frac{BO}{BE}$ , and $\frac{CO}{CF}$ . Since the height of C to AO has the same length as the height of C to AD, we have that the ratio of the area of AOC to the area of ADC is $\frac{AO}{AD}$ . Similarly, the ratio of the area of AOB to the area of ADB is also $\frac{AO}{AD}$ . Since these two area ratios are both equal to $\frac{AO}{AD}$ , we have the ratio $\frac{AOC+AOB}{ABC}=\frac{AO}{AD}$

Doing this for vertices B and C, we get that $\frac{BOC+BOA}{ABC}=\frac{BO}{BE}$ and $\frac{COA+COB}{ABC}=\frac{CO}{CF}$ .

Adding these three equations, we have that $\frac{CO}{CF}+\frac{BO}{BE}+\frac{AO}{AD}=\frac{2AOC+2AOB+2BOC}{ABC}$ . Since AOB+AOC+BOC=ABC, and AO, BO, CO are all radii of the circumcircle, we have that $2=R\times (\frac{1}{AD}+\frac{1}{BE}+\frac{1}{CF})$ , so our desired value is equal to $\frac{2}{R}=80$

A circumcircle is drawn for triangle ABC. By using angle in same segments are equal and angle in semi circle is 90 degrees we get that angle BAD = (90-C) and angle DAC = (90-B) Let us take side AB=c,BC=a,AC=b. and let [ABC] represent the area of triangle ABC

Now, [ABC]={1/2 (c) (AD) (Sin(90-C))}+{(1/2) (b) (AD) (Sin(90-B))} On Simplifying we get (1/AD)=(1/2) {(c CosC)+(b CosB)}/[ABC] Similarly, (1/BE)=(1/2) {(c CosC)+(a CosA)}/[ABC] (1/CF)=(1/2) {(a CosA)+(b*CosB)}/[ABC]

Adding these three equations we get, (1/AD)+(1/BE)+(1/CF)={(a CosA)+(b CosB)+(c CosC)}/[ABC] = {(2RSinA CosA)+(2RSinB CosB)+(2RSinC CosC)}/[ABC] ={R(Sin2A+Sin2B+Sin2C)}/[ABC] ={4R (SinA) (SinB) (SinC)}/[ABC] ={4 (R^2) (SinA) (SinB) (SinC)}/{R [ABC]} ={2 [ABC]}/{R [ABC]}
=2/R =2/0.025
=80

When triangle ABC is a right angle triangle, wlog angleBAC=90, B=F, O=D, C=E, 1/AD+1/BE+1/CF=1/0.025+1/(2x0.025)+1/(2x0.025)=80. Let [ABC] be the area of triangle ABC. Let G,H,I be the projections of O on BC, AC, AB respectively. When triangle ABC is acute, [ABC]=1/2xADxBDx(sin angleADB)+1/2xADxDCx(sin angle ADC)=1/2xADxBCx(sin angleADB)=1/2xADxBCxOG/OD=1/2xADxBCxOG/(AD-R) Rearranging, 1/AD={[ABC]-1/2xBCxOG}/{[ABC]xR} Similarly, 1/BE={[ABC]-1/2xACxOH}/{[ABC]xR} and 1/CF={[ABC]-1/2xABxOI}/{[ABC]xR} Adding the three equations, 1/AD+1/BE+1/CF={3[ABC]-[ABC]}/{[ABC]xR}=2/R=2/0.025=80 When triangle ABC is obtuse, wlog angleBAC>90, [ABC]=1/2xADxBDx(sin angleADB)+1/2xADxDCx(sin angleADC)=1/2xADxBCx(sin angleADB)=1/2xADxBCx(sin angleGDO)=1/2xADxBCxGO/OD=1/2xADxBCxGOx(R-AD) Rearranging, 1/AD={[ABC]+1/2xBCxGO}/{[ABC]xR} [ABC]=1/2xBExFAx(sin angleAEB)-1/2xBExFCx(sin angleAEB)=1/2xBExACx(sin angleAEB)=1/2xBExACxOH/OE=1/2xBExACxOH/(BE-R) Rearranging, 1/BE={[ABC]-1/2xACxOH}/{[ABC]xR} Similarly, 1/CF={[ABC]-1/2xABxOI}/{[ABC]xR} Adding the three equations, 1/AD+1/BE+1/CF={3[ABC]-[ABC]}/{[ABC]xR}=2/R=2/0.025=80

Let $[PQRS]$ denote the area of the figure $PQRS$ .

We have $\frac {AO}{AD} = \frac {[AOB]}{[ADB]} = \frac {[AOC]}{[ADC]}$ , thus $\frac { [AOB]+[AOC] } {[ABC]} = \frac{AO}{AD} \left(\frac{[ADB]+[ADC]}{[ABC]}\right) = \frac{AO}{AD}$ . Similarly, $\frac {BO}{BE} = \frac {[BOC]+[BOA] }{[ABC]}$ and $\frac {CO}{CF} = \frac { [COA]+[COB] } {[ABC] }$ .

Adding the three equations and substituting $AO=BO=CO=R=0.025$ , we get $\frac {R}{AD} + \frac {R}{BE} + \frac {R}{CF} = 2$ . Thus $\frac {1}{AD} + \frac {1}{BE} + \frac {1}{CF} = \frac {2}{0.025} = 80$ .

Given Circumradius = 0.025. It is clear that circumradius R = two third of AD, So AD = half of (thrice R) = 0.0375. As the triangle is circumscibed by the circle, AD = BE = CF = 0.0375. So 1/AD+1/BE+1/CF = 3(1/0.0375) = 80

You can assume that ABC is a right isosceles triangle. Then A corresponds to F, O corresponds to E, and C corresponds to D. O is the midpoint of AC and is equidistant from the three vertices of the triangle. The answer is thus 1/(0.025 2)+1/(0.025)+1/(0.025 2)=20+40+20=80.

Due to the way the answers are given here, it is clear that the answer is the same for any such triangle ABC, and therefore I choose ABC to be equilateral. All of AD, BE and CF are then equal. Let the side of the triangle be a. By the theorem of Pythagoras, the triangle's height AD is a sqrt(3)/2, and its area = base height/2 = a * a * sqrt(3)/4. The triangle BCO has the same base a and the height OD and the area 1/3 of ABC, since BCO, CAO and ABO are congruent and together make up ABC. OD is therefore 1/3 of AD, and AO is 2/3 of AD. AO = R, so 1/AD = 1/(3 AO/2) = 2/(3 AO) = 2/(3*0.025). The sum we want is three times this = 2/0.025 = 80.

As it is true for every triangle, it is also true for equilateral triangle. So O is circumcentre as well as circumradius. By trigonometry sin 30 = R/AO 2R = AO AD = AO+R = 3R AD=BE=CF=3R (true in equilateral triangle) 1/AD + 1/BE + 1/CF = 1/3R + 1/3R + 1/3R = 1/R = 80

We can check a special case, when ABC is equilateral, and easily find that AD=BE=CF=3/80, since AO=BO=CO=(2/3)AD=1/40. Plugging in values gives the desired result.