On Area percentage -3 (with bonus challenge!)

Not go on the title, in this discussion I will not deal with percentage but instead measure of area

In the below diagram \(O\) is the center of the circle with tangents \(\overline{BD}\) and \(\overline{AE}\)

The area of AOB=12OD2tan(AOB)sec(AOB)\triangle AOB=\frac{1}{2}\overline{OD}^2\tan(\angle AOB)\sec(\angle AOB)

Proof:

Let OD=x;AOB=θ\overline{OD}=x;\angle AOB=\theta

In AEO\triangle AEO tan(θ)=AEx\tan(\theta)=\dfrac{\overline{AE}}{x} AE=xtan(θ)........[1]\overline{AE}=x\tan(\theta)........[1] As AEO=90°\angle AEO=90^\degree, applying Pythagorus theorem x2+AE2=AO2x^2+\overline{AE}^2=\overline{AO}^2 AO2=x2+x2tan2(θ)=x2(1+tan2(θ))=x2sec2(θ)\overline{AO}^2=x^2+x^2\tan^2(\theta)=x^2(1+\tan^2(\theta))=x^2\sec^2(\theta) AO=xsec(θ)........[2]\overline{AO}=x\sec(\theta)........[2]

Now in AEO\triangle AEO and BDO\triangle BDO AEO=90°=BDO\angle AEO=90^\degree=\angle BDO AOE=θ=BOD\angle AOE=\theta=\angle BOD OE=x=OD\overline{OE}=x=\overline{OD} From RHS criterion of congruence AEOBDO\triangle AEO\cong\triangle BDO AO=BO........[3]\therefore \overline{AO}=\overline{BO}........[3]

Area of ABO=12BO×AE\triangle ABO=\frac{1}{2}\overline{BO}\times\overline{AE}

From [3][3] and this

Area of ABO=12AO×AE\triangle ABO=\frac{1}{2}\overline{AO}\times\overline{AE}

From [1],[2][1],[2] and this

Area of ABO=12xsec(θ)×xtan(θ)=12x2tan(θ)sec(θ)\triangle ABO=\frac{1}{2}x\sec(\theta)\times x\tan(\theta)=\frac{1}{2}x^2\tan(\theta)\sec(\theta)


Bonus

In the above diagram if you know the value of R1,R2R_1,R_2 and R3R_3, then find a formula to find the area of the red\red{red} region.

Note:

  • I myself don't know the answer to the bonus problem.

Note by Zakir Husain
3 days, 5 hours ago

No vote yet
1 vote

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Centres of circles are joined,then we can find the area of the formed triangle by heron's formula and then we find each angle by cosine rule and area of each sector.Subtracting area of sectors from area of triangle gives our result.

@Zakir Husain,I will try to find shorter method.

Kriti Kamal - 3 days, 4 hours ago

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Or try writing it out as one formula and try to factorise & simplify :)

Jeff Giff - 3 days, 2 hours ago

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If someone gets the answer with solution of bonus, please share! Thanks!

Mahdi Raza - 2 days, 16 hours ago

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I got it Where,s=r1+r2+r3

@Mahdi Raza,@Zakir Husain,@Jeff Giff

Kriti Kamal - 2 days, 5 hours ago

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Incredibly! long formula

Zakir Husain - 2 days, 5 hours ago

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Ya,the earlier solution is longer than it.I'll try to short the formula

Kriti Kamal - 2 days, 5 hours ago

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Can you describe the parts of this formula? Did you "use Heron's formula and then try to subtract the sectors"... If so I think there is a small mistake in Heron's formula, it's: s(sr1)(sr2)(sr3)\sqrt{s (s-r_{1})(s-r_{2})(s-r_{3})}. Anyways please describe how you got till this formula. Thanks!

Mahdi Raza - 2 days, 5 hours ago

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@Mahdi Raza,check my solution

Kriti Kamal - 2 days, 3 hours ago

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Wait…but the triangle’s sides aren’t r1,r2,r3r_1,r_2,r_3, and it is equilateral (tangent of circle from same point)

Jeff Giff - 2 days, 4 hours ago

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It might be the smallest among the radii of the circles, but we need more proof. :/

Jeff Giff - 2 days, 4 hours ago

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@Jeff Giff, @Mahdi Raza, @Kriti Kamal - the red area is not a triangle so don't confuse, it is a shape made of three arcs not three sides.

Zakir Husain - 2 days, 4 hours ago

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Yes, but I meant the smallest triangle containing the red area :)

Jeff Giff - 2 days, 3 hours ago

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I have considered them as arcs not as linesegmets :-)

Kriti Kamal - 2 days, 3 hours ago

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Oic. My solution is different. I thought about the smallest triangle containing the red area :)

Jeff Giff - 2 days, 3 hours ago

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So my solution is sizeofsizeof\trianglecontaining red area sizeofARCH1,2,3-sizeofARCH1,2,3

Jeff Giff - 2 days, 3 hours ago

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Maybe I am wrong :D Sorry for being unhelpful :|

Jeff Giff - 2 days, 3 hours ago

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See my solution..

Kriti Kamal - 2 days, 3 hours ago

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Here,is my solution.Please comment if you find any mistake.Sorry for delay,it happens because tommorow my hands were burn by fire.

@Zakir Husain,@Mahdi Raza,@Jeff Giff,

Kriti Kamal - 2 days, 3 hours ago

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A good explanation!. Any other formula? @Justin Travers, @jordi curto, @Yajat Shamji, @Kumudesh Ghosh, @Alak Bhattacharya, @Vinayak Srivastava, @Aryan Sanghi, @Marvin Kalngan.

Zakir Husain - 2 days, 1 hour ago

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The angles can be calculated in sin inverse and cos inverse function

Kriti Kamal - 2 days, 1 hour ago

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@Zakir Husain,can i send another version of solution

Kriti Kamal - 2 days, 1 hour ago

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@Kriti Kamal Sure!

Zakir Husain - 2 days, 1 hour ago

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The area of the triangle can be found using Zakir’s formula above:

12r12tanAsecA\frac12r_1^2\tan\angle A\sec\angle A

Or any xx and θ\theta pair with the relation above.

Jeff Giff - 2 days, 1 hour ago

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@Jeff Giff You are assuming that the perpendicular of AB\overline{AB} at point DD will touch the point CC, the same assumption you have to make to the perpendicular of AC\overline{AC} and point BB. Then only you can apply my formula to get the area of ABC\triangle ABC. You first have to prove that these assumptions are true for all cases. Or you can build a new formula for this special case separately.

Zakir Husain - 2 days, 1 hour ago

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Aha! @Zakir Husain, your formula can be simplified as 12x2sinθ\frac12x^2\sin \theta, since tanθsecθ=tanθ1cosθ=sinθcosθcosθ=sinθ\tan \theta \sec \theta=\tan \theta \frac{1}{\cos \theta}=\frac{\frac{\sin \theta}{\cos \theta}}{\cos \theta}=\sin \theta :)

Jeff Giff - 2 days ago

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@Jeff Giff tanθsecθ=sinθcosθ×1cosθ=sinθcos2θ\tan\theta\sec\theta=\dfrac{\sin\theta}{\cos\theta}\times\dfrac{1}{\cos\theta}=\red{\dfrac{\sin\theta}{\cos^2\theta}}

Zakir Husain - 2 days ago

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@Zakir Husain Oops!

Jeff Giff - 2 days ago

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Slight improvement: express angles in radians. Then, the formula becomes sr1r2r3cycr12tan1r2r3sr1\sqrt{sr_1r_2r_3}-\sum _{cyc}r_1^2\tan^{-1}\sqrt{\frac{r_2r_3}{sr_1}}

Jeff Giff - 2 days, 1 hour ago

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I know it.but generally i use my most of work in degrees

Kriti Kamal - 2 days, 1 hour ago

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