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Twin-of-Q-3943-Three-circles-are-drawn-in-a-plane-in-such-a-way-that-a-closed-region-is-produced-which-is-not-included-in-any-of-the-circles-Determine-the-area-of-this-region-Circles-Radii-Cente

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Question Number 4036 by Rasheed Soomro last updated on 27/Dec/15
Twin of Q#3943 Three circles are drawn in a plane in such a way that a closed region is produced which is not included in any of the circles. Determine the area of this region. Circles ∣ Radii ∣ Centers _(−) A r_1 C_1 B r_2 C_2 C r_3 C_3 Conditions: r_1 +r_2 ≥C_1 C_2 r_2 +r_3 ≥C_2 C_3 r_3 +r_1 ≥C_3 C_1
$$\mathrm{Twin}\:\mathrm{of}\:\mathrm{Q}#\mathrm{3943} \\ $$$$\mathrm{Three}\:\mathrm{circles}\:\mathrm{are}\:\mathrm{drawn}\:\mathrm{in}\:\mathrm{a}\:\mathrm{plane} \\ $$$$\mathrm{in}\:\mathrm{such}\:\mathrm{a}\:\mathrm{way}\:\mathrm{that}\:\mathrm{a}\:\mathrm{closed}\:\mathrm{region} \\ $$$$\mathrm{is}\:\mathrm{produced}\:\mathrm{which}\:\mathrm{is}\:\mathrm{not}\:\mathrm{included} \\ $$$$\mathrm{in}\:\mathrm{any}\:\mathrm{of}\:\mathrm{the}\:\mathrm{circles}.\:\mathrm{Determine}\:\mathrm{the} \\ $$$$\mathrm{area}\:\mathrm{of}\:\mathrm{this}\:\mathrm{region}. \\ $$$$\underset{−} {\mathrm{Circles}\:\mid\:\mathrm{Radii}\:\mid\:\mathrm{Centers}\:} \\ $$$$\:\:\:\:\:\:\mathrm{A}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{r}}_{\mathrm{1}} \:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{C}_{\mathrm{1}} \:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\mathrm{B}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{r}}_{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{C}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\mathrm{C}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{r}}_{\mathrm{3}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{C}_{\mathrm{3}} \\ $$$$\mathrm{Conditions}: \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{r}}_{\mathrm{1}} +\boldsymbol{\mathrm{r}}_{\mathrm{2}} \geqslant\mathrm{C}_{\mathrm{1}} \mathrm{C}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{r}}_{\mathrm{2}} +\boldsymbol{\mathrm{r}}_{\mathrm{3}} \geqslant\mathrm{C}_{\mathrm{2}} \mathrm{C}_{\mathrm{3}} \\ $$$$\:\:\:\:\:\boldsymbol{\mathrm{r}}_{\mathrm{3}} +\boldsymbol{\mathrm{r}}_{\mathrm{1}} \geqslant\mathrm{C}_{\mathrm{3}} \mathrm{C}_{\mathrm{1}} \\ $$
Commented by Yozzii last updated on 27/Dec/15
Commented by Yozzii last updated on 27/Dec/15
Area of region GHI is required.
$${Area}\:{of}\:{region}\:{GHI}\:{is}\:{required}. \\ $$
Commented by Rasheed Soomro last updated on 27/Dec/15
Yes. Exactly! In other cases two or all the three circles are tangent to one another.
$$\mathrm{Yes}.\:\mathrm{Exactly}! \\ $$$$\mathrm{In}\:\mathrm{other}\:\mathrm{cases}\:\mathrm{two}\:\mathrm{or}\:\:\mathrm{all}\:\mathrm{the}\:\mathrm{three} \\ $$$$\mathrm{circles}\:\mathrm{are}\:\mathrm{tangent}\:\mathrm{to}\:\mathrm{one}\:\mathrm{another}. \\ $$
Commented by Yozzii last updated on 27/Dec/15
Commented by Yozzii last updated on 27/Dec/15
Disregard the explicit measurements of angles given in the diagram. I only required the image to delineate the important angles for the calculation.
$${Disregard}\:{the}\:{explicit}\:{measurements} \\ $$$${of}\:{angles}\:{given}\:{in}\:{the}\:{diagram}.\: \\ $$$${I}\:{only}\:{required}\:{the}\:{image}\:{to}\:{delineate} \\ $$$${the}\:{important}\:{angles}\:{for}\:{the}\:{calculation}. \\ $$
Commented by Yozzii last updated on 27/Dec/15
Let A_1 be the area of △AEC. By the question we know the lengths AE,EC & CA. So, Heron′s formula could be used to find A_1 . A_1 =(√(s(s−CA)(s−AE)(s−EC))) s=((AE+EC+CA)/2). Let θ_1 =∠ACE,θ_2 =∠CEA and θ_3 =∠EAC. By cosine rule AE^2 =EC^2 +CA^2 −2(EC)(CA)cosθ_1 ⇒θ_1 =cos^(−1) {((EC^2 +CA^2 −AE^2 )/(2(EC)(CA)))}. In like fashion, θ_2 can be found and θ_3 =π−θ_1 −θ_2 . Let r_1 ,r_2 ,r_3 be the radii of the circles with centres C,E and A respectively. The area A_2 of the sector MCL (from diagram) is given by A_2 =(1/2)r_1 ^2 θ_1 . Similarly, the area A_3 of the sector OEN is A_3 =(1/2)r_2 ^2 θ_2 and the area A_4 of the sector KAJ is A_4 =(1/2)r_3 ^2 θ_3 . Now, the value of s=Σ_(i=2) ^4 A_i includes duplication of pairwise overlapping of circles. To find the correct value of s to subtract from A_1 to give the required area, work s by parts. Starting with A_2 , A_2 +A_3 considers the overlap of circles C and E two times. ⇒suitable area, s_1 =A_2 +A_3 −Area(LTN). Next, A_2 +A_3 +A_4 considers the overlap between circles A and E, and also A and C,twice. So, truly s=Σ_(r=2) ^4 A_r −Area(LTN)−Area(OKR)−Area(MBJ). Mr. Jain′s answer to question 3877 can help you find Areas LTN,OKR andMBJ, once you half each result for area found using his method directly. Afterwards, Area(RBT)=A_1 −s.
$${Let}\:{A}_{\mathrm{1}} \:{be}\:{the}\:{area}\:{of}\:\bigtriangleup{AEC}.\:{By}\:{the} \\ $$$${question}\:{we}\:{know}\:{the}\:{lengths}\:{AE},{EC} \\ $$$$\&\:{CA}.\:{So},\:{Heron}'{s}\:{formula}\:{could} \\ $$$${be}\:{used}\:{to}\:{find}\:{A}_{\mathrm{1}} .\: \\ $$$${A}_{\mathrm{1}} =\sqrt{{s}\left({s}−{CA}\right)\left({s}−{AE}\right)\left({s}−{EC}\right)} \\ $$$${s}=\frac{{AE}+{EC}+{CA}}{\mathrm{2}}. \\ $$$${Let}\:\theta_{\mathrm{1}} =\angle{ACE},\theta_{\mathrm{2}} =\angle{CEA}\:{and}\:\theta_{\mathrm{3}} =\angle{EAC}. \\ $$$${By}\:{cosine}\:{rule}\: \\ $$$${AE}^{\mathrm{2}} ={EC}^{\mathrm{2}} +{CA}^{\mathrm{2}} −\mathrm{2}\left({EC}\right)\left({CA}\right){cos}\theta_{\mathrm{1}} \\ $$$$\Rightarrow\theta_{\mathrm{1}} ={cos}^{−\mathrm{1}} \left\{\frac{{EC}^{\mathrm{2}} +{CA}^{\mathrm{2}} −{AE}^{\mathrm{2}} }{\mathrm{2}\left({EC}\right)\left({CA}\right)}\right\}. \\ $$$${In}\:{like}\:{fashion},\:\theta_{\mathrm{2}} \:{can}\:{be}\:{found} \\ $$$${and}\:\theta_{\mathrm{3}} =\pi−\theta_{\mathrm{1}} −\theta_{\mathrm{2}} . \\ $$$${Let}\:{r}_{\mathrm{1}} ,{r}_{\mathrm{2}} ,{r}_{\mathrm{3}} \:{be}\:{the}\:{radii}\:{of}\:{the}\:{circles} \\ $$$${with}\:{centres}\:{C},{E}\:{and}\:{A}\:{respectively}. \\ $$$${The}\:{area}\:{A}_{\mathrm{2}} \:{of}\:{the}\:{sector}\:{MCL}\:\left({from}\:{diagram}\right) \\ $$$${is}\:{given}\:{by}\:{A}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}{r}_{\mathrm{1}} ^{\mathrm{2}} \theta_{\mathrm{1}} .\:{Similarly}, \\ $$$${the}\:{area}\:{A}_{\mathrm{3}} \:{of}\:{the}\:{sector}\:{OEN}\:{is}\:{A}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{2}}{r}_{\mathrm{2}} ^{\mathrm{2}} \theta_{\mathrm{2}} \\ $$$${and}\:{the}\:{area}\:{A}_{\mathrm{4}} \:{of}\:{the}\:{sector}\:{KAJ}\:{is} \\ $$$${A}_{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{2}}{r}_{\mathrm{3}} ^{\mathrm{2}} \theta_{\mathrm{3}} .\:{Now},\:{the}\:{value}\:{of}\: \\ $$$${s}=\underset{{i}=\mathrm{2}} {\overset{\mathrm{4}} {\sum}}{A}_{{i}} \:{includes}\:{duplication}\:{of}\: \\ $$$${pairwise}\:{overlapping}\:{of}\:{circles}.\:{To} \\ $$$${find}\:{the}\:{correct}\:{value}\:{of}\:{s}\:{to}\:{subtract} \\ $$$${from}\:{A}_{\mathrm{1}} \:{to}\:{give}\:{the}\:{required}\:{area},\:{work} \\ $$$${s}\:{by}\:{parts}.\:{Starting}\:{with}\:{A}_{\mathrm{2}} ,\:{A}_{\mathrm{2}} +{A}_{\mathrm{3}} \\ $$$${considers}\:{the}\:{overlap}\:{of}\:{circles}\:{C}\:{and}\:{E} \\ $$$${two}\:{times}.\:\Rightarrow{suitable}\:{area}, \\ $$$${s}_{\mathrm{1}} ={A}_{\mathrm{2}} +{A}_{\mathrm{3}} −{Area}\left({LTN}\right). \\ $$$${Next},\:{A}_{\mathrm{2}} +{A}_{\mathrm{3}} +{A}_{\mathrm{4}} \:{considers}\:{the}\: \\ $$$${overlap}\:{between}\:{circles}\:{A}\:{and}\:{E},\:{and}\:{also}\:{A}\:{and}\:{C},{twice}. \\ $$$${So},\:{truly}\:{s}=\underset{{r}=\mathrm{2}} {\overset{\mathrm{4}} {\sum}}{A}_{{r}} −{Area}\left({LTN}\right)−{Area}\left({OKR}\right)−{Area}\left({MBJ}\right). \\ $$$${Mr}.\:{Jain}'{s}\:{answer}\:{to}\:{question}\:\mathrm{3877}\:{can}\:{help} \\ $$$${you}\:{find}\:{Areas}\:{LTN},{OKR}\:{andMBJ}, \\ $$$${once}\:{you}\:{half}\:{each}\:{result}\:{for}\:{area}\: \\ $$$${found}\:{using}\:{his}\:{method}\:{directly}. \\ $$$$ \\ $$$${Afterwards},\: \\ $$$${Area}\left({RBT}\right)={A}_{\mathrm{1}} −{s}. \\ $$
Commented by Rasheed Soomro last updated on 27/Dec/15
GREAT ! YOU HAVE DONE SUCCESSFULLY!
$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathbb{G}\mathcal{REA}\mathbb{T}\:! \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathbb{Y}\mathcal{OU}\:\:\:\mathbb{H}\mathcal{AVE}\:\mathbb{D}\mathcal{ONE}\:\:\mathbb{S}\mathcal{UCCESSFULLY}! \\ $$

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