Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

Question Number 128303 by mr W last updated on 06/Jan/21

Commented by mr W last updated on 06/Jan/21

$$findsidelengthsoftheequilateral$$$$triangleABCintermsofradiiR$$$$andr.$$

Commented by mr W last updated on 06/Jan/21

$$resultis$$$$s=\sqrt{3}(R+r)+\sqrt{3{(R+r)}^2-8Rr}$$

Answered by mr W last updated on 06/Jan/21

Commented by mr W last updated on 07/Jan/21

$${AE}^2={(s-\sqrt{3}R)}^2+R^2=s^2-2\sqrt{3}Rs+4R^2$$$${AF}^2={(s-\sqrt{3}r)}^2+r^2=s^2-2\sqrt{3}rs+4r^2$$$${EF}^2={(s-\sqrt{3}R-\sqrt{3}r)}^2+{(R-r)}^2=s^2-2\sqrt{3}(R+r)+4(R^2+r^2+Rr)$$$${EF}^2={AE}^2+{AF}^2-2\times AE\times AF\times \cos 30°$$$${(s-\sqrt{3}R-\sqrt{3}r)}^2+{(R-r)}^2={(s-\sqrt{3}R)}^2+R^2+{(s-\sqrt{3}r)}^2+r^2-\sqrt{3[{(s-\sqrt{3}R)}^2+R^2][{(s-\sqrt{3}r)}^2+r^2]}$$$$s^2-2\sqrt{3}(R+r)+4(R^2+r^2+Rr)=s^2-2\sqrt{3}Rs+4R^2+s^2-2\sqrt{3}rs+4r^2-\sqrt{3(s^2-2\sqrt{3}Rs+4R^2)(s^2-2\sqrt{3}rs+4r^2)}$$$$s^2-4Rr=\sqrt{3(s^2-2\sqrt{3}Rs+4R^2)(s^2-2\sqrt{3}rs+4r^2)}$$$${(s^2-4Rr)}^2=3(s^2-2\sqrt{3}Rs+4R^2)(s^2-2\sqrt{3}rs+4r^2)$$$$s^4-3\sqrt{3}(R+r)s^3+2(3R^2+3r^2+11Rr)s^2-12\sqrt{3}Rr(R+r)s+16R^2{r}^2=0$$$$9{\left(\frac{s}{\sqrt{3}}\right)}^4-27(R+r){\left(\frac{s}{\sqrt{3}}\right)}^3+6(3R^2+3r^2+11Rr){\left(\frac{s}{\sqrt{3}}\right)}^2-36Rr(R+r)\left(\frac{s}{\sqrt{3}}\right)+16R^2{r}^2=0$$$$lett=\frac{s}{\sqrt{3}}$$$$\Rightarrow 9t^4-27(R+r)t^3+6(3R^2+3r^2+11Rr)t^2-36Rr(R+r)t+16R^2{r}^2=0$$$$\Rightarrow [3t^2-6(R+r)t+8Rr][3t^2-3(R+r)t+2Rr]=0$$$$\Rightarrow t=R+r+\frac{1}{\sqrt{3}}\sqrt{3{(R+r)}^2-8Rr}$$$$\Rightarrow s=\sqrt{3}(R+r)+\sqrt{3{(R+r)}^2-8Rr}$$$$example:$$$$R=3,r=2$$$$s=\sqrt{3}(3+2)+\sqrt{3\times {(3+2)}^2-8\times 3\times 2}$$$$=5\sqrt{3}+3\sqrt{3}$$$$=8\sqrt{3}✓$$

Answered by MJS_new last updated on 06/Jan/21

$$\mathrm{setting}s=1$$$$\measuredangle BAD=\alpha ;0<\alpha <60°;k=\tan \alpha ;0<k<\sqrt{3}$$$$R=\frac{\sqrt{3}k+1-\sqrt{k^2+1}}{2(k+\sqrt{3})}$$$$r=\frac{2-\sqrt{k^2+1}}{2(k+\sqrt{3})}$$$$\frac{R}{r}=v$$$$\mathrm{I}\mathrm{get}$$$$v=\frac{k^2-2\sqrt{3}k-1-(\sqrt{3}k-1)\sqrt{k^2+1}}{k^2-3}$$$$\iff$$$$k=\frac{\sqrt{3}(1-2v)+(v-1)\sqrt{3v^2-2v+3}}{v^2-2v-2}$$$$\mathrm{from}\mathrm{here}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{easy}\mathrm{to}\mathrm{calculate}\mathrm{backwards}$$$$\mathrm{from}\mathrm{given}R,r$$

Commented by mr W last updated on 06/Jan/21

$$answeriscorrectsir!eveni{don}^{\prime }t$$$$knowhowyougotk=...vfromv=...k.$$

Commented by MJS_new last updated on 06/Jan/21

$$\mathrm{solving}v=\frac{k^2-2\sqrt{3}k-1-(\sqrt{3}k-1)\sqrt{k^2+1}}{k^2-3}\mathrm{for}$$$$k\mathrm{is}\mathrm{easy}\mathrm{because}\mathrm{the}{4}^{\mathrm{th}}\mathrm{degree}\mathrm{we}\mathrm{get}\mathrm{after}$$$$\mathrm{transforming}/\mathrm{squaring}/\mathrm{transforming}\mathrm{has}$$$$\mathrm{the}2\left(\mathrm{false}\right)\mathrm{solutions}k=\pm \sqrt{3}\mathrm{which}\mathrm{I}\mathrm{found}$$$$\mathrm{after}\mathrm{a}\mathrm{lot}\mathrm{of}\mathrm{trying}...$$

Commented by mr W last updated on 06/Jan/21

$$nowisee.igaveupbeforeihadreally$$$$tried.$$

Answered by ajfour last updated on 07/Jan/21

Commented by ajfour last updated on 07/Jan/21

$$\tan \theta =m$$$$x=\frac{R}{m},y=mr$$$$s-z=R\sqrt{3}\Rightarrow z=s-R\sqrt{3}..\left(i\right)$$$$s+y-(x+z)=r\sqrt{3}$$$$BD+DO+OF+FC=s$$$$R\sqrt{3}+\frac{R}{m}+mr+r\sqrt{3}=s...\left(ii\right)$$$$\frac{PH}{AH}=\frac{R}{z}=\tan (60°-\theta )...\left(iii\right)$$$$\frac{r}{z+x-y}=\tan (\theta -30°)....\left(iv\right)$$$$equatingz+x-yfrom\left(iii\right),\left(iv\right):$$$$\frac{R}{\tan (60-\theta )}+\frac{R}{m}-mr=\frac{r}{\tan (\theta -30)}$$$$\frac{R(1+m\sqrt{3})}{\sqrt{3}-m}+\frac{R}{m}-mr=\frac{r(m+\sqrt{3})}{m\sqrt{3}-1}$$$$\Rightarrow$$$$\frac{R[m+m^2\sqrt{3}+\sqrt{3}-m]}{m(\sqrt{3}-m)}=\frac{r[m^2\sqrt{3}+\sqrt{3}]}{m\sqrt{3}-1}$$$$\Rightarrow \frac{R}{m(\sqrt{3}-m)}=\frac{r}{m\sqrt{3}-1}$$$$\Rightarrow mR\sqrt{3}-R=mr\sqrt{3}-m^{2}r$$$${rm}^2+m(R-r)\sqrt{3}-R=0$$$$m=-\frac{(R-r)\sqrt{3}}{2r}+\sqrt{\frac{3{(R-r)}^2}{4r^2}+\frac{R}{r}}$$$$\frac{R}{m^2}-\frac{(R-r)\sqrt{3}}{m}-r=0$$$$\frac{1}{m}=\frac{(R-r)\sqrt{3}}{2R}+\sqrt{\frac{3{(R-r)}^2}{4R^2}+\frac{r}{R}}$$$$substitutingin\left(ii\right)$$$$s=\sqrt{3}(R+r)+\frac{R}{m}+mr$$$$s=\sqrt{3}(R+r)+\frac{(R-r)\sqrt{3}}{2}$$$$+\sqrt{\frac{3{(R-r)}^2}{4}+rR}$$$$-\frac{(R-r)\sqrt{3}}{2}+\sqrt{\frac{3{(R-r)}^2}{4}+rR}$$$$\Rightarrow$$$$s=\sqrt{3}(R+r)+\sqrt{3{(R-r)}^2+4rR}$$

Commented by mr W last updated on 07/Jan/21

$$perfectlysolved!thanksalotsir!$$

Commented by ajfour last updated on 07/Jan/21

$$gotitcorrectnow,Sir.$$

Commented by Tawa11 last updated on 06/Nov/21

$$\mathrm{Great}\mathrm{sir}.$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com