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Question Number 25602 by behi.8.3.4.17@gmail.com last updated on 11/Dec/17

Commented by behi.8.3.4.17@gmail.com last updated on 11/Dec/17

A circle passes through mid points of sides of triangle AB^△ C . find: 1)its radius in term of:a,b,c. 2)distance of its center from:A,B,C. 3)distance from its centre to centerpoint of out circle of: AB^△ C.

$$\boldsymbol{{A}}\:\boldsymbol{{circle}}\:\boldsymbol{{passes}}\:\boldsymbol{{through}}\: \\ $$$$\boldsymbol{{mid}}\:\boldsymbol{{points}}\:\boldsymbol{{of}}\:\boldsymbol{{sides}}\:\boldsymbol{{of}} \\ $$$$\boldsymbol{{triangle}}\:\boldsymbol{{A}}\overset{\bigtriangleup} {\boldsymbol{{B}C}}\:. \\ $$$$\mathrm{find}: \\ $$$$\left.\mathrm{1}\right)\mathrm{its}\:\mathrm{radius}\:\mathrm{in}\:\mathrm{term}\:\mathrm{of}:\boldsymbol{{a}},\boldsymbol{{b}},\boldsymbol{{c}}. \\ $$$$\left.\mathrm{2}\right)\mathrm{distance}\:\mathrm{of}\:\mathrm{its}\:\mathrm{center}\:\mathrm{from}:\mathrm{A},\mathrm{B},\mathrm{C}. \\ $$$$\left.\mathrm{3}\right)\mathrm{distance}\:\mathrm{from}\:\mathrm{its}\:\mathrm{centre}\:\mathrm{to} \\ $$$$\mathrm{centerpoint}\:\mathrm{of}\:\mathrm{out}\:\mathrm{circle}\:\mathrm{of}:\:\mathrm{A}\overset{\bigtriangleup} {\mathrm{B}C}. \\ $$

Answered by mrW1 last updated on 12/Dec/17

DE=(a/2)=a′ DF=(b/2)=b′ EF=(c/2)=c′ r=((a′b′c′)/(4A_(DEF) ))=((a′b′c′)/(4(√(s′(s′−a′)(s′−b′)(s′−c′))))) s′=((a′+b′+c′)/2)=((a+b+c)/4)=(s/2) ⇒r=((abc)/(8(√(s(s−a)(s−b)(s−c)))))

$${DE}=\frac{{a}}{\mathrm{2}}={a}' \\ $$$${DF}=\frac{{b}}{\mathrm{2}}={b}' \\ $$$${EF}=\frac{{c}}{\mathrm{2}}={c}' \\ $$$${r}=\frac{{a}'{b}'{c}'}{\mathrm{4}{A}_{{DEF}} }=\frac{{a}'{b}'{c}'}{\mathrm{4}\sqrt{{s}'\left({s}'−{a}'\right)\left({s}'−{b}'\right)\left({s}'−{c}'\right)}} \\ $$$${s}'=\frac{{a}'+{b}'+{c}'}{\mathrm{2}}=\frac{{a}+{b}+{c}}{\mathrm{4}}=\frac{{s}}{\mathrm{2}} \\ $$$$\Rightarrow{r}=\frac{{abc}}{\mathrm{8}\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}} \\ $$

Commented by behi.8.3.4.17@gmail.com last updated on 12/Dec/17

thank you so much dear MrW1. Do you have any solutions for#2,#3? please!

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{dear}\:\mathrm{MrW1}. \\ $$$$\mathrm{Do}\:\mathrm{you}\:\mathrm{have}\:\mathrm{any}\:\mathrm{solutions}\:\mathrm{for}#\mathrm{2},#\mathrm{3}? \\ $$$$\mathrm{please}! \\ $$

Commented by mrW1 last updated on 12/Dec/17

to Q3: the circle above is the so called nine− point circle. it is half so large as the circumcircle: r=(R/2) with R=radius of circumcircle and R=((abc)/(4(√(s(s−a)(s−b)(s−c))))) The distance between the nine−point circle center and the circumcenter is ΔL=(1/2)(√(9R^2 −a^2 −b^2 −c^2 ))

$${to}\:{Q}\mathrm{3}: \\ $$$${the}\:{circle}\:{above}\:{is}\:{the}\:{so}\:{called}\:{nine}− \\ $$$${point}\:{circle}.\:{it}\:{is}\:{half}\:{so}\:{large}\:{as}\:{the} \\ $$$${circumcircle}:\:{r}=\frac{{R}}{\mathrm{2}}\:{with} \\ $$$${R}={radius}\:{of}\:{circumcircle}\:{and} \\ $$$${R}=\frac{{abc}}{\mathrm{4}\sqrt{{s}\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}} \\ $$$${The}\:{distance}\:{between}\:{the}\:{nine}−{point} \\ $$$${circle}\:{center}\:{and}\:{the}\:{circumcenter} \\ $$$${is}\: \\ $$$$\Delta{L}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{9}{R}^{\mathrm{2}} −{a}^{\mathrm{2}} −{b}^{\mathrm{2}} −{c}^{\mathrm{2}} } \\ $$

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