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Let-l-1-l-2-l-3-be-a-set-of-three-nonparallel-lines-that-all-meet-at-one-point-R-in-the-plane-Investigate-conditions-on-such-that-R-is-the-centroid-of-some-triangle-

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Question Number 6564 by Yozzii last updated on 03/Jul/16
Let φ={ l_1 , l_2 , l_3 } be a set of three, nonparallel lines that all meet at one point R in the plane. Investigate conditions on φ such that R is the centroid of some triangle.
$${Let}\:\phi=\left\{\:{l}_{\mathrm{1}} \:,\:{l}_{\mathrm{2}} \:,\:{l}_{\mathrm{3}} \:\right\}\:{be}\:{a}\:{set}\:{of}\:{three},\:{nonparallel} \\ $$$${lines}\:{that}\:{all}\:{meet}\:{at}\:{one}\:{point}\:{R}\:{in}\: \\ $$$${the}\:{plane}.\:{Investigate}\:{conditions}\:{on}\:\phi \\ $$$${such}\:{that}\:{R}\:{is}\:{the}\:{centroid}\:{of}\:{some}\:{triangle}. \\ $$$$ \\ $$$$ \\ $$
Commented by Yozzii last updated on 03/Jul/16
Equivalently, prove/disprove that there exists a triangle whose centroid is that of the point of intersection of any three concurrent nonparallel lines in the plane.
$${Equivalently},\:{prove}/{disprove}\:{that} \\ $$$${there}\:{exists}\:{a}\:{triangle}\:{whose}\:{centroid} \\ $$$${is}\:{that}\:{of}\:{the}\:{point}\:{of}\:{intersection} \\ $$$${of}\:{any}\:{three}\:{concurrent}\:{nonparallel}\:{lines} \\ $$$${in}\:{the}\:{plane}. \\ $$
Commented by Rasheed Soomro last updated on 03/Jul/16
Not related to above Sharing: See intmath.com if you haven′t seen already. Subscribe for intmath newsletter. Perhaps there is something of interst for you.
$${Not}\:{related}\:{to}\:{above} \\ $$$${Sharing}:\:{See}\:\mathrm{intmath}.\mathrm{com}\:{if}\:{you}\:{haven}'{t}\: \\ $$$${seen}\:{already}.\:{Subscribe}\:{for}\:{intmath}\:{newsletter}. \\ $$$${Perhaps}\:{there}\:{is}\:\:{something}\:{of}\:{interst}\:{for}\:{you}. \\ $$
Commented by Yozzii last updated on 03/Jul/16
Interesting website. Thanks!
$${Interesting}\:{website}.\:{Thanks}! \\ $$
Commented by Rasheed Soomro last updated on 05/Jul/16
Commented by Rasheed Soomro last updated on 05/Jul/16
A median is divided at the centroid in ratio 2:1 from the vertex of the triangle. p,q and r are three given concurrent lines cutting at the point O. take A and D on any of three lines(say p here) in such a way that ∣AO∣ : ∣OD∣ = 2 : 1 Now if A is a vertex of the required triangle and D is midpoint of its (A′s) opposite side, then AD is a median of the triangle. We have to prove now that there exist points B and C on lines r and q, such that D is midpoint of BC i-e ∣BD∣=∣DC∣ and B, D,C are collinear. continue
$${A}\:{median}\:{is}\:{divided}\:{at}\:{the}\:{centroid}\:{in} \\ $$$${ratio}\:\mathrm{2}:\mathrm{1}\:{from}\:{the}\:{vertex}\:{of}\:{the}\:{triangle}. \\ $$$$ \\ $$$$\:{p},{q}\:{and}\:{r}\:{are}\:{three}\:{given}\:{concurrent}\:{lines} \\ $$$${cutting}\:{at}\:{the}\:{point}\:\mathrm{O}. \\ $$$${take}\:\mathrm{A}\:{and}\:\:\mathrm{D}\:{on}\:{any}\:{of}\:{three}\:{lines}\left({say}\:{p}\:{here}\right) \\ $$$${in}\:{such}\:{a}\:{way}\:{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\mathrm{AO}\mid\::\:\mid\mathrm{OD}\mid\:=\:\mathrm{2}\::\:\mathrm{1} \\ $$$${Now}\:\:{if}\:\:\mathrm{A}\:{is}\:{a}\:{vertex}\:{of}\:{the}\:{required}\:{triangle}\:{and}\:\:\mathrm{D}\:{is} \\ $$$${midpoint}\:{of}\:\:{its}\:\left(\mathrm{A}'{s}\right)\:{opposite}\:{side},\:{then}\:\mathrm{AD}\:{is} \\ $$$${a}\:{median}\:{of}\:{the}\:{triangle}. \\ $$$${We}\:{have}\:{to}\:{prove}\:{now}\:{that}\:{there}\:{exist}\:{points}\:\mathrm{B}\:{and}\:\mathrm{C} \\ $$$${on}\:{lines}\:{r}\:{and}\:{q},\:{such}\:{that}\:\mathrm{D}\:{is}\:{midpoint}\:{of}\:\mathrm{BC} \\ $$$${i}-{e}\:\mid\mathrm{BD}\mid=\mid\mathrm{DC}\mid\:\:{and}\:\:\mathrm{B},\:\mathrm{D},\mathrm{C}\:{are}\:{collinear}. \\ $$$${continue} \\ $$
Answered by Rasheed Soomro last updated on 06/Jul/16
Commented by Rasheed Soomro last updated on 07/Jul/16
Analytical Approach Let one of three given concurrent lines(named p) is taken as y-axis and the point of their concurrency is taken as origin of coordiate system. Since p is along y-axis its equation is p: x=0 Also q and r pass through O (0,0) their equatons will be of tbe form y=mx Let q: y=m_q x and r: y= m_r x [m_q and m_r are slopes] Let a triangle ABC is such that (i) A and D are on line p (y-axis) (ii) ∣AO∣ : ∣OD∣ = 2 : 1 (iii) D is midpoint of BC ∴ AD is a median of the triangle and O is its centroid. Assume unit of coordinate system equal to ∣OD∣ then in the diagram above D=(0,−1) and A=(0,2) Let B=(b_1 ,b_2 ) and C=(c_1 ,c_2 ) ∵ D(0,−1) is midpoint of BC ∴ ((b_1 +c_1 )/2)=0 ∧ ((b_2 +c_2 )/2)=−1 Or c_1 =−b_1 ∧ c_2 =−2−b_2 Or C=(c_1 , c_2 )=(−b_1 , −2−b_2 ) If B and C are on given lines r and q respectively Then coordinates of B and C will satisfy the equations of r and q respectively r: y= m_r x ⇒ b_2 =m_r b_1 .........................(i) q: y=m_q x ⇒ −2−b_2 =−m_q b_1 ...............(ii) From (i) and (ii) −2=(m_r −m_q )b_1 ⇒ b_1 =((−2)/(m_r −m_q )) b_2 =m_r (((−2)/(m_r −m_q )))=((−2m_r )/(m_r −m_q )) B=(b_1 ,b_2 )=(((−2)/(m_r −m_q )) , ((−2m_r )/(m_r −m_q )))...................(iii) C=(−b_1 , −2−b_2 )=((2/(m_r −m_q )) , ((2m_r )/(m_r −m_q ))−2) C=((2/(m_r −m_q )),((2m_q )/(m_r −m_q ))).................................(iv) So from (i) and (iv) if m_r ≠m_q , B and C are on r and q respectively. So finally If all the three concurrent lines have different slopes there exist a triangle whose centroid is the point of concurrency of given lines
$$\mathcal{A}{nalytical}\:\mathcal{A}{pproach} \\ $$$${Let}\:{one}\:{of}\:{three}\:{given}\:{concurrent}\:{lines}\left({named}\:{p}\right) \\ $$$$\:\:\:{is}\:{taken}\:{as}\:{y}-{axis}\:{and}\:{the}\:{point}\:{of}\:{their}\:{concurrency} \\ $$$$\:\:{is}\:{taken}\:{as}\:{origin}\:{of}\:{coordiate}\:{system}. \\ $$$${Since}\:{p}\:{is}\:{along}\:{y}-{axis}\:{its}\:{equation}\:{is} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{p}:\:\:{x}=\mathrm{0} \\ $$$${Also}\:{q}\:{and}\:{r}\:{pass}\:{through}\:\mathrm{O}\:\left(\mathrm{0},\mathrm{0}\right)\:{their}\:{equatons} \\ $$$${will}\:{be}\:{of}\:{tbe}\:{form}\:\:{y}={mx} \\ $$$${Let}\:\:\:\:\:\:\:\:\:\:\:\:\:\:{q}:\:\:{y}={m}_{{q}} {x}\:\:\:\:{and}\:\:\:\:\:{r}:\:\:\:{y}=\:{m}_{{r}} {x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{m}_{{q}} \:\:{and}\:\:{m}_{{r}} \:\:{are}\:{slopes}\right] \\ $$$$\:\:\: \\ $$$${Let}\:{a}\:{triangle}\:\mathrm{ABC}\:{is}\:{such}\:{that} \\ $$$$\left({i}\right)\:\mathrm{A}\:{and}\:\mathrm{D}\:{are}\:{on}\:{line}\:{p}\:\left({y}-{axis}\right) \\ $$$$\left({ii}\right)\:\mid\mathrm{AO}\mid\::\:\mid\mathrm{OD}\mid\:=\:\mathrm{2}\::\:\mathrm{1} \\ $$$$\left({iii}\right)\:\mathrm{D}\:{is}\:{midpoint}\:{of}\:\:\mathrm{BC} \\ $$$$\:\:\:\:\therefore\:\:\mathrm{AD}\:{is}\:{a}\:{median}\:{of}\:{the}\:{triangle}\:{and}\:\mathrm{O}\:{is}\:{its} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{centroid}. \\ $$$${Assume}\:\:\boldsymbol{{unit}}\:\:{of}\:\:{coordinate}\:{system}\:{equal}\:{to}\:\mid\mathrm{OD}\mid \\ $$$${then}\:{in}\:\:{the}\:{diagram}\:{above}\:\:\mathrm{D}=\left(\mathrm{0},−\mathrm{1}\right)\:{and}\:\:\mathrm{A}=\left(\mathrm{0},\mathrm{2}\right) \\ $$$${Let}\:\mathrm{B}=\left({b}_{\mathrm{1}} ,{b}_{\mathrm{2}} \right)\:{and}\:\:\mathrm{C}=\left({c}_{\mathrm{1}} ,{c}_{\mathrm{2}} \right) \\ $$$$\because\:\:\mathrm{D}\left(\mathrm{0},−\mathrm{1}\right)\:\:{is}\:{midpoint}\:{of}\:\mathrm{BC} \\ $$$$\therefore\:\:\:\:\:\frac{{b}_{\mathrm{1}} +{c}_{\mathrm{1}} }{\mathrm{2}}=\mathrm{0}\:\:\wedge\:\:\frac{{b}_{\mathrm{2}} +{c}_{\mathrm{2}} }{\mathrm{2}}=−\mathrm{1} \\ $$$$\:\:\:\:{Or}\:\:\:\:\:\:{c}_{\mathrm{1}} =−{b}_{\mathrm{1}} \:\:\wedge\:\:{c}_{\mathrm{2}} =−\mathrm{2}−{b}_{\mathrm{2}} \\ $$$$\:\:\:\:{Or}\:\:\:\mathrm{C}=\left({c}_{\mathrm{1}} \:,\:{c}_{\mathrm{2}} \right)=\left(−{b}_{\mathrm{1}} \:,\:−\mathrm{2}−{b}_{\mathrm{2}} \right) \\ $$$$ \\ $$$${If}\:\:\mathrm{B}\:{and}\:\mathrm{C}\:{are}\:{on}\:{given}\:\:{lines}\:{r}\:{and}\:\:{q}\:\:{respectively} \\ $$$${Then}\:{coordinates}\:{of}\:\mathrm{B}\:{and}\:\mathrm{C}\:{will}\:{satisfy}\:{the}\:{equations} \\ $$$${of}\:\:{r}\:\:{and}\:\:\:{q}\:\:\:{respectively} \\ $$$${r}:\:\:\:{y}=\:{m}_{{r}} {x}\:\:\Rightarrow\:{b}_{\mathrm{2}} ={m}_{{r}} {b}_{\mathrm{1}} …………………….\left({i}\right) \\ $$$${q}:\:\:{y}={m}_{{q}} {x}\:\:\Rightarrow\:\:−\mathrm{2}−{b}_{\mathrm{2}} =−{m}_{{q}} {b}_{\mathrm{1}} ……………\left({ii}\right) \\ $$$${From}\:\left({i}\right)\:\:{and}\:\:\left({ii}\right) \\ $$$$\:\:\:\:\:−\mathrm{2}=\left({m}_{{r}} −{m}_{{q}} \right){b}_{\mathrm{1}} \:\Rightarrow\:{b}_{\mathrm{1}} =\frac{−\mathrm{2}}{{m}_{{r}} −{m}_{{q}} } \\ $$$$\:\:\:\:{b}_{\mathrm{2}} ={m}_{{r}} \left(\frac{−\mathrm{2}}{{m}_{{r}} −{m}_{{q}} }\right)=\frac{−\mathrm{2}{m}_{{r}} }{{m}_{{r}} −{m}_{{q}} } \\ $$$$\mathrm{B}=\left({b}_{\mathrm{1}} ,{b}_{\mathrm{2}} \right)=\left(\frac{−\mathrm{2}}{{m}_{{r}} −{m}_{{q}} }\:,\:\frac{−\mathrm{2}{m}_{{r}} }{{m}_{{r}} −{m}_{{q}} }\right)……………….\left({iii}\right) \\ $$$$\mathrm{C}=\left(−{b}_{\mathrm{1}} \:,\:−\mathrm{2}−{b}_{\mathrm{2}} \right)=\left(\frac{\mathrm{2}}{{m}_{{r}} −{m}_{{q}} }\:,\:\:\frac{\mathrm{2}{m}_{{r}} }{{m}_{{r}} −{m}_{{q}} }−\mathrm{2}\right) \\ $$$$\mathrm{C}=\left(\frac{\mathrm{2}}{{m}_{{r}} −{m}_{{q}} },\frac{\mathrm{2}{m}_{{q}} }{{m}_{{r}} −{m}_{{q}} }\right)……………………………\left({iv}\right) \\ $$$${So}\:{from}\:\:\left({i}\right)\:\:{and}\:\:\:\left({iv}\right)\: \\ $$$${if}\:\:\:{m}_{{r}} \neq{m}_{{q}} \:\:,\:\mathrm{B}\:{and}\:\mathrm{C}\:{are}\:{on}\:{r}\:{and}\:\:{q}\:{respectively}. \\ $$$${So}\:\:{finally}\:\:\:{If}\:\:\:{all}\:{the}\:{three}\:{concurrent}\:{lines}\:\:{have}\: \\ $$$$\boldsymbol{{different}}\:\boldsymbol{{slopes}}\:\:{there}\:{exist}\:{a}\:\:{triangle}\:{whose} \\ $$$${centroid}\:\:{is}\:{the}\:{point}\:{of}\:{concurrency}\:\:{of}\:\:{given}\:{lines}\: \\ $$$$ \\ $$$$ \\ $$
Commented by Yozzii last updated on 07/Jul/16
Thanks! Nice approach.
$${Thanks}!\:{Nice}\:{approach}. \\ $$

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