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Question Number 19293 by Tinkutara last updated on 08/Aug/17

Let AC be a line segment in the plane and B a point between A and C. Construct isosceles triangles PAB and QBC on one side of the segment AC such that ∠APB = ∠BQC = 120° and an isosceles triangle RAC on the other side of AC such that ∠ARC = 120°. Show that PQR is an equilateral triangle.

$$\mathrm{Let}\:{AC}\:\mathrm{be}\:\mathrm{a}\:\mathrm{line}\:\mathrm{segment}\:\mathrm{in}\:\mathrm{the}\:\mathrm{plane} \\ $$$$\mathrm{and}\:{B}\:\mathrm{a}\:\mathrm{point}\:\mathrm{between}\:{A}\:\mathrm{and}\:{C}. \\ $$$$\mathrm{Construct}\:\mathrm{isosceles}\:\mathrm{triangles}\:{PAB}\:\mathrm{and} \\ $$$${QBC}\:\mathrm{on}\:\mathrm{one}\:\mathrm{side}\:\mathrm{of}\:\mathrm{the}\:\mathrm{segment}\:{AC} \\ $$$$\mathrm{such}\:\mathrm{that}\:\angle{APB}\:=\:\angle{BQC}\:=\:\mathrm{120}°\:\mathrm{and} \\ $$$$\mathrm{an}\:\mathrm{isosceles}\:\mathrm{triangle}\:{RAC}\:\mathrm{on}\:\mathrm{the}\:\mathrm{other} \\ $$$$\mathrm{side}\:\mathrm{of}\:{AC}\:\mathrm{such}\:\mathrm{that}\:\angle{ARC}\:=\:\mathrm{120}°. \\ $$$$\mathrm{Show}\:\mathrm{that}\:{PQR}\:\mathrm{is}\:\mathrm{an}\:\mathrm{equilateral} \\ $$$$\mathrm{triangle}. \\ $$

Commented by ajfour last updated on 08/Aug/17

Commented by ajfour last updated on 08/Aug/17

PQ^2 =(a+c)^2 +(((c−a)^2 )/3) =(4/3)(a^2 +c^2 +ac) QR^2 =a^2 +(((2c+a)^2 )/3) =(4/3)(a^2 +c^2 +ac) RP^2 =c^2 +(((2a+c)^2 )/3) =(4/3)(a^2 +c^2 +ac) ⇒ PQ=QR=RP , hence △PQR is equilateral .

$$\:\mathrm{PQ}^{\mathrm{2}} =\left(\mathrm{a}+\mathrm{c}\right)^{\mathrm{2}} +\frac{\left(\mathrm{c}−\mathrm{a}\right)^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}}{\mathrm{3}}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} +\mathrm{ac}\right) \\ $$$$\:\mathrm{QR}^{\mathrm{2}} =\mathrm{a}^{\mathrm{2}} +\frac{\left(\mathrm{2c}+\mathrm{a}\right)^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}}{\mathrm{3}}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} +\mathrm{ac}\right) \\ $$$$\:\mathrm{RP}^{\mathrm{2}} =\mathrm{c}^{\mathrm{2}} +\frac{\left(\mathrm{2a}+\mathrm{c}\right)^{\mathrm{2}} }{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{4}}{\mathrm{3}}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} +\mathrm{ac}\right) \\ $$$$\Rightarrow\:\mathrm{PQ}=\mathrm{QR}=\mathrm{RP}\:,\:\mathrm{hence}\:\bigtriangleup\mathrm{PQR} \\ $$$$\mathrm{is}\:\mathrm{equilateral}\:. \\ $$

Commented by Tinkutara last updated on 08/Aug/17

Thank you very much Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

Answered by Tinkutara last updated on 03/Dec/17

Commented by Tinkutara last updated on 03/Dec/17

This was also a solution. I have a doubt in 3rd line below the figure. Why ∠KPQ=∠PKL?

$${This}\:{was}\:{also}\:{a}\:{solution}.\:{I}\:{have}\:{a}\:{doubt} \\ $$$${in}\:\mathrm{3}{rd}\:{line}\:{below}\:{the}\:{figure}.\:{Why} \\ $$$$\angle{KPQ}=\angle{PKL}? \\ $$

Commented by ajfour last updated on 03/Dec/17

PQLK is an isosceles trapezium(√) Its base angles ∠KPQ is thus equal to ∠PKL (with base PK ).

$${PQLK}\:{is}\:{an}\:{isosceles}\:{trapezium}\sqrt{} \\ $$$${Its}\:{base}\:{angles}\:\angle{KPQ}\:{is}\:{thus} \\ $$$${equal}\:{to}\:\angle{PKL}\:\left({with}\:{base}\:{PK}\:\right). \\ $$

Commented by Tinkutara last updated on 03/Dec/17

Why isosceles?

$${Why}\:{isosceles}? \\ $$

Commented by ajfour last updated on 03/Dec/17

K i suppose (in the solution posted by you) is reflection of P in AC. L is reflection of Q in AC.

$${K}\:{i}\:{suppose}\:\left({in}\:{the}\:{solution}\:{posted}\right. \\ $$$$\left.{by}\:{you}\right)\:{is}\:{reflection}\:{of}\:{P}\:{in}\:{AC}. \\ $$$${L}\:{is}\:{reflection}\:{of}\:{Q}\:{in}\:{AC}. \\ $$

Commented by Tinkutara last updated on 03/Dec/17

Thank you Sir!

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Sir}! \\ $$

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