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Question Number 64203 by Prithwish sen last updated on 15/Jul/19

M is the midpoint of segment AB. Prove that,for every point P in space, ∣PM∣ ≤ ((∣PA∣ +∣PB∣)/2)

$$\mathrm{M}\:\mathrm{is}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:\mathrm{segment}\:\mathrm{AB}.\:\mathrm{Prove} \\ $$$$\mathrm{that},\mathrm{for}\:\mathrm{every}\:\mathrm{point}\:\mathrm{P}\:\mathrm{in}\:\mathrm{space}, \\ $$$$\mid\mathrm{PM}\mid\:\leqslant\:\frac{\mid\mathrm{PA}\mid\:+\mid\mathrm{PB}\mid}{\mathrm{2}} \\ $$

Commented by Prithwish sen last updated on 16/Jul/19

Sir,isn′t PB or PA will be the diagonl if triangle PAB transform into a parallalogram.

$$\mathrm{Sir},\mathrm{isn}'\mathrm{t}\:\mathrm{PB}\:\mathrm{or}\:\mathrm{PA}\:\mathrm{will}\:\mathrm{be}\:\mathrm{the}\:\mathrm{diagonl}\:\mathrm{if}\: \\ $$$$\mathrm{triangle}\:\mathrm{PAB}\:\mathrm{transform}\:\mathrm{into}\:\mathrm{a}\:\mathrm{parallalogram}. \\ $$

Commented by MJS last updated on 16/Jul/19

draw AB and M now choose a point P and its mirror P′ so that M is also the mid of PP′ ⇒ AB is one diagonal and PP′ is the other one ∣PP′∣=2∣PM∣≤∣PA∣+∣PB∣

$$\mathrm{draw}\:{AB}\:\mathrm{and}\:{M} \\ $$$$\mathrm{now}\:\mathrm{choose}\:\mathrm{a}\:\mathrm{point}\:{P}\:\mathrm{and}\:\mathrm{its}\:\mathrm{mirror}\:{P}'\:\mathrm{so}\:\mathrm{that} \\ $$$${M}\:\mathrm{is}\:\mathrm{also}\:\mathrm{the}\:\mathrm{mid}\:\mathrm{of}\:{PP}' \\ $$$$\Rightarrow \\ $$$${AB}\:\mathrm{is}\:\mathrm{one}\:\mathrm{diagonal}\:\mathrm{and}\:{PP}'\:\mathrm{is}\:\mathrm{the}\:\mathrm{other}\:\mathrm{one} \\ $$$$\mid{PP}'\mid=\mathrm{2}\mid{PM}\mid\leqslant\mid{PA}\mid+\mid{PB}\mid \\ $$

Commented by Prithwish sen last updated on 16/Jul/19

∵ PAP′ is a triangle. Ok sir thank you.

$$\because\:\mathrm{PAP}'\:\mathrm{is}\:\mathrm{a}\:\mathrm{triangle}.\:\mathrm{Ok}\:\mathrm{sir}\:\mathrm{thank}\:\mathrm{you}. \\ $$

Commented by MJS last updated on 16/Jul/19

you′re welcome sorry I have no possibility to add drawings at the moment...

$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome} \\ $$$$\mathrm{sorry}\:\mathrm{I}\:\mathrm{have}\:\mathrm{no}\:\mathrm{possibility}\:\mathrm{to}\:\mathrm{add}\:\mathrm{drawings} \\ $$$$\mathrm{at}\:\mathrm{the}\:\mathrm{moment}... \\ $$

Commented by Prithwish sen last updated on 16/Jul/19

No sir you have already done a lot. I appreciate your effort Thank you very much again.

$$\mathrm{No}\:\mathrm{sir}\:\mathrm{you}\:\mathrm{have}\:\mathrm{already}\:\mathrm{done}\:\mathrm{a}\:\mathrm{lot}.\:\mathrm{I}\:\mathrm{appreciate} \\ $$$$\mathrm{your}\:\mathrm{effort}\:\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{again}. \\ $$

Answered by MJS last updated on 16/Jul/19

think of a triangle sides AB, PA, PB now make it a parallelogram obviously 2∣PM∣ which is one diagonal must be shorter than the sum of two sides

$$\mathrm{think}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle} \\ $$$$\mathrm{sides}\:{AB},\:{PA},\:{PB} \\ $$$$\mathrm{now}\:\mathrm{make}\:\mathrm{it}\:\mathrm{a}\:\mathrm{parallelogram} \\ $$$$\mathrm{obviously}\:\mathrm{2}\mid{PM}\mid\:\mathrm{which}\:\mathrm{is}\:\mathrm{one}\:\mathrm{diagonal} \\ $$$$\mathrm{must}\:\mathrm{be}\:\mathrm{shorter}\:\mathrm{than}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{two}\:\mathrm{sides} \\ $$

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