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In-a-chess-board-number-of-unit-squares-with-1-one-vertex-common-2-2-vertices-common-3-2-sides-common-




Question Number 178454 by SLVR last updated on 16/Oct/22
In a chess board number of unit squares  with 1)one vertex common?  2)2 vertices common??  3)2 sides common??
$${In}\:{a}\:{chess}\:{board}\:{number}\:{of}\:{unit}\:{squares} \\ $$$$\left.{with}\:\mathrm{1}\right){one}\:{vertex}\:{common}? \\ $$$$\left.\mathrm{2}\right)\mathrm{2}\:{vertices}\:{common}?? \\ $$$$\left.\mathrm{3}\right)\mathrm{2}\:{sides}\:{common}?? \\ $$
Answered by SLVR last updated on 16/Oct/22
kindly help me
$${kindly}\:{help}\:{me} \\ $$
Commented by mr W last updated on 16/Oct/22
nevertheless we can not answer a  question like “how many squares  on a chess board have a common  vertex?”. just think about it. it is not  clear what the question asks.
$${nevertheless}\:{we}\:{can}\:{not}\:{answer}\:{a} \\ $$$${question}\:{like}\:“{how}\:{many}\:{squares} \\ $$$${on}\:{a}\:{chess}\:{board}\:{have}\:{a}\:{common} \\ $$$${vertex}?''.\:{just}\:{think}\:{about}\:{it}.\:{it}\:{is}\:{not} \\ $$$${clear}\:{what}\:{the}\:{question}\:{asks}. \\ $$
Commented by mr W last updated on 16/Oct/22
i think your question is not clear.  at most 4 unit squares have a common  vertex. at most 2 unit squres have 2  common vertexex. at most 2 unit  squares have a common side.
$${i}\:{think}\:{your}\:{question}\:{is}\:{not}\:{clear}. \\ $$$${at}\:{most}\:\mathrm{4}\:{unit}\:{squares}\:{have}\:{a}\:{common} \\ $$$${vertex}.\:{at}\:{most}\:\mathrm{2}\:{unit}\:{squres}\:{have}\:\mathrm{2} \\ $$$${common}\:{vertexex}.\:{at}\:{most}\:\mathrm{2}\:{unit} \\ $$$${squares}\:{have}\:{a}\:{common}\:{side}. \\ $$
Commented by SLVR last updated on 16/Oct/22
Sir..it is my luck ..you are answering  ....it is not unit squares..  but sqares of any size..typo..  kidly forgive?my mistake..
$${Sir}..{it}\:{is}\:{my}\:{luck}\:..{you}\:{are}\:{answering} \\ $$$$….{it}\:{is}\:{not}\:{unit}\:{squares}.. \\ $$$${but}\:{sqares}\:{of}\:{any}\:{size}..{typo}.. \\ $$$${kidly}\:{forgive}?{my}\:{mistake}.. \\ $$
Answered by Acem last updated on 16/Oct/22
1) One vertex common: on four board edges   Each vertex is formed of 2 squares   Num_(1 ver.com) = (n−1)×4_(edge) = 28 vertices ;n=8   if you want number of squares multiply by 2    2) Vertices common: inside board   Each vertex is formed of 4 squares   Num_(vertices com) = (n−1)^2 = 49 vertices_(of each 4 squ.)      3) Sides common: inside board   Each side is formed of 2 squares   Num_(sides com) = 2n(n−1)= 112 sides_(of each 2 squ.)
$$\left.\mathrm{1}\right)\:{One}\:{vertex}\:{common}:\:{on}\:{four}\:{board}\:{edges} \\ $$$$\:{Each}\:{vertex}\:{is}\:{formed}\:{of}\:\mathrm{2}\:{squares} \\ $$$$\:{Num}_{\mathrm{1}\:{ver}.{com}} =\:\left({n}−\mathrm{1}\right)×\mathrm{4}_{{edge}} =\:\mathrm{28}\:\boldsymbol{{vertices}}\:;{n}=\mathrm{8} \\ $$$$\:{if}\:{you}\:{want}\:{number}\:{of}\:{squares}\:{multiply}\:{by}\:\mathrm{2} \\ $$$$ \\ $$$$\left.\mathrm{2}\right)\:{Vertices}\:{common}:\:{inside}\:{board} \\ $$$$\:{Each}\:{vertex}\:{is}\:{formed}\:{of}\:\mathrm{4}\:{squares} \\ $$$$\:{Num}_{{vertices}\:{com}} =\:\left({n}−\mathrm{1}\right)^{\mathrm{2}} =\:\mathrm{49}\:\boldsymbol{{vertices}}_{{of}\:{each}\:\mathrm{4}\:{squ}.} \\ $$$$\: \\ $$$$\left.\mathrm{3}\right)\:{Sides}\:{common}:\:{inside}\:{board} \\ $$$$\:{Each}\:{side}\:{is}\:{formed}\:{of}\:\mathrm{2}\:{squares} \\ $$$$\:{Num}_{{sides}\:{com}} =\:\mathrm{2}{n}\left({n}−\mathrm{1}\right)=\:\mathrm{112}\:\boldsymbol{{sides}}_{{of}\:{each}\:\mathrm{2}\:{squ}.} \\ $$$$ \\ $$

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