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Question Number 74087 by mr W last updated on 19/Nov/19

(Q73828) prove that no cube exists whose corners are located on all faces of an other cube.

$$\left({Q}\mathrm{73828}\right) \\ $$$${prove}\:{that}\:{no}\:{cube}\:{exists}\:{whose}\:{corners} \\ $$$${are}\:{located}\:{on}\:{all}\:{faces}\:{of}\:{an}\:{other}\:{cube}. \\ $$

Commented by mr W last updated on 18/Nov/19

here is my proof: the inner cube has 8 corners, but the outer cubes has only 6 faces. that means two of its corners must lie on the same face of the outer cube. say these two corners of the inner cube are 1 and 2. they lie on the face ABCD of the outer cube.

$${here}\:{is}\:{my}\:{proof}: \\ $$$${the}\:{inner}\:{cube}\:{has}\:\mathrm{8}\:{corners},\:{but}\:{the} \\ $$$${outer}\:{cubes}\:{has}\:{only}\:\mathrm{6}\:{faces}. \\ $$$${that}\:{means}\:{two}\:{of} \\ $$$${its}\:{corners}\:{must}\:{lie}\:{on}\:{the}\:{same}\:{face} \\ $$$${of}\:{the}\:{outer}\:{cube}.\:{say}\:{these}\:{two}\:{corners} \\ $$$${of}\:{the}\:{inner}\:{cube}\:{are}\:\mathrm{1}\:{and}\:\mathrm{2}.\:{they}\:{lie} \\ $$$${on}\:{the}\:{face}\:{ABCD}\:{of}\:{the}\:{outer}\:{cube}. \\ $$

Commented by ajfour last updated on 18/Nov/19

think over sir- it shouldn′t mean that. (question said on each face of outer cube there is an inner cube′s corner) later i added at least, still two corners may or maynot touch the outer cube walls..

$${think}\:{over}\:{sir}-\:{it}\:{shouldn}'{t}\: \\ $$$${mean}\:{that}.\:\left({question}\:\:{said}\:{on}\:{each}\:\right. \\ $$$${face}\:{of}\:{outer}\:{cube}\:{there}\:{is}\:{an}\:{inner} \\ $$$$\left.{cube}'{s}\:{corner}\right)\:\:\: \\ $$$${later}\:{i}\:{added}\:{at}\:{least},\:{still}\:{two} \\ $$$${corners}\:{may}\:{or}\:{maynot}\:{touch} \\ $$$${the}\:{outer}\:{cube}\:{walls}.. \\ $$

Commented by mr W last updated on 18/Nov/19

Commented by mr W last updated on 18/Nov/19

so we have an edge 12 from the inner cube in the plane ABCD. other corners of the inner cube are numbered as shown. we know these edges are parallel: 34//56//78//12. their projections in the plane EFGH are also parallel. it′s easy to see that the edge 56 must lie in the opposite face EFGH from the outer cube. (but this is not relevant for the proof.) face 2468 and face 1357 from the inner cube are perpendicular to the edge 12. (clear) let′s image that we cut the outer cube along these two faces from the inner cube. both cut sections are perpendicular to the face ABCD and EFGH. the shape of both cut sections is rectangle.

$${so}\:{we}\:{have}\:{an}\:{edge}\:\mathrm{12}\:{from}\:{the}\:{inner} \\ $$$${cube}\:{in}\:{the}\:{plane}\:{ABCD}. \\ $$$${other}\:{corners}\:{of}\:{the}\:{inner}\:{cube}\:{are} \\ $$$${numbered}\:{as}\:{shown}. \\ $$$${we}\:{know}\:{these}\:{edges}\:{are}\:{parallel}: \\ $$$$\mathrm{34}//\mathrm{56}//\mathrm{78}//\mathrm{12}. \\ $$$${their}\:{projections}\:{in}\:{the}\:{plane}\:{EFGH} \\ $$$${are}\:{also}\:{parallel}. \\ $$$${it}'{s}\:{easy}\:{to}\:{see}\:{that}\:{the}\:{edge}\:\mathrm{56}\:{must} \\ $$$${lie}\:{in}\:{the}\:{opposite}\:{face}\:{EFGH}\:{from} \\ $$$${the}\:{outer}\:{cube}.\:\left({but}\:{this}\:{is}\:{not}\:{relevant}\right. \\ $$$$\left.{for}\:{the}\:{proof}.\right) \\ $$$${face}\:\mathrm{2468}\:{and}\:{face}\:\mathrm{1357}\:{from}\:{the} \\ $$$${inner}\:{cube}\:{are}\:{perpendicular}\:{to}\:{the} \\ $$$${edge}\:\mathrm{12}.\:\left({clear}\right) \\ $$$${let}'{s}\:{image}\:{that}\:{we}\:{cut}\:{the}\:{outer}\:{cube} \\ $$$${along}\:{these}\:{two}\:{faces}\:{from}\:{the}\:{inner} \\ $$$${cube}.\:{both}\:{cut}\:{sections}\:{are}\:{perpendicular} \\ $$$${to}\:{the}\:{face}\:{ABCD}\:{and}\:{EFGH}. \\ $$$${the}\:{shape}\:{of}\:{both}\:{cut}\:{sections}\:{is} \\ $$$${rectangle}. \\ $$

Commented by ajfour last updated on 18/Nov/19

but Sir i believe that two left out corners can just be inside and not touch the outer cube, such an orientation, might even result...(here 2 can be bit below the ceiling and 5 slightly above the floor ..

$${but}\:{Sir}\:{i}\:{believe}\:{that}\:{two}\:{left} \\ $$$${out}\:{corners}\:{can}\:{just}\:{be}\:{inside} \\ $$$${and}\:{not}\:{touch}\:{the}\:{outer}\:{cube}, \\ $$$${such}\:{an}\:{orientation},\:{might}\:{even} \\ $$$${result}...\left({here}\:\mathrm{2}\:{can}\:{be}\:{bit}\:{below}\right. \\ $$$${the}\:{ceiling}\:{and}\:\mathrm{5}\:{slightly}\:{above} \\ $$$${the}\:{floor}\:.. \\ $$

Commented by mr W last updated on 18/Nov/19

Commented by mr W last updated on 18/Nov/19

since the face 2468 is an inscribed square of the rectangle, the rectangle must be also a square with side length a. that means the both cut sections must be of same size.

$${since}\:{the}\:{face}\:\mathrm{2468}\:{is}\:{an}\:{inscribed} \\ $$$${square}\:{of}\:{the}\:{rectangle},\:{the}\:{rectangle} \\ $$$${must}\:{be}\:{also}\:{a}\:{square}\:{with}\:{side}\:{length}\:{a}. \\ $$$${that}\:{means}\:{the}\:{both}\:{cut}\:{sections}\:{must} \\ $$$${be}\:{of}\:{same}\:{size}. \\ $$

Commented by mr W last updated on 18/Nov/19

Commented by mr W last updated on 18/Nov/19

we get the length of the cut section of (√2)a−s and heigth of a. (√2)a−s=a ⇒s=((√2)−1)a ...(i) since square of length s is an inscribed square of the square with side length a, s≥(a/(√2)) ...(ii) we can not find a value of s which fulfills both (i) and (ii). that means an inner cube doesn′t exist which touches all 6 faces of an other cube.

$${we}\:{get}\:{the}\:{length}\:{of}\:{the}\:{cut}\:{section}\:{of} \\ $$$$\sqrt{\mathrm{2}}{a}−{s}\:{and}\:{heigth}\:{of}\:{a}. \\ $$$$\sqrt{\mathrm{2}}{a}−{s}={a} \\ $$$$\Rightarrow{s}=\left(\sqrt{\mathrm{2}}−\mathrm{1}\right){a}\:\:\:...\left({i}\right) \\ $$$${since}\:{square}\:{of}\:{length}\:{s}\:{is}\:{an}\:{inscribed} \\ $$$${square}\:{of}\:{the}\:{square}\:{with}\:{side}\:{length}\:{a}, \\ $$$${s}\geqslant\frac{{a}}{\sqrt{\mathrm{2}}}\:\:\:\:...\left({ii}\right) \\ $$$${we}\:{can}\:{not}\:{find}\:{a}\:{value}\:{of}\:{s}\:{which} \\ $$$${fulfills}\:{both}\:\left({i}\right)\:{and}\:\left({ii}\right). \\ $$$${that}\:{means}\:{an}\:{inner}\:{cube}\:{doesn}'{t} \\ $$$${exist}\:{which}\:{touches}\:{all}\:\mathrm{6}\:{faces}\:{of}\:{an} \\ $$$${other}\:{cube}. \\ $$

Commented by ajfour last updated on 18/Nov/19

you might not be correct here Sir, when you believe- ′′if such an inner cube exists, two of its corners must lie on the same face of the outer cube.′′

$${you}\:{might}\:{not}\:{be}\:{correct}\:{here}\:{Sir}, \\ $$$${when}\:{you}\:{believe}- \\ $$$$''{if}\:{such}\:{an}\:{inner}\:{cube}\:{exists},\:{two}\:{of} \\ $$$${its}\:{corners}\:{must}\:{lie}\:{on}\:{the}\:{same}\:{face} \\ $$$${of}\:{the}\:{outer}\:{cube}.'' \\ $$$$ \\ $$$$ \\ $$

Commented by mr W last updated on 18/Nov/19

ajfour sir: the original question requested that on each of the 6 faces of larger cube there is at least one corner from the inner cube. that means the 8 corners of the inner cube must touch all 6 faces of the outer cube.

$${ajfour}\:{sir}: \\ $$$${the}\:{original}\:{question}\:{requested}\:{that} \\ $$$${on}\:{each}\:{of}\:{the}\:\mathrm{6}\:{faces}\:{of}\:{larger}\:{cube} \\ $$$${there}\:{is}\:{at}\:{least}\:{one}\:{corner}\:{from}\:{the} \\ $$$${inner}\:{cube}.\:{that}\:{means}\:{the}\:\mathrm{8}\:{corners} \\ $$$${of}\:{the}\:{inner}\:{cube}\:{must}\:{touch}\:{all}\:\mathrm{6} \\ $$$${faces}\:{of}\:{the}\:{outer}\:{cube}. \\ $$

Commented by mr W last updated on 18/Nov/19

if not all corners of the inner cube should be on the faces of the outer cube, the situation is quite different.

$${if}\:{not}\:{all}\:{corners}\:{of}\:{the}\:{inner}\:{cube} \\ $$$${should}\:{be}\:{on}\:{the}\:{faces}\:{of}\:{the}\:{outer} \\ $$$${cube},\:{the}\:{situation}\:{is}\:{quite}\:{different}. \\ $$

Commented by ajfour last updated on 18/Nov/19

let us not be deceived by the lack of clarity in diagram, and instead focus on how the question is worded, let us both review it and Sir please see the diagram believing that 2 is below ceiling and not touching the back face/wall and at the same time 5 is above the floor, and not touching the front face; (this orientation even might not exist but until it is proved with analysis, i imagined this was possible with a little range of values of s. Hence asked for s_(min) .

$${let}\:{us}\:{not}\:{be}\:{deceived}\:{by}\:{the} \\ $$$${lack}\:{of}\:{clarity}\:{in}\:{diagram}, \\ $$$${and}\:{instead}\:{focus}\:{on}\:{how}\:{the} \\ $$$${question}\:{is}\:{worded},\:{let}\:{us}\:{both} \\ $$$${review}\:{it}\:{and}\:{Sir}\:{please}\:{see}\:{the} \\ $$$${diagram}\:{believing}\:{that}\:\mathrm{2}\:{is}\:{below} \\ $$$${ceiling}\:{and}\:{not}\:{touching}\:{the} \\ $$$${back}\:{face}/{wall}\:{and}\:{at}\:{the}\:{same} \\ $$$${time}\:\mathrm{5}\:{is}\:{above}\:{the}\:{floor},\:{and} \\ $$$${not}\:{touching}\:{the}\:{front}\:{face}; \\ $$$$\left({this}\:{orientation}\:{even}\:{might}\right. \\ $$$${not}\:{exist}\:{but}\:{until}\:{it}\:{is}\:{proved} \\ $$$${with}\:{analysis},\:{i}\:{imagined}\:{this} \\ $$$${was}\:{possible}\:{with}\:{a}\:{little}\:{range} \\ $$$${of}\:{values}\:{of}\:\boldsymbol{{s}}.\:{Hence}\:{asked}\:{for} \\ $$$$\boldsymbol{{s}}_{{min}} . \\ $$

Commented by ajfour last updated on 18/Nov/19

the situation is quite different/ much interesting then, thank you Sir, my solution attempt is split in two posts. First in the original question, and then in a later post as a system of simultaneous equations, (Q.74024) please give a glance Sir..

$${the}\:{situation}\:{is}\:{quite}\:{different}/ \\ $$$${much}\:{interesting}\:{then},\:{thank} \\ $$$${you}\:{Sir},\:{my}\:{solution}\:{attempt} \\ $$$${is}\:{split}\:{in}\:{two}\:{posts}.\:{First} \\ $$$${in}\:{the}\:{original}\:{question},\:{and} \\ $$$${then}\:{in}\:{a}\:{later}\:{post}\:{as}\:{a}\:{system} \\ $$$${of}\:{simultaneous}\:{equations}, \\ $$$$\left({Q}.\mathrm{74024}\right) \\ $$$${please}\:{give}\:{a}\:{glance}\:{Sir}.. \\ $$

Commented by mr W last updated on 19/Nov/19

now i understand your solutions. yet it′s not easy to find s.

$${now}\:{i}\:{understand}\:{your}\:{solutions}.\:{yet} \\ $$$${it}'{s}\:{not}\:{easy}\:{to}\:{find}\:{s}. \\ $$

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