You are in a room which is 300 wide x 300 long x 300 tall in cm. Your job is to mark a couple of spots on one of the (non-window) walls. Spot 1: about 73 cm high on the side of the wall (x and y coordinates of 300, 73) within 1 cm error. Spot 2: about 145cm high in the middle (lengthwise) of the wall (x and y coordinates of 150, 145). Within 1 cm error. In the room you have a heavy Table 60 wide x 80 tall x 90 long in cm. It has wheels (included in dimensions). There is also a Stick that is 130 cm tall. You look around the room for any other things that can help you. There is a used pencil, a wall clock, couple of LED lights (one lamp on the Table and the other in the center of the ceiling), a couple of books (unknown sizes) but no paper of any kind, the door and the window. You can use the things that were in the room but you cannot use anything you brought with you for actual measurement (like mobile phone). However you can get any internet info from the phone. You can only mark (with the pencil) on the wall, not anywhere else. Can you locate those spots with MINIMUM number of markings on the wall? I think you should do it in only 2 markings per spot if you are careful. (In my opinion spot 1 should be easy). Note: Please don’t break your back trying to lift the table. Also don’t use your own height for measurement!

Room you are in

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    $\begingroup$ a) The perspective in this image is wrong in a number of places and makes the measurements confusing. b) This looks a lot like a homework question (complete with scanned textbook image). $\endgroup$ Mar 17, 2017 at 14:17
  • $\begingroup$ I had no problem understanding and benefiting from the picture, having read its description. But if parts of this puzzle are indeed from another source please include a mention of that, Deepak Mahulikar, and I hope you are not dismayed by awkwardly unproductive comments. $\endgroup$
    – humn
    Mar 17, 2017 at 17:19
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    $\begingroup$ Not really dismayed but thanks for the support humn. I myself have not seen anything similar before. I am personally fascinated by measurement related puzzles that use different scientific principles. If you look at the "Create a 3 inch measurement" puzzle you might know what I mean. I hope others will understand this puzzle and at least give it a go $\endgroup$
    – DrD
    Mar 17, 2017 at 19:56
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    $\begingroup$ Unlike the wall clock the Table Lamps are mostly wired. You can take some liberty with that. $\endgroup$
    – DrD
    Mar 18, 2017 at 13:19

2 Answers 2


You can use the items in the room to create a measuring aid:

By tearing out a page of the book and folding the paper in two, then fold each half of the paper in two and repeat until the folds are clearly less than 1cm apart, you have created a ruler with $2^n$, e.g. 32 or 64 markings. We can now use this ruler to measure any of the objects with known length, e.g. the 60cm side of the table. If the table is $3\frac{7}{32}$ pages wide we can use the calculator on the phone or a calculator service on the internet to determine that the book page is about 18.64cm long and the $\frac{1}{32}$ folds would be abut 0.58cm apart. This can all be done without creating any markings on the wall.

Then use this tool to create the marking in the corner:

You have now a new convenient measure unit, let's call it a 'page' and you can easily convert 73cm into $3\frac{29}{32}$ pages, which is at least within the 1cm allowed tolerance. By starting at the floor and flipping the page up along the wall, it should be feasible to mark a spot close enough to 73cm above ground to fulfil the requirements. Hence, only one marking is required.

... and the marking in the middle of room:

If you now push the table into the same corner with the 60cm side touching the wall, on which the markins should be made, the left edge of the table will be 60cm out on the wall from the room's corner. By rotating the table 90° clockwise (as seen from above) and keeping the said edge touching the wall, the left edge of the table's long side will now be an additional 90cm, in total 150cm away from each corner and mark the middle of the room. You can now, with the aid of the table's edge and the stick, draw a vertical straight line from the floor and at least 145cm high. Now, just as in the corner, convert 145cm to $7\frac{25}{32}$ pages and starting at the floor, use the book page to measure a point 145cm above ground. Here, two markings are required, the vertical line and the 145cm height marking.

  • $\begingroup$ This is definitely OOB thinking. Wow. Jarnjbo, I entirely focused on the Table, the stick first. $\endgroup$
    – DrD
    Mar 18, 2017 at 13:21

These solutions mark the desired spots just barely within 1 cm of their target heights by treating the lamps as point sources and by treating the book and table as rectangular boxes.   As groundwork, here is a quick inventory of readily available distances that may be obtained through diagonals.

  $\small\sqrt{ 60^2{+}\,80^2\strut}~ = ~ 100 ~ ~$cm      $\small\sqrt{ 130^2{-}\,90^2\strut}~\approx~ 93.8 ~$cm      $\small\sqrt{ 130^2{+}\,60^2\strut}~\approx~ 143.2 $ cm
  $\small\sqrt{ 60^2{+}\,90^2\strut}~\approx~ 108.2 ~$cm     $\small\sqrt{ 130^2{-}\,80^2\strut}~\approx~ 102.5 ~$cm     $\small\sqrt{ 130^2{+}\,80^2\strut}~\approx~ 152.6 ~$cm
  $\small\sqrt{ 80^2{+}\,90^2\strut}~\approx~ 120.4 ~$cm     $\small\sqrt{ 130^2{-}\,60^2\strut}~\approx~ 115.3 $ cm     $\small\sqrt{ 130^2{+}\,90^2\strut}~\approx~ 158.1 ~$cm


  a. Begin with the table in the room’s front right corner.

b. Place the book on the floor so that one corner of it marks the table’s free corner, which is now a $\small \sqrt{60^2{+}\,90^2 \strut} = {\scriptsize\sim}108.2$ cm diagonal away from the room’s corner.

c. Move the table out of the way.

  d. Place one end of the stick on the floor at the book’s marker corner and lean the other end into the room’s corner. The top of the stick now marks a height of $\small \sqrt{130^2-108.2^2 \strut} = {\scriptsize\sim}72.1~$cm, just within 1 cm of the target 73 cm.

Back story 1.   To form a right triangle with a 130 cm hypotenuse and a 73 cm leg, the perfect distance from the corner to the bottom end of the stick would be $\small \sqrt{130^2-73^2 \strut} = {\scriptsize\sim}107.6~$cm.   Only one diagonal from the groundwork list was close enough to try, and it worked.


  a. Begin with the table’s 90 cm side flush against the rear wall.

b. Slide the table along the wall until the stick exactly reaches diagonally along the floor from the room’s rear right corner to the desk’s front right corner, 60 cm out from the rear wall.

c. Place the book on the floor so that it is flush with both the rear wall and the table’s right wall.   The left edge of the book now marks a distance $\small \sqrt{130^2-60^2 \strut} = {\scriptsize\sim}115.3 ~$cm along the wall from the room’s rear right corner.

d. Move the table so that its 60 cm side is flush with the right wall and so that its 90 cm side casts no shadow on the floor from the ceiling-centered overhead lamp.

e. Place the stick on the floor along that no-shadow side of the table.   The stick now marks a line parallel to the rear wall exactly 150 cm from the room’s rear right corner.

  f.  Slide the table leftward along the stick.

g. Visually align the table’s left side with the book’s marker edge, leaving a gap of $\small {\scriptsize\sim}115.3 - 90 = {\scriptsize\sim}25.3~$cm between the right side of the desk and the wall.

h. Stand the stick along the front right vertical edge of the table.

i.   Adjust the desk lamp so that its bulb is on the front left corner of the table top and shines toward the top of the stick, which is 90 cm to the right and 50 cm higher.

  j.   The top of the stick’s shadow now marks a height of  $\small 130 + \dfrac{\raise-.5ex{{\scriptsize\sim}25.3}}{90} {\scriptsize\times \,} 50 ~ = ~ { \scriptsize\sim }144.1~$cm above the floor, just within 1 cm of the target 145 cm.

Back story 2.   Where can the target distance of 150 cm from the back wall be found? The ceiling lamp!, which can also provide a perpendicular line from the right wall thanks to the table’s perpendicular vertical sides.
  Hoping to cast the stick’s shadow 145 cm up the wall, the two simplest alignments along the desk’s top edges are 60 cm and 90 cm away from corners where the desk lamp might be positioned. As the stick is 50 cm taller than the table, and SPOT 2 is 15 cm higher than the stick’s top, the perfect distances of the lamp from the wall would be  $\small 60 + \dfrac{\raise-.5ex{60}}{50} {\scriptsize\times \,} 15 \, = \, 78~$cm and  $\small 90 + \dfrac{\raise-.5ex{90}}{50} {\scriptsize\times \,} 15 \, = \, 117~$cm.
  Again, only one diagonal from the groundwork list was close enough to try, and it worked.

  • $\begingroup$ What a great thinking humn. Clue : Think about the other LED light also! $\endgroup$
    – DrD
    Mar 21, 2017 at 13:12
  • $\begingroup$ I suspect that you had different approaches in mind for both spots, @Deepak Mahulikar, and more accurate. These approaches came out close enough and simple enough to post in any case. $\endgroup$
    – humn
    Mar 21, 2017 at 13:18
  • $\begingroup$ humn, The length of a shadow is dictated by the Geometry as you were thinking. The shadow of the stick standing at a partcular spot from the Ceiling light can give a very specific measurement $\endgroup$
    – DrD
    Mar 22, 2017 at 13:48

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