Math Brain Teaser: Concentric Shapes or The Unkindest Cut of All, Part 2 of 2

If you missed Part 1, also writ­ten by puz­zle mas­ter Wes Car­roll, you can start there and then come back here to Part 2.

Con­cen­tric Shapes:
The Unkind­est Cut of All, Part 2 of 2

Dif­fi­cul­ty: HARDER
Type: MATH (Spa­tial)
Vitruvian Man

Imag­ine a square with­in a cir­cle with­in a square. The cir­cle just grazes each square at exact­ly four points. Find the ratio of the area of the larg­er square to the smaller.

In this puz­zle you are work­ing out many of the same skills as in Part I: spa­tial visu­al­iza­tion (occip­i­tal lobes), mem­o­ry (tem­po­ral lobes), log­ic (frontal lobes), plan­ning (frontal lobes), and hypoth­e­sis gen­er­a­tion (frontal lobes).

Two to one.

Draw the small­er square’s diag­o­nal to see that the the small­er square’s diag­o­nal is the diam­e­ter of the cir­cle. Divide the larg­er square into two equal rec­tan­gu­lar halves to see that the larg­er square’s side is also the diam­e­ter of the cir­cle. This means that the small­er square’s diag­o­nal equals the larg­er square’s side. (Or, if you pre­fer, sim­ply rotate the inner square by 45 degrees.) As we’ve seen in the ear­li­er puz­zle “The Unkind­est Cut Of All,” the area of the small­er square is half that of the larg­er, mak­ing the ratio two to one.


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  1. stefan on May 13, 2007 at 9:18

    I mis-read the instruc­tions to be the oppo­site: an out­er cir­cle, a square inside that and anoth­er cir­cle inside the square, and calc the area dif­fer­ence between the *cir­cles*. Per­haps I looked at the pic­ture instead of read the instructions!

    Any­way, I solved it with trig (draw a right tri­an­gle with one point in the cen­ter, anoth­er to a cor­ner of the square and anoth­er to a near side of the square). The area of each cir­cle can be derived from the length of the short­er side.

    Inter­est­ing­ly, it comes out to the same answer: the out­er cir­cle is twice the area of the small­er one!

  2. Caroline on May 13, 2007 at 11:30

    I think you cre­at­ed a more dif­fi­cult brain teas­er! Nice solu­tion. Now we just need to strength­en those read­ing atten­tion skills! ;-) (sor­ry if the pic­ture was mis­lead­ing — I just like the pic­ture and thought it went with the theme, if not the specifics)

  3. T ROY on July 14, 2007 at 11:17



  4. Peter Cooper on May 13, 2008 at 7:30

    I did it by say­ing that the out­er square’s area is x^2. The inter­nal square’s diag­o­nal length must be x, so its per­pen­dic­u­lar length must be the square root of x^2 / 2 (Pythagorus). Squar­ing that undoes the square root, so you get x^2 com­pared to x^2 / 2. Fac­tor­ing down the x^2 to 1, we get 1/0.5.. or 2/1.

  5. whoop on May 14, 2008 at 6:40

    Its easy if you real­ize that the radius of the cir­cle is both half the diag­o­nal of the small­er square and half the side val­ue of the larg­er square.

  6. easyas123 on September 4, 2008 at 2:40

    quite easy, i like those kind of problems

  7. Egon_Freeman on November 6, 2008 at 12:48

    I arrived at this con­clu­sion through “graz­ing both squares in exact­ly 4 points” — cor­ners of the small­er one, and the ‘cen­ters’ of walls of the big­ger one.

    Pri­mar­i­ly through “spa­tial rea­son­ing”, rotat­ed the inner square, and arrived at a point stat­ed in the pre­vi­ous puzzle.

    10 sec­onds. Once again, change it to “Easy” (esp. if some­one solved, or at least read, the pre­vi­ous puzzle).

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