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The New Coffee Room

  1. TNCR
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  3. Puzzle time: Geometry

Puzzle time: Geometry

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  • jon-nycJ jon-nyc

    :::

    1/2.

    By questionology i assume it has a single answer, in other words it is invariant as to the size of the two semicircles. Take the limit as one circle grows to the radius of the outer circle and the small circle shrinks to zero.

    :::

    KlausK Online
    KlausK Online
    Klaus
    wrote on last edited by
    #4

    @jon-nyc said in Puzzle time: Geometry:

    :::

    1/2.

    By questionology i assume it has a single answer, in other words it is invariant as to the size of the two semicircles. Take the limit as one circle grows to the radius of the outer circle and the small circle shrinks to zero.

    :::

    That’s a clever answer. But what’s the mathematical reason for the invariance?

    1 Reply Last reply
    • jon-nycJ jon-nyc

      :::

      1/2.

      By questionology i assume it has a single answer, in other words it is invariant as to the size of the two semicircles. Take the limit as one circle grows to the radius of the outer circle and the small circle shrinks to zero.

      :::

      taiwan_girlT Offline
      taiwan_girlT Offline
      taiwan_girl
      wrote on last edited by
      #5

      @jon-nyc I actually guessed this in my mind. No real research, just a guess by trying to "eye" it. LOL

      1 Reply Last reply
      • jon-nycJ Offline
        jon-nycJ Offline
        jon-nyc
        wrote on last edited by jon-nyc
        #6

        Note I'm using the underline to indicate line segment, I don't know how to put an 'overline' across two letter.

        Screen Shot 2021-02-05 at 6.34.48 AM.png

        You can picture that red triangle revolving around center point C and the two semicircles would grow and shrink (in opposite directions) from radius R to radius 0.

        Thank you for your attention to this matter.

        1 Reply Last reply
        • KlausK Online
          KlausK Online
          Klaus
          wrote on last edited by Klaus
          #7

          Interesting, but why are the ATB and ACB angles necessarily 90 degrees? Don't you presume your solution by making that assmption?

          1 Reply Last reply
          • jon-nycJ Offline
            jon-nycJ Offline
            jon-nyc
            wrote on last edited by jon-nyc
            #8

            Man you're picky. I marked enough angles that should be obvious.

            I have to add some labels so new photo.

            Screen Shot 2021-02-05 at 9.34.24 AM.png

            Angle ATD is 90, since ATE and ETD are 45 degrees. That means ATB is 90 since they sum to 180.

            It's crowded in there but point P is the intersection touched by the base of the letter itself. Angle PAT I've defined as theta. That means the angle APT is 180 - 90 - theta, or just 90-theta.

            That means BPC is also 90-theta.

            Angle PBC is theta because triangles ACH and BCP are congruent.

            That means ACB is 180 - (90-theta) - theta or 90 degrees.

            Bitch.

            Thank you for your attention to this matter.

            KlausK 1 Reply Last reply
            • jon-nycJ jon-nyc

              Man you're picky. I marked enough angles that should be obvious.

              I have to add some labels so new photo.

              Screen Shot 2021-02-05 at 9.34.24 AM.png

              Angle ATD is 90, since ATE and ETD are 45 degrees. That means ATB is 90 since they sum to 180.

              It's crowded in there but point P is the intersection touched by the base of the letter itself. Angle PAT I've defined as theta. That means the angle APT is 180 - 90 - theta, or just 90-theta.

              That means BPC is also 90-theta.

              Angle PBC is theta because triangles ACH and BCP are congruent.

              That means ACB is 180 - (90-theta) - theta or 90 degrees.

              Bitch.

              KlausK Online
              KlausK Online
              Klaus
              wrote on last edited by
              #9

              @jon-nyc said in Puzzle time: Geometry:

              Angle ATD is 90, since ATE and ETD are 45 degrees. That means ATB is 90 since they sum to 180.

              I think you need to invoke Thales's theorem to argue that ATD is 90.

              Angle PBC is theta because triangles ACH and BCP are congruent.

              But why are these triangles congruent? It certainly looks like it, but this is again not obvious unless you already assume a part of the solution.

              AxtremusA 1 Reply Last reply
              • KlausK Online
                KlausK Online
                Klaus
                wrote on last edited by Klaus
                #10

                Here's a solution that doesn't involve any angles. Radius of upper circle is a and of lower circle is b and of the outer circle r.

                57550d93-1bc7-4811-8fe9-72c8b580c5bd-image.png

                The segment marked with "b-a" has length b-a by the intersecting chords theorem.

                Now, by Pythagoras, (2a)^+(2b)^2=(2r)^2, or a^2+b^2=r^2.

                The sum of the areas of the semi-circles is 1/2 pi a^2 + 1/2 pi b^2, or pi/2 (a^2+b^2).
                Substituting in the equation from above, we get that the combined area is
                pi/2 r^2.

                Since r^2*pi = 1, we get that the area is 1/2.

                AxtremusA 1 Reply Last reply
                • KlausK Klaus

                  @jon-nyc said in Puzzle time: Geometry:

                  Angle ATD is 90, since ATE and ETD are 45 degrees. That means ATB is 90 since they sum to 180.

                  I think you need to invoke Thales's theorem to argue that ATD is 90.

                  Angle PBC is theta because triangles ACH and BCP are congruent.

                  But why are these triangles congruent? It certainly looks like it, but this is again not obvious unless you already assume a part of the solution.

                  AxtremusA Away
                  AxtremusA Away
                  Axtremus
                  wrote on last edited by
                  #11

                  @klaus said in Puzzle time: Geometry:

                  @jon-nyc said in Puzzle time: Geometry:

                  Angle ATD is 90, since ATE and ETD are 45 degrees. That means ATB is 90 since they sum to 180.

                  I think you need to invoke Thales's theorem to argue that ATD is 90.

                  Not necessary. Angle ATE and Angle DTE being 45º each is easy to derive from the observation that Angle AET and Angle DET are 90º each. Add up Angle ATE and Angle DTE you get 90º for Angle ATD.

                  1 Reply Last reply
                  • KlausK Online
                    KlausK Online
                    Klaus
                    wrote on last edited by
                    #12

                    Yes, that's right.

                    1 Reply Last reply
                    • KlausK Klaus

                      Here's a solution that doesn't involve any angles. Radius of upper circle is a and of lower circle is b and of the outer circle r.

                      57550d93-1bc7-4811-8fe9-72c8b580c5bd-image.png

                      The segment marked with "b-a" has length b-a by the intersecting chords theorem.

                      Now, by Pythagoras, (2a)^+(2b)^2=(2r)^2, or a^2+b^2=r^2.

                      The sum of the areas of the semi-circles is 1/2 pi a^2 + 1/2 pi b^2, or pi/2 (a^2+b^2).
                      Substituting in the equation from above, we get that the combined area is
                      pi/2 r^2.

                      Since r^2*pi = 1, we get that the area is 1/2.

                      AxtremusA Away
                      AxtremusA Away
                      Axtremus
                      wrote on last edited by
                      #13

                      @klaus said in Puzzle time: Geometry:

                      Here's a solution that doesn't involve any angles. Radius of upper circle is a and of lower circle is b and of the outer circle r.

                      57550d93-1bc7-4811-8fe9-72c8b580c5bd-image.png

                      The segment marked with "b-a" has length b-a by the intersecting chords theorem.

                      Now, by Pythagoras, (2a)^+(2b)^2=(2r)^2, or a^2+b^2=r^2.

                      The sum of the areas of the semi-circles is 1/2 pi a^2 + 1/2 pi b^2, or pi/2 (a^2+b^2).
                      Substituting in the equation from above, we get that the combined area is
                      pi/2 r^2.

                      Since r^2*pi = 1, we get that the area is 1/2.

                      Yeap, that's the solution that requires Thales theorem for you to establish that the "hypothenuse" has to cut through the middle of the circle and thus has the length 2r. Without this you would not be able to relate "r" to "a" and 'b".

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