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Inscribing a Square in a Triangle


Date: 10/13/2000 at 01:15:32
From: Alexandra
Subject: Geometry - Triangle

How can you inscribe a square in a scalene triangle? I have no idea 
how to start.


Date: 10/13/2000 at 11:13:20
From: Doctor Rob
Subject: Re: Geometry - Triangle

Thanks for writing to Ask Dr. Math, Alexandra.

Let the triangle be ABC, and let AB be the longest side. Then <A and 
<B are acute. We will construct a square one of whose sides lies along 
AB and whose other two corners lie on AC and BC.

Pick a point D on AC, and drop a perpendicular from D to AB, meeting 
it at point E. Mark point F on AB (possibly extended) such that 
DE = DF, and D is between A and F. Erect a perpendicular to AB at F. 
On the same side of AB as D, mark point G such that FG = DE. Connect D 
and G, and you have a square DEFG.

                               C
                              _o
                            _-  \
                          _-     \
                        _-        \
                      _-           \
                    _-              \
                  _-       G         \ D
                _-          o---------o
              _-            |         |\
            _-              |         | \
          _-                |         |  \
        _-                  |         |   \
     B o--------------------o---------o----o A
                            F         E

Now draw line AG. This is the locus of the upper left corner of all 
squares with a side along AB and upper right corner on AC. Let the 
point of intersection of AG and BC be H. Then H is the upper left 
corner of the inscribed square lying on BC.


                               C
                              _o
                            _-  \
                          _-     \
                        _-        \
                    H _-           \
                    _o._            \
                  _-    `-. G        \ D
                _-         `o---------o
              _-            | `-.     |\
            _-              |    `-.  | \
          _-                |       `-|  \
        _-                  |         |`-.\
     B o--------------------o---------o----o A
                            F         E

Drop a perpendicular from H to AB, meeting it at point I. Draw a line 
through H parallel to AB, and intersecting AC at K. K is the upper 
right corner of the inscribed square lying on AC. Drop a perpendicular 
from K to AB, meeting it at J. IJ is the side of the inscribed square 
lying along AB. Then HIJK is the required inscribed square. (Clearly 
HIJK is a rectangle. You can prove that HK = HI by using similar 
triangles and the fact that FG = DG, so HIJK is a square.)

                               C
                              _o
                            _-  \
                          _-     \
                        _-        \
                    H _-           \ K
                    _o._------------o
                  _- |  `-. G       |\ D
                _-   |     `o-------+-o
              _-     |      | `-.   | |\
            _-       |      |    `-.| | \
          _-         |      |       |-|  \
        _-           |      |       | |`-.\
     B o-------------o------o-------o-o----o A
                     I      F       J E

Why did we choose AB so that <A and <B were acute?

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   


Date: 10/13/2000 at 14:02:14
From: Doctor Floor
Subject: Re: Geometry - Triangle

Hi, Alexandra,

I saw the beautiful answer that Dr. Rob has given, but since I took a 
different approach I thought you might like my reply as an addition.

Let us consider ABC, angles A and B acute, with an inscribed square 
PQRS with side PQ on AB:

                               C
                              _o
                            _- |\
                          _-   | \
                        _-     |  \
                    S _-       |   \ R
                    _o---------+----o
                  _- |         |    |\ 
                _-   |         |    | \
              _-     |         |    |  \
            _-       |         |    |   \
          _-         |         |    |    \
        _-           |         |    |     \
     A o-------------o---------o----o------o B
                     P         D    Q

CD is the perpendicular altitude from C. For the length let CD = h. 
Let SC = q*AC, so that S divides AC in the ratio CS:SA = q : 1-q.

Since SR is parallel to AB, we know that ABC and SRC are similar. So 
we have the equal ratios CS:CA = CR:CB = SR:AB. We conclude SR = q*AB.

On the other hand CD is parallel to SP, so we have similar triangles 
ADC and APS. And from AS = (1-q)*AC we find in the same way as above 
that SP = (1-q)*CD = (1-q)*h.

Of course in a square SR = SP, or 

     q*AB=(1-q)*h
     q*AB = h - q*h
     q*(AB+h) = h
     q = h/(AB+h)     

and thus 
 
     1-q = (AB+h)/(AB+h) - h/(AB+h) = AB/(AB+h)

But that gives us that CS:SA = q:1-q = h:AB, and S divides AC in the 
ratio of the altitude CD and the side AB. That gives a simple 
construction:

                             C
                            _o
                          _- |\`-.
                        _-   | \  `-.
                      _-     |  \    `-.
                  S _-       |   \ R    `-.
                  _o---------+----o        `-.
                _- |`-.      |    |\          `-.
              _-   |   `-.   |    | \            `-.
            _-     |      `-.|    |  \              `-.
          _-       |         |-.  |   \                `-.
        _-         |         |  `-|    \                  `-.
      _-           |         |    |`-._ \                    `-._
   A o-------------o---------o----o------o------------------------o
                   P         D    Q      B                        X

Let X be on AB extended, so that BX = CD. Then let S be the point 
where the parallel to XC through B meets AC. Then S divides AC in the 
same ratio as B divides AX, and thus in the desired AB:h. The rest of 
the construction is straightforward.

I hope you liked this addition. If you have more questions, just write 
back.

Best regards,
- Doctor Floor, The Math Forum
  http://mathforum.org/dr.math/   


Date: 10/13/2000 at 18:22:48
From: Alexandra
Subject: Re: Geometry - Triangle

Thank you for your reply.

I had trouble with the proof, not realizing which triangles we are 
proving similar. I ended up proving two pairs of similar triangles. 
One side of the smaller square is on one triangle in each pair, and 
one side of the inscribed square is on the other triangle in each 
pair.

I think that you must have <A and <C acute because otherwise the 
perpendicular drawn to AC will fall outside the triangle and not allow 
you to find the locus which locates the top left-hand point of the 
inscribed square.

Thank you for your help,
Alexandra


Date: 10/16/2000 at 09:26:03
From: Doctor Rob
Subject: Re: Geometry - Triangle

Thanks for writing back.

Your similar triangle pairs were just the ones of which I was 
thinking. Yes, your reason for <A and <C acute is correct. Good work! 
Obviously you are grasping the subject pretty well.

- Doctor Rob, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Constructions
High School Geometry
High School Triangles and Other Polygons

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