A Geometry Problem

The internet, and especially Youtube, are a vast resource for mathematics now. You find all sorts of problems, explanations, and tutorials, usually done by very talented people. I like to dig around in that pile and soon find some pearls like the following one. The problem has been presented by a buy named Presh Talwalkar. Some of them I knew already, but this one was new to me.

The problem is to determine the area of the quadrilateral with the question mark. Of course, no angles or side lengths are given. It took me a while to figure out a simple solution.

The first idea is to put all this in algebraic terms. To make things simpler, we would put the lower left corner to (0,0) and the lower right one to (0,a), and the third one to (b,c). The points on the left and right sides of the triangle can then be determined using algebraic geometry (depending on a,b,c) and thus the area in question. This looks tedious but is a really nice exercise. Do not forget the determinant formula for the area of a triangle. Let’s do it!

Euler Math Toolbox (EMT) happens to have a geometry package which can use Maxima to compute symbolically with geometric objects. It turns out to get more involved than expected. Trying to do this by hand seems to be really hard work.

>load geometry
 Numerical and symbolic geometry.
>A &= [0,0]; B &= [a,0]; C &= [b,c];
>P1 &= u*C+(1-u)*A; P2 &= v*C+(1-v)*B;
>&solve(areaTriangle(A,P2,B)=6,v), P2v &= expand(P2 with %)
 
                                    12
                               [v = ---]
                                    a c
 
 
                           12 b   12      12
                          [---- - -- + a, --]
                           a c    c       a
 
>&solve(areaTriangle(A,P1,B)=7,u), P1u &= expand(P1 with %)
 
                                    14
                               [u = ---]
                                    a c
 
 
                                14 b  14
                               [----, --]
                                a c   a
 
>lv &= lineThrough(A,P2v); 
>lu &= lineThrough(P1u,B); 
>D &= lineIntersection(lv,lu)
 
                      2
                   7 a  c + 84 b - 84 a     84 c
                  [--------------------, -----------]
                       13 a c - 84       13 a c - 84
 
>sol &= solve(areaTriangle(A,D,B)=4,[a,b,c])[1]  ...
>  with [%rnum_list[1]=s,%rnum_list[2]=t]
 
                             168
                        [a = ---, b = s, c = t]
                             5 t
 
>&expand(areaTriangle(A,C with sol,P2v with sol)) - 3
 
                                   39
                                   --
                                   5
 
>&B with sol
 
                                 168
                                [---, 0]
                                 5 t
 
>&C with sol
 
                                 [s, t]
 
>&P1u with sol
 
                                5 s  5 t
                               [---, ---]
                                12   12
 
>&P2v with sol
 
                             108   5 s  5 t
                            [--- + ---, ---]
                             5 t   14   14
 

The problem is that there is more than one solution. The triangle is not determined by those three areas! However, the area in question is always the same. This is quite a surprising fact of we start algebraically as above. On the other hand, it is not surprising if we assume that the problem can be solved at all. For any transformation (x,y) -> (kx,y/k) keeps the areas intact. Moreover, any transformation (x,y) -> (x+c,y) keeps the given areas intact. So if we have one solution, we can effectively get another solution where C is placed anywhere.

I took the values into C.a.R. (with sliders for s and t, and B,C,D,P1,P2 according to the expressions above) and here is one such triangle.

 

The computations seem to work, and moving the sliders shows that 7.8 is indeed the solution to our problem.

Isn’t there an easier and more intuitive solution? I have not watched the linked video yet. But upon discussion with a friend, the solution occurred to me. I will tell you Have a look at the following sketch.

The trick is to look at the triangle ABP1 and to see BP1 as its base side. Then divide the triangle into ADP1 and ABD. Those triangles have the same height upon BP1 and since the area is height times base over 2, we get (solving for the height)

\(\dfrac{4}{DB} = \dfrac{3}{DP1}\)

Now the same can be done in the CBP1. Thus

\(\dfrac{y+2}{DB} = \dfrac{x}{DP1}\)

Thus

\(\dfrac{y+2}{4} = \dfrac{x}{3}\)

Now we do the same with the triangles over the baseline AP2.  We get in the same way

\(\dfrac{x+3}{4} = \dfrac{y}{2}\)

Now we have two equations for x and y. The solution is

\(x = \dfrac{21}{5}, \quad y = \dfrac{18}{5}\)

So

\(x+y = \dfrac{39}{5}\)

To find such solutions is a matter of practice, combined with trial and error, plus a grain of stubbornness. The brain needs to have enough math tricks to look for, simply. Then you need to look around if anything looks familiar to you.

18. Juni 2019 von mga010
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