Two Tilted Rectangles

I found this problem from the wonderful Math with bad Drawings site in my news feed. They cite it from Carolina Shearer. Her twitter account contains more such nice problems. It is the kind of problem which seems only adequate for advanced students. Sometimes, you can solve them by looking at it in the right way. Most of the time, you will start a lengthy computation. Often, you will notice in retrospect that the solution was quite easy and you could have guessed it.

For this problem, I did not see the solution. The problem, of course, is to fit two equal rectangles into a square in the shown manner. For a start, try to find a reason why the point 1/2 on the lower side solves the problem. You can use the general fact that a=b+c in the figure below.

I admit that I did the computations using Euler Math Toolbox. That works. However, there is an elementary fact that can be used to prove a+b=c. For a hint, consult the following image. If the green lines are equal the red lines are equal.

Euclidea App

Geometry is not as popular nowadays as it was. So I was very surprised to learn about a game that features geometry puzzles. It poses problems of Euclidean geometry in the plane, e.g. constructing the tangent to a circle. To solve the problem fully, goals have to be completed. One goal (L) is to do it with the minimum number of steps with all tools available (which include composite tools like the middle perpendicular). Another goal (E) is to do it with the minimum number of elementary tools, which are circles and lines. The third goal (V) is to construct variations of the solution, like reflections of the solution. The goal (E) is often very difficult even for elementary constructions. I show two examples below.

The app is done by some Russian programmers. I could find no information on the authors at all. Actually, there are lots of Russian programmers that make money with their skills in different areas. Besides gaming, an example is flight simulation where Russia programmers made some of the best addons. The programmers are not necessarily skilled geometry geeks. There are old books buried somewhere in libraries that contain lots of geometry problems and solutions. Those are the reminders of the old times when geometry was still popular.

The app is extremely nicely done and shows dedication and skill with user interfaces. Some details could need perfection and I will talk about that later. The app is free. But you can buy more problems. It is a nice touch that you get these for free if you have solved all the previous goals. However, I am considering to donate to these programmers.

Here is an example of a problem. I present the problem with my own program C.a.R. This program could also pose problems with restricted toolsets, called assignments. However, there is no game feature and I never cared about the number of steps taken. It was a program to help to teach geometry in schools.

Your job is to construct the green line given the black lines. In the app, the given objects are not fat and the user cannot set colors. That makes it more difficult than it should be to keep track of the progress.

Of course, it is quite easy to do the job with two given tools, the line tool, and the perpendicular. Moreover, every geometry student should know the following solution.

It takes 5 elementary steps. It is a typical mathematical solution because it is based on another solution to another problem. Mathematics often does that. We know how to construct the middle perpendicular. Thus we know how to construct a perpendicular to a line through a point on the line. It remains to see the tangent as perpendicular to the radius. But the goal was to do it with three elementary steps, not 5. My second solution was the following.

It takes 4 elementary steps still and is based on another well-known geometry fact, the Thales circle. Observe the brown circle (2) to see why this works. It is still not the optimal solution. I now tell you a secret: You can find all solutions on Youtube. Nowadays, going to Youtube is the first step that students take.

The following is the solution with three steps. Unfortunately, I was given that solution before I could find it. Why it works is by no means obvious.

The only proof I could find is a computation with angles.

By basic equivalences, we get that the angles with equal color are equal. By the circle chord theorem, the angle of a chord of a circle with respect to a point on the circle is half the angle with respect to the midpoint of the circle, thus black=2*blue. We also have black+red=90°. Thus black+2*blue=90°.

There should be a simpler proof to this, but I have not yet found it.

All in all, this is an extremely well-done app. I would like to use colors while constructing. I would also like to use the given segments as lines. This is important in the following example, where you have to construct the center of the inscribed circle with only 6 elementary steps. Note that the usual construction takes 4 elementary steps for each angle bisector.

That is already very complicated. The first step is to construct one angle bisector using the blue circles. The second bisector is found with the mysterious red circle which is constructed by one of the blue circles. Its second intersection with the same blue circle yields the second angle bisector. I found this by trial and error. After all, you can construct the angle bisector in C.a.R. so see where it should be. In the app, you can show the targets too. Did I tall you that you can also find the solution on Youtube?

Does anybody have an easy proof?

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 guy named Presh Talwalkar. Some of his problems 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 if 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. 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.

Ellipse Geometry – a Problem

I found the following nice problem in my Facebook account. Facebook, however, is a miracle to me and I am always unable to find a posting a second time. Unless you answer something silly like „Following“ it is lost. The best I could find was this link.

The problem is to prove:

The intersections of two perpendicular tangents to an ellipse form a circle.

Of course, this can be computed by Analytic Geometry and I carry this out below. It is no fun, however. E.g., you can find the radius of the circle if the claim is correct, take a point on the circle, compute the two tangents to the ellipse and check that they are perpendicular.

Let us find a more geometrical solution! At first, there is a well known „construction“ of ellipses folding a circular paper. You can do it with actual paper. Cut a circular paper. Mark a point A inside the circle. Then fold the paper, so that the boundary lies on A. I.e., the point P is reflected along the folding line to A. The line is the middle perpendicular of AP. If you do that often enough the folds will outline the green ellipse. There are even videos on Youtube showing this.

For the proof, you need that ellipses are the set of points where the sum of distances to A and M is constant. In our case, it is „obviously“ constantly equal to the radius of the circle. We have the following.

The set of all middle perpendiculars to a fixed point A and points P on a circle is the set of tangents to an ellipse.

Now you know how to get two perpendicular tangents. In the following construction, I did just that. The four blue lines are four middle perpendiculars. They are constructed by selecting a point P on the black circle, the line PA and a rectangular to that line. Then the blue lines are the four middle perpendiculars on A and the four intersections of our green lines with the circle.

We have to prove that

• the midpoint Z on AM is the center of the rectangle
• and that the rectangle has the same diameter, independent on the choice of P on the black circle.

Then the corners of the blue rectangle will always be on the same circle.

Now, we stretch the blue rectangle by a factor 2 from the center at A. The point Z will be stretched to the point M then. The blue rectangle will become a rectangle through the intersections as in the following construction.

It is now „quite obvious“ that M is the center of the blue rectangle. Thus Z was the center of the 1/2 times smaller rectangle. This proves our first claim.

For the second claim, we need to show that the length of the diagonal of the large blue rectangle does not depend on the choice of the red point. The diagonal has the length

$$\sqrt{c^2+d^2}$$

by Pythagoras. Now we have another claim.

The sum of the squares of the lengths of each two perpendicular secants of a fixed circle that meet in a fixed point inside the circle is constant.

To prove that have a look at the dashed rectangle with diagonal AM. Using this rectangle and Pythagoras you can „easily“ express the diagonal of the blue rectangle by the length of AM and the radius of the circle. This proves the second claim.

The images in this posting have been done with C.a.R. (Compass and Ruler), a Java program developed by the author. It allows beautiful exports of images and the automatic creation of polar sets for sets of lines. That feature was used to compute the ellipse in the second image. The first ellipse was done using the two focal points. C.a.R. has also ellipses defined by 5 points, or even by an equation or a parameterization.

I promised to compute the problem using Analytic Geometry. I am using the Computer Algebra system Maxima via my Euler Math Toolbox for this.

The first method computes two perpendicular tangents to the ellipse with the equation

$$x^2 + c^2 y^2 = 1$$

To find a tangent perpendicular to a vector v, we can maximize the expression

$$v_1 x + v_2 y$$

on the ellipse using the method of Lagrange. If the vector v has norm 1 the value of this maximum will be the distance of the tangent from the origin. The Lagrange equations fot his are

$$v_1 = 2 \lambda x, \quad v_2 = 2 \lambda c^2 y, \quad x^2+c^2 y^2 = 1$$

After solving this, we get

$$v_1 x + v_2 y = 2 \lambda (x^2+c^2y^2) = 2 \lambda$$

Then we do the same with the orthogonal vector, i.e., we maximize

$$-v_2 x + v_1 y$$

We then show that the sum of squares of these two values is constant.

In EMT and Maxima, this is the following code.

>&solve([v1=2*la*x,v2=2*la*c^2*y,x^2+c^2*y^2=1],[x,y,la]); ...
>  la1 &= la with %[1]

2    2   2
sqrt(v2  + c  v1 )
------------------
2 c

>&solve([-v2=2*la*x,v1=2*la*c^2*y,x^2+c^2*y^2=1],[x,y,la]);  ...
>  la2 &= la with %[1]

2   2     2
sqrt(c  v2  + v1 )
------------------
2 c

>&factor(la1^2+la2^2)

2         2     2
(c  + 1) (v2  + v1 )
--------------------
2
4 c


Thus the circle of intersections has the radius

$$r = \dfrac{\sqrt{1+c^2}}{c} = \sqrt{1 + \dfrac{1}{c^2}}$$

This is confirmed by the special case of tangents parallel to the axes.

There are several other methods. One is to construct the tangents using tangents to a circle. For this, the ellipse needs to be stretched by 1/c into y-direction. It will then become a circle. We need to compute points on the image of our circle with radius r under this mapping, then take the tangents to the mapped ellipse and map back. Below is the plan of such a construction.

The computations are very involved, however.

Another method to find both tangents is the following: We look at all lines through a given point (e.g. determined by their slope) and find the ones that intersect the ellipse only once. The product of the two slopes that solve this problem should be 1. Again, this is a complicated computation.

A Geometry Problem with Three Circles and a Point

I friend gave me a geometric problem which turns out to boil down to the figure above. We have three arbitrary circles C1, C2, C3. Now we construct the three green lines through the two intersection points of each pair of these circles. We get lines g11, g12, g13. These lines intersect in one point. Why?

As you know, the argument for the middle perpendicular lines on the sides of a triangle goes like this: Each middle perpendicular is the set of points which have the same distance to two of the corners. So if we intersect two of them in P then d(P,A)=d(P,B) and d(P,B)=d(P,C), which implies d(P,A)=d(P,C). As usual, d(P,A) denotes the distance from P to A. Thus P is also on the third middle perpendicular. Note that we need that P is on the middle perpendicular on AB if and only if d(P,A)=d(P,B).

A similar argument is possible for the angle bisectors. These rays are the set of points with equal distance to two sides. For the heights, such an argument is not available. The standard proof goes by constructing a bigger triangle where the heights are middle perpendiculars. By the way, this proof stops working in Non-Euclidean hyperbolic geometry, where the fact still holds.

Can we make up a proof similar to these proofs for our problem? It turns out that this is indeed possible. The correct value is the following:

$$f(P,C) = d(P,M_C)^2-r_C^2$$

where r(C) is the radius of the circle C, and M(C) is its center. To complete the proof, we only need to show that the line through the intersection of two circles C1 and C2 is the set of all points P such that f(P,C1)=f(P,C2). Then the proof is as easy as the proofs above.

There are several ways to see this.

We could use a result that I call chord-secant-tangent theorem which deals with products of distances of a point on a secant or chord to the circle. But it is possible to get a proof using the Pythagoras only. In the image above we have

$$d(P,Q)^2+d(M,Q)^2 = d(P,M)^2, \quad d(S,Q)^2 + d(M,Q)^2 = r^2$$

Thus

$$f(P,C) = d(P,M)^2 – r^2 = d(P,Q)^2-d(S,Q)^2$$

where C is the circle. Now, if we have two intersection circles, the right-hand side of the equation is the same for both circles, and thus also the left-hand side.

We have seen that f(P,C1)=f(P,C2) for all points on the green line.

But we have to prove the converse too. For this, we observe that D(P,C1)=D(P,C2) implies that P is on a circle around M(C1) and on another circle around M(C2). The two circles meet only in points on the green line.

There is also another way to see that f(P,C1)=f(P,C2) defines the green line. If you work out this equation analytically, you see that it is equivalent to an equation for a line. I leave that to you to check.

Note that there is a second situation where the result does hold too.

In this case, we need f(P,C) for P inside C. It will be negative, but f(P,C1)=f(P,C2) still holds for all points on the line through the intersection, and even if P is on the circles.

There is the following special situation.

It can be seen as a limit case of the previous situation. But it can also be proved by observing that all the tangents have the same length between the intersecting point and the tangent point.

Here is another situation.

The green lines are the sets of points such that f(P,C1)=f(P,C2) for two of the circles. It is quite interesting to construct these lines. I leave that to the reader.

A Geometric Problem

I found the following problem on my Google+ account: In the image below, the total length of the blue line segments is equal to the total length of the green line segments. The segments form an angle of 60° among each other.

After a while of thinking, I found a nice proof of this phenomenon. It is probably the standard proof, and it also yields a lot of generalizations.

We assume the black dot at (0,0) in the plane and denote the direction of the green line segments by vectors by v1, v2, and v3, all of them with length 1. Then

$$v_1+v_2+v_3=0$$

If we denote the lengths of the green segments with a1, a2 and a3, we have that

$$a_1v_1, \quad a_2v_2, \quad a_3a_3$$

are the endpoints of the segments. Now, let c be the center of the circle. We thus have

$$\|c-a_kv_k\| = \|c+b_kv_k\|$$

for k=1,2,3 if we denote the lengths of the blue segments with b1, b2, b3. Squaring this and using a bit of vector algebra, we get

$$– a_k (c \cdot v_k) + a_k^2 = b_k (c \cdot v_k) + b_k^2.$$

Here, the dot denotes the scalar product of two vectors. Thus

$$a_k – b_k = c \cdot v_k$$

Summing up for k=1,2,3 we get our conclusion

$$a_1+a_2+a_3 = b_1 + b_2 + b_3.$$

This generalizes to finitely many vectors which sum up to 0. E.g., we can take the sides of the following pentagram as vectors.

The corresponding result can be seen in the next figure.

Of course, a special case is 10 line segments with equal angle 36° between them.