Enigma 1313 - Triangles

Enigma 1313 - Triangles New Scientist Magazine, 30 October 2004. by Keith Austin.

Draw a triangle ABC. On the side AB mark the point P such that AP=(2/5)AB, on BC mark Q such that BQ=(2/5)BC and on CA mark R such that CR=(2/5)CA.

Draw the lines AQ, BR and CP. Call the point where AQ and BR cross X, the point where BR and CP cross Y and the point where CP and AQ cross Z.

If you did the appropriate calculations you would find that the area of triangle XYZ is 1/19 of the area of triangle ABC.

I went through the whole procedure above again, but, this time with each occurrence of the number 2/5 replaced by the number k, which is between 0 and 1/2. This time I found that the area of triangle XYZ was 1/37 of the area of triangle ABC.

What was the number k?

Ciao,

should be 147 for both k^2 and k This gives k=3/7. In general k is a simple fraction for 1/d where d is (3*x^2+1)/4 and x is 4m+/-1

I started with vectors and paled after the first 8 lines! It succumbs to Menelaus' theorem fairly quickly.

Regards

original post as spoiler space (although evil googlegroups 2 will automatically hide this grrrrr)

I like this one! Even tho I can't use by beloved Excel

Set A as our origin. Define vector AB = i and vector AC = j as our two basis vectors. If ABC is a proper triangle then i and j are not parallel and so make a useful basis for the plane. This means that for any two points P and Q, if P = ai + bj and Q = ci + dj, P=Q <=> (a=c AND b=d), which will prove useful

In this post I can't use the usual handwriting conventions to denote a vector (an arrow over a line segment or a twiddle under a letter) so a little care will be required when reading. I will refer to points by capital letters and use the lowercase letter to denote the position vector of the point (the vector from A to the point).

We go straight to the general case where the magic ratio (initially 2/5) is denoted by k.

With terminology out of the way we can proceed. Let us first establish the positions of the points of interest.

By definition, a = 0, b = i and c = j.

p = k.AB = ki q = b + k.BC = i + k(b-c) = i + k(j-i) r = (1-k).AC = (1-k)j

To find x, we note that since X is a point on RB, then for some real m (between 0 and 1, although we don't use that)

x = r + m.RB = r + m(b-r) = (1-k)j + m(i - (1-k)j) = mi + (1-k-m+mk)j = mi + (1-k)(1-m)j

But X is also a point on CP so for some real n

x = c + n.CP = c + n(p-c) = j + n(ki - j) = kni + (1-n)j

Therefore we must have

m = kn and (1-m)(1-k) = 1-n

Substituting former in latter,

(1-kn)(1-k) = 1-n 1-k-kn+nk^2 = 1-n n(1-k+k^2) = k

n = k/(1-k+k^2)

so

m = k^2/(1-k+k^2)

[Sanity check: when k = 2/5 this gives n = 10/19 and m = 4/19 which looks vaguely like my diagram, so ok]

So letting g = 1-k+k^2, we have

x = k^2/g i + (1-k)^2/g j

Now reusing m and n we find y in a similar fashion:

y = mq = m(1-k)i + mkj

and

y = c + n.CP = c + n(p-c) = j + n(ki - j) = kni + (1-n)j

so

m(1-k) = kn and mk = 1-n

giving

m(1-k) = k(1-mk) m-mk = k-mk^2 m(1-k+k^2) = k m = k/g

so

y = k(1-k)/g i + k^2/g j

Reusing m and n once more we find z:

z = mq = m(1-k)i + mkj

and

z = r + n.RB = r + n(b-r) = (1-k)j + n(i - (1-k)j) = ni + (1-k-n+nk)j = ni + (1-k)(1-n)j

so

m(1-k) = n and mk = (1-k)(1-n)

giving

mk = (1-k)(1-m+mk) mk = 1-m+mk-k+mk-mk^2 m-mk+mk^2 = 1-k m = (1-k)/g

so

z = (1-k)^2/g i + k(1-k)/g j

All that manipulation and I fall at the last... sigh... forgot the meta-constraint, Enigmas have 'simple' answers!

I don't know any proper geometry tho

SPOILER

Let the co-ordinates of A, B, and C be (b, c), (0, 0), and (a, 0) respectively. The triangle's area is then S = a*c/2.

Using analytic geometry:

The co-ordinates of the intersection points on the edges are then: P: (1-k)*b, (1-k)*c) Q: (k*a, 0) R: (1-k)*a+k*b, k*c)

The lines are then given by: AQ: y = (c/(b-k*a))*x-k*a*c/(b-k*a) BR: y = (k*c/((1-k)*a+k*b)*x CP: y = ((1-k)*c/((1-k)*b-a))*x-(1-k)*a*c/((1-k)*b-a)

The co-ordinates of the intersection points inside are then: X: (k*((1-k)*a+k*b)/(1-k+k^2), k^2*c/(1-k+k^2)) Y: (1-k)*((1-k)*a+k*b)/(1-k+k^2), (1-k)*k*c/(1-k+k^2)) Z: ((k^2*a+(1-k)^2*b)/(1-k+k^2), (1-k)^2*c/(1-k+k^2))

The area of the triangle XYZ is then:

A = (2*k-1)^2*a*c/(2*(1-k+k^2))

Thus The ratio is A/C = (2*k-1)^2/(1-k+k^2).

If the ratio is 1/37, solving for k yields k = 3/7.

Please reply to drgmayer at hotmail dot com

I also used the equations of the lines, but since the solution was clearly going to work for any given triangle, I used (0,0), (1,0) and (0,1) as my vertices.

Your version is neater, and I'm pleased to say, gave the same answer that I got.

I was a bit worried that I'd got an Enigma wrong when I saw Ian's horror solution, but as John Jones discovered, it looks like Ian fluffed one bit of the arithmetic (as can so easily happen).

Nice work, and elegantly concise as usual.

Ciao,