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Question:

Let ABC be an arbitrary triangle with vertices A, B, C. On the side AB, draw a point D so that the length of AD is half of the length of DB, i.e. D is 1/3 of the way along the line segment AB, and closer to A. Similarly, draw a point E on the side BC so that the length of BE is half of the length of EC. Finally, draw a point F on the side CA so that the length of CF is half of the length of FA.

Now draw line segments (called cevians) AE, BF, CD. These three cevians will form a small triangle. Prove that the area of the small triangle is one seventh the area of the big triangle ABC.

Submit your answers to the UCSD Math Advising Office, located in room 2313 of the Applied Physics and Mathematics building in Muir College, or email your solutions to the Math Advising Officer.

Answer:

Allow me to prove a more general statement:
Instead of 1/2, let the ratio AD/DB = BE/EC = CF/FA be 1/x, where x is an non-negative real number.
Then the area of the small triangle is (x-1)"/(x" + x + 1) times the area of the big triangle ABC.

Proof:

Let the small triangle formed by the cevians have vertices P, Q and R,
where P is the intersection of AE and CD, Q is the intersection of BF and EA,
and R is the intersection of CD and FB. Assume our construction lies in a cartesian (x,y)-plane,
so that, for example, point A has coordinates (x_a, y_a). With this notation, it is easy to describe
first the line through A and B, and then the line orthogonal to that one and through C,
the first in terms of x_a, y_a, x_b, and y_b, and the last in terms of those and x_c and y_c,
to construct the height of triangle ABC and finally calculate its area,

|x_a * (y_c - y_b) + x_b*(y_a - y_c) + x_c * (y_b - y_a)|/2.

In the same style, by deriving the equations of the lines through A and E,
through B and F and through C and D, after having found that

D = ((x*x_a + x_b)/(x + 1), (x*y_a + y_b)/(x+1)),

E = ((x*x_b + x_c)/(x+1) , (x*y_a + y_b)/(x + 1)), and

F = ((x*x_c + x_a)/(x + 1), (x*y_a + y_b)/(x+1)),

we find that

P = ((x"x_a + x*x_b + x_c)/(x" + x + 1), (x"y_a + xy_b + y_c)/(x" + x + 1)),

Q = ((x"x_b + x*x_c + x_a)/(x" + x + 1), (x"y_b + xy_c + y_a)/(x" +x + 1)) and

R = ((x"x_c + x*x_a + x_b)/(x" +x + 1), (x"y_c + x*y_a + y_b)/(x" + x + 1)),

to conclude that, after a lot of boring and trivial algebra, the area of triangle PQR is:

((x-1)")/(x" + x + 1) * (|x_a(y_c - y_b) + x_b(y_a - y_c) + x_c(y_b - y_a)|/2),

which proves the proposition. Indeed, substituting x = 2 yields 1/7.

I'm not a big fan of geometry and i''m sure there's an elegant solution without a straightforward
approach using coordinate geometry. I'll wait 'til non-geometrical problems.
--- Henry Shin