Angular Bisector and Tangent

Given is an integer sided triangle $ABC$ with $BC \le AC \le AB$.$k$ is the angular bisector of angle $ACB$.$m$ is the tangent at $C$ to the circumscribed circle of $ABC$.$n$ is a line parallel to $m$ through $B$.
The intersection of $n$ and $k$ is called $E$.

How many triangles $ABC$ with a perimeter not exceeding $100\,000$ exist such that $BE$ has integral length?

This is a relatively advanced problem incorporating geometry and number theory, and it requires familiarity with some non-rudimentary concepts. Let us proceed with the solution.

The main trick of this problem is to notice that it’s related to the Angle Bisector Theorem, which tells us that the ratio of the lengths of the two segments created by an angle bisector of a triangle is equal to the ratio of the lengths of the opposite sides. In this case, it tells us that:


Now, let us apply some Euclidean geometry and notice that triangle ABC is similar to triangle EBC due to the parallel lines and the alternate interior angles. This gives us:


We can see that we have the same ratio on the right side of both equations, so we equate the left sides:


Multiplying through by AC yields:

Get rid of BC from both sides and we find that AC=AB.

So, our triangle is isosceles and we can denote the sides as AB = AC = x and BC = y.

The perimeter condition translates to 2x + y ≤ 100,000.

E is a point on the circumscribed circle of ∆ABC, and since AB = AC, the segment BE is not only tangent to the circumcircle at point B, but is in fact a radius of the circumcircle. Therefore, the length of BE, being a radius, is the same length as the circumradius.

There’s a formula that allows us to express the circumradius R of a triangle in terms of the sides:

R = abc/4K

where a, b, c are the sides of a triangle and K is its area. For an isosceles triangle, this formula simplifies to:

R = x^2 / 2y

Since the length of BE=R must be an integer, this means that x^2 must be a multiple of 2y. The perimeter condition gives us a range for possible values y can take:

1 ≤ y ≤ 50,000

We now simply count all pairs (x,y) such that x^2 is a multiple of 2y and satisfies the perimeter condition. For each y, x^2 ≥ 2y so x ≥ √(2y). And also, since 2x + y ≤ 100,000, we have x ≤ (100,000 – y)/2.

The count of all possible triangles comes down to how many integer x are there in the interval [√(2y), (100,000 – y)/2] for each y.

The implementation of this method would involve programming and would provide the solution for the number of triangles satisfying these conditions.

More Answers:
Panaitopol Primes
Pseudo-Fortunate Numbers
Lenticular Holes

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