Set the set right!

Geometry Level 4

{ A } \{A\} is the largest possible set consisting of numbers selected from { 1 , 2 , 3 , … , 2015 } \{1, 2, 3,\ldots,2015\} such that any two elements in { A } \{A\} can define the lengths of an isosceles triangle, each playing the role of either base or the two equal sides.

If the largest and smallest elements in { A } \{A\} are selected to form the lengths of an isosceles triangle (with the largest element as the measure of the equal sides), what is the difference between the circumference of the circle which can be inscribed in this triangle and the the largest number in { A } \{A\} (round to nearest whole number)?

This problem was adapted from the 2016 SMO Junior Round 2.


The answer is 438.

Teng Chou Nathan Mar by Teng Chou Nathan Mar , 21, Singapore

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Let the sides of the isosceles triangle be of length m, m, n. By triangle inequality,

m + m > > n

2m > > n

Since any two elements in {A} can define the lengths of an isosceles triangle, we consider the special case when m is the smallest element in the {A} and n is the largest element in {A}. Then for {A} to be the largest, we need to have m and n maximized. Thus n is 2015. It follows that

m > > 2015 2 \frac{2015}{2} ⟹ \implies m = 1008

Thus the largest possible size of {A} is 2015 - 1008 + 1 = = 1008, where {A} = {1008, 1009…2014, 2015}. Then the isosceles triangle will have sides of length 1008 and 2015.

Note that the sides of triangle ABC are tangent to the inscribed circle, and that the radius meets any tangent at right angles. Let r be the radius of the inscribed circle. Therefore, we can derive

Area of triangle ABC

= 1 2 \frac{1}{2} (2015 + 2015 + 1008)r

= 2519r

By Pythagoras’ Theorem,

Height of triangle ABC

= 3806209 \sqrt{3806209}

Area of triangle ABC

= 1 2 \frac{1}{2} × 3806209 \sqrt{3806209} × 1008

= 504 × 3806209 \sqrt{3806209}

Thus 2519r = 504 × 3806209 \sqrt{3806209} . Solving for r then calculating 2 π \pi r, we get circumference of circle as 2453, to the nearest whole number.

Hence we have the answer as 2453 - 2015 = 438.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...