Competition problem — Geometry — Maths Olympiad Prep

Geometry Difficulty 6.4 National Olympiad Prove it Italy

Problem:

Let $ABC$ be an acute triangle and $H$ the foot of the altitude relative to vertex $A$ . We color every point $P$ interior to the triangle in this way: red, if the vertex closest to $P$ is $A$ ; green, if the vertex closest to $P$ is $B$ ; blue, if the vertex closest to $P$ is $C$ . Knowing that $AH=35$ , $BH=21$ and $CH=15$ , what is the measure of the area formed by the red points?

Solution

Solution:

The answer is $198$ . Let us denote by $L, M, N$ the midpoints of sides $AB, BC, CA$ , and by $O$ the circumcenter of $ABC$ (the point of intersection of the perpendicular bisectors of the sides of the triangle). By a well-known property of the perpendicular bisector of a segment, the set of red points is the quadrilateral $ALON$ , whose area we must therefore determine.

We have $LN = BM = MC = (BH + HC)/2 = (21 + 15)/2 = 18$ . From the property of the circumcenter we have $OA^2 = OC^2$ ; letting $K$ be the foot of the perpendicular drawn from $O$ to $AH$ , by the Pythagorean theorem, we have $OK^2 + KA^2 = OM^2 + BM^2$ . Setting $KA = x$ , and observing that $OM = AH - KA = 35 - x$ , we obtain the equation $3^2 + x^2 = (35 - x)^2 + 18^2$ , from which $9 + x^2 = 1225 + x^2 - 70x + 324$ , that is $70x = 1540$ and finally $x = 22$ .

Since $ALON$ can be decomposed into two triangles having the same base $LN$ and sum of heights equal to $KA$ , the required area will be $18 \times 22 / 2 = 198$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement translated into English from it; metadata (topic, difficulty) added by this project.