HMMT 二月 2019 · 团队赛 · 第 7 题
HMMT February 2019 — Team Round — Problem 7
题目详情
- [ 50 ] A convex polygon on the plane is called wide if the projection of the polygon onto any line in the 1 same plane is a segment with length at least 1. Prove that a circle of radius can be placed completely 3 inside any wide polygon.
解析
- [ 50 ] A convex polygon on the plane is called wide if the projection of the polygon onto any line in the
1
same plane is a segment with length at least 1. Prove that a circle of radius can be placed completely
3
inside any wide polygon.
Proposed by: Shengtong Zhang
Solution 1.
Lemma. for any polygon including its boundary, there exists a largest circle contained inside it.
Proof. Its easy to see that for any circle inside the polygon, it can be increased in size until it is tangent
to at least three sides of the polygon. Then for any three sides of the polygon, there is only one circle
tangent to all three, so there are only finitely many possibilities. Therefore there exists a largest one.
Alternatively, one can show that the space of valid ( x, y, r ) such that the circle with center ( x, y ) and
radius r is compact, e.g. by showing the complement is open and that the complement is open. Then
the map ( x, y, r ) → r is continuous and therefore has a maximum.
Now, take the largest circle. It clearly must be tangent to three sides. If the circle lies inside the
triangle made by the three lines, we can expand the polygon to that triangle and solve it for the
triangle instead. Otherwise, we have the following diagram:
Here the circle is an excircle of the triangle ABC made by the lines AB, AD, and BE . (Note that AD
and AB dont have to be consecutive sides of the polygon, but the ones in between dont really matter.)
Then since the circle is an excircle, we can consider a homothety at C with power 1 + , which sends
the circle to a slightly larger circle which does not touch line AB. If this homothety causes the circle
to leave the polygon for small enough , it must be because the circle was initially tangent to another
line ` , for which it would be an incircle of the triangle made by ` and lines AD , BE , bringing us back
to the first case.
Thus we can reduce to a case where we have a triangle with each height at least 1, and we want to
2 K 2 K 2 K
show the inradius is at least 1 / 3. Let K be the area of the triangle, so the heights , , are all
a b c
at least 1. Then the inradius r satisfies
K 2 K 2 K 1
r = = ≥ = ,
s a + b + c 2 K + 2 K + 2 K 3
as desired.
Solution 2. Consider the center of mass G . We will use the notion of support lines for convex shapes.
( Support lines are the lines that touches the shape but does not cut through it.) If a circle centered at
G with radius 1 / 3 cannot be contained inside the polygon, then there exist a point P on the boundary
that GP < 1 / 3. Let ` be the support line passing through p , ` be the line parallel to
and passing 1 2 1 ′ through G , and ` be the other support line that is parallel to ` , touching the polygon at P . Suppose 3 1 ′ ′ ′ ′ ` intersects the polygon at A and B . Extend P A and P B , intersecting ` at A and B . Then, if we 2 1 consider the two parts of the polygon thatdivides the polygon into, we have: 2 ′ ′ • the part of the polygon that contains P contains the triangle P AB ; ′ ′ • the part of the polygon that contains P is contained in the quadrilateral AA B B . ′ ′ ′ ′ Then we conclude that the center of mass G of the triangle P A B lies between ` and ` , which by 2 1 1 ′ assumption is less than away from ` . However, because the height from P to ` is at least 1, the 1 1 3 1 ′ distance from G to ` is at least , so we have a contradiction. Therefore no such P exists and the 1 3 circle can be placed inside the polygon.