Kelly Shigemoto
Math 300
Half-Planes, Angles and Triangles: A Proof of Convex Sets
Given the definition that a convex set is a set in which if points A and B are in the set, then so is the entire segment AB, we want to show that a half-plane, the interior of an angle and the interior of a triangle are all convex sets. We also intend to show that the exterior of a triangle is not a convex set.
Proof:
Given a half-plane p bounded by a line l, and two points A and B lying in that half-plane, we claim that all points of segment AB lie in that given half-plane p .
Proof:
For every point A and every point B lying in the half-plane p bounded by line l, and for any point P on l, we know that P is not equal to A, P is not equal to B and that ~(A * P * B) by the definition of a half-plane. We also know that A and B are on the same side of l. Let T be an element of segment AB. Then one of the following must hold:
If A = T, then T is in the half-plane p .
If B = T, then T is in the half-plane p .
If A * T * B, suppose T lies in the opposite half-plane (RAA hypothesis).
Then T lies on the opposite side of l from A and B by the Same Side Lemma. If T and A are on opposite sides of l then segment TA intersects l at a point C such that T * C * A. If A * C * T and A * T * B, by Betweenness Proposition 3.3, A * C * B and C * T * B. If A * C * B, then A and B are on opposite sides of l by the Same Side Lemma. But A and B are on the same side of l so T must lie in p (RAA conclusion). Since every point T that is an element of segment AB lies in p , a half-plane is a convex set.
To prove that the interior of an angle is a convex set, we consider the definition of the interior of an angle, which is the intersection of two half-planes. We have already shown that a half-plane is a convex set. To show that the interior of an angle is a convex set, we must show that the intersection of two half-planes is also a convex set. To do this, we consider the half-plane, P , bounded by line, l, and the half-plane, S , bounded by the line, m, which is not equal to l. Let the point of intersection of lines l and m be the point A. We know that there are points C and B on l such that C * A * B by Betweenness Axiom 2. Similarly, by the same axiom, we know that there are points E and F on line m such that E * A * F.
We then consider the angle Ð BAF as the intersection of half-planes P and S . Let T be a point in the interior of Ð BAF, that is, in the intersection of half-planes P and S . Additionally, let U be a point in the interior of Ð BAF that is not equal to T. Then if T and U are interior to Ð BAF, then T and U are in the intersection of P and S . We need to show that all points of segment TU, that is, all points X such that T * X * U, are elements of both P and S .
Given that X is an arbitrary point such that T * X * U, either X lies in the interior of the intersection of P and S , or it does not lie in the intersection of P and S .
If X lies in the intersection of P and S , then all points of segment X lie in the intersection of P and S and the interior of the angle formed by the intersection of the two half-planes is a convex set.
If X does not lie in the intersection of P and S , then X is not interior to Ð BAF. If X is not interior to Ð BAF, then X is not on the same side of line AB as F. And if T and F are on the same side of line AB (by the definition of the interior of an angle), then T and X are on opposite sides of line AB by Betweenness Axiom 4. If T and X are on opposite sides of line AB and T * X * U, then T and U are on opposite sides of line AB. However, if U is on the opposite side of line AB than T, then U is not interior to Ð BAF, which contradicts the hypothesis. Therefore, it is not the case that X does not lie in the intersection of T>P and S . So X must lie in the intersection of P and S , and every point of segment TU, with endpoints T and U interior to Ð BAF, is an element of the interior to Ð BAF. Hence, the interior of an angle is a convex set.
To determine whether or not the interior of a triangle is a convex set, let triangle D PQR be an arbitrary triangle and A and B be distinct points in the interior of D PQR. Also, let T be an element of segment AB. Then either T = A, T = B or A * T * B.
If T = A, then T is in the interior of D PQR.
If T = B, then T is in the interior of D PQR.
If A * T * B, we need to show that T is in the interior of Ð PQR, Ð QRP, and Ð RPQ. We have already proven that T is in the interior of Ð PQR. Similar arguments will show that T is interior to Ð QRP by substituting line QR for line QP and line RP for line QR, and using points Q and P for points P and R, respectively. Substituting line RP for line QP and line PQ for line QR in the first proof, and using points R and Q for points P and R will show that T is interior to Ð RPQ. Since T is in the interior of all three angles, any segment AB and all points T such that A < /FONT>* T * B are in the interior of a triangle. Therefore, the interior of a triangle is a convex set.
To determine whether or not the exterior of a triangle is a convex set, we consider D PQR and distinct points A and B exterior to D PQR. Let A be on the opposite side of line PQ from R. Let r be a ray emanating from A that intersects PQ between points P and Q. Then r also intersects either PR or QR by Betweenness Proposition 3.9. If B is a point on r exterior to D PQR on the opposite side of line PR from Q, we claim that there is a point T that is an element of segment AB interior to D PQR.
Let the point of intersection of PQ and r be called C , and the point of intersection of PR and r be called U . Either T = A, T = C , T = U , T = B, A * T * C , C * T * U , of U * T * B.
T = A, the T is exterior to D PQR.
If T = C , then T lies on triangle D PQR and T is not in the exterior of the triangle by the definition of exterior.
Since there exists a point T where T is not exterior to D PQR, the exterior of a triangle is not a convex set.
Besides the interiors of angles, interiors of triangles and half planes, rays and segments are also convex sets. These two claims can be easily proven by showing that any two points, E and F, in a ray (or segment) form a segment such that every point of segment EF lies in that given ray (or segment).
Convex regions become useful in projective geometry, in which all lines meet at some point, by giving another method for finding a point of intersection that does not fall on the sheet of paper that the diagram is drawn on. A convex set is considered as a model of descriptive geometry, which also provides a model for projective geometry (Coxeter, 1965). Coxeter gives a method for constructing a line between two coplanar lines that do not meet in a given convex region.
Bibliography
Coxeter, Harold Scott Macdonald, Projective Geometry, New York: Springer-Verlag, 1987.