MATH 5375 Homework 3: Exploring Convexity and Geometric Proofs

This assignment solution for MATH 5375 Homework 3 addresses several key concepts in geometry, including convexity, half-planes, and triangle interiors. The first question utilizes Hilbert’s Proposition to demonstrate that a point within a triangle lies in the intersection of half-planes defined by its sides. The second question proves that the intersection of two convex sets is also convex, using the definition of convexity and half-planes. The third question shows that the interior of a triangle is always a convex set, relying on the characterization of the interior and the convexity of angles. Finally, the fourth question identifies which of the provided figures can be drawn without lifting the pencil or retracing a line, focusing on graph traversal concepts. This document is available on Desklib, where students can find more solved assignments and study resources.

Q1)
P
C
E
A D B
K T
Consider triangle ∆ABC shown above.
Let P be the half plane of line AB in which point C lies, let K be the half plane of line BC in which point A
lies, and let L be the half plane of line CA in which point B lies. By definition, the interior domain of the
triangle ∆ABC is the intersection P ∩ K ∩ T.
Using Hilbert’s Proposition, there exists a point D lying between A and B. using the same reasoning,
there exists a point E lying between C and D. claiming that E ∈ P ∩ K ∩ T, which is the interior of triangle
∆ABC. Indeed, point E lies in the same half plane P of line AB as point C. Indeed, the segment EC lies
entire in P since D ∗ E ∗ C. The points B, D and E lie in the same half plane T of line CA. Indeed, because
of A ∗ D ∗ B and D ∗ E ∗ C, segments DB and DE lie entirely in T.
Similarly, since segments AD and DE lie in the half plane K, we see that points A, D and E lie in the same
half plane K of line BC.
Q2)
Let assume that S is a convex having the two points A and B being in S
Lemma
If S is convex, T is convex, and then S ∩ T is convex
Let π be a half plane bounded by the line l
Let A and b be two points in the plane π.
For any point P in the plane such that P ≠ B the point (1 – t)P + tQ is in T
Note that A and B are on the same side of the linel.
Let T be an element of the line segment AB

If A = T, then T is in half plane π
If B = T, then T is in half plane π
Thus every point T is an element of segment AB in π.
The π is a convex set
Q3.
The interior of triangle is always a convex set
Proof:
Denote the triangle as ∆, and the interior of the boundary of ∆ as int(∆)
From boundary of polygon is Jordan curve, it follows that the boundary of ∆ is equal to imag of a Jordan
curve, so int(∆) is well defined.
Denote the vertices of ∆ as A1, A2, A3 for i ∈{1 ,2 , 3 }, put j = 1mod3 +1, k = (i + 1)mod3 +1, and:
Ui = {Ai +st(Aj – Ai)+(1 – s)t(Ak – Ai): s∈(0..1), t∈ R> 0 }
Suppose that ¿Ai is an integral in ∆ , it follows from definition of polygon that ¿Ai cannot be zero or
straight. Then ¿Ai is larger than a straight angle, which is impossible by sum angles of triangle equals two
right angles.
It follows that ¿Ai is convex
From characterization of interior of triangle, it follows that
Int(∆) = ¿ i=1 ¿ 3U i
From interior of convex angle is convex set, it follows for i∈ { 1 ,2 , 3 } that Ui is a convex set.

Q4.
Figure A and C
Figure A
B
A C
E D
The sequence of drawing will be
A-B-C-DA-E-C
Figure C
B
E A C
G
F D

The sequence of drawing A-B-C-D-A-E-F-C-G-D-F