MAT 312 Grand Canyon University Week 8 Classical Geometry Questions

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funuenz13

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Mathematics

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MAT 312

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Grand Canyon University

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MAT

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Week 8 HW MAT 312 GCU Book Classical Geometry: Euclidean, Transformational, Inversive, and Projective: 5.5.1: In the regular octagon shown below, is the area of the shaded region larger or smaller than the total area of the unshaded regions? 5.5.2: P is a point inside square ABCD. Show that [𝐴𝑃𝐵] + [𝐷𝑃𝐶] = [𝐴𝑃𝐷] + [𝐵𝑃𝐶]. Week 8 HW MAT 312 GCU 5.5.3: A square is divided into five nonoverlapping rectangles, with four of the rectangles completely surrounding the fifth rectangle, as shown in the diagram. The outer rectangles are the same area. Prove that the inner rectangle is a square. 5.5.9: Seven children share a square pizza whose crust may be considered to consist only of the perimeter. Show how they make straight cuts to divide the pizza into seven pieces such that all pieces have the same amount of pizza and the same amount of crust. 15.3.8: Given 𝐶(𝑂, 𝑟) and ordinary points A, B, C, and D on a ray through O, inverting 𝐴′ , 𝐵 ′ , 𝐶 ′ , 𝑎𝑛𝑑 𝐷′ , respectively, show that (𝐴𝐵, 𝐶𝐷) = (𝐴′ 𝐵 ′ , 𝐶 ′ 𝐷′ ) ; that is, that the cross ratio is invariant under inversion. Hint: Use the distance formula 𝐴′ 𝐵 ′ = 𝑟2 𝑂𝐴.𝑂𝐵 𝐴𝐵. Week 8 HW MAT 312 GCU The book Foundations of Geometry: 10.1.2: Let S be a set of points in the plane. A line 𝑙 is called a line of symmetry for S if 𝑝𝑙(𝑆) = 𝑆. Find all lines of symmetry for each of the following. a) b) c) d) An isosceles triangle. An equilateral triangle. A regular polygon. A circle.
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Week 8 HW
MAT 312 GCU
Book Classical Geometry: Euclidean, Transformational, Inversive, and Projective:

5.5.1:
In the regular octagon shown below, is the area of the shaded region larger or smaller than the
total area of the unshaded regions?

Answer:
If we partition the shaded and unshaded regions in the manner shown in the figure below, we notice
that both shaded and unshaded regions have the same area.

Week 8 HW
MAT 312 GCU
5.5.2: P is a point inside square ABCD. Show that [𝐴𝑃𝐵] + [𝐷𝑃𝐶] = [𝐴𝑃𝐷] + [𝐵𝑃𝐶].
Answer:
A

Px

B

Py
P
D

C

From the figure above, note that:
[𝐴𝑃𝐵] + [𝐷𝑃𝐶] =

𝐴𝐵 × 𝐴𝑃𝑦 𝐶𝐷 × 𝑃𝑦 𝐷
+
2
2

Since 𝐴𝐵 = 𝐶𝐷:
[𝐴𝑃𝐵] + [𝐷𝑃𝐶] =

𝐴𝐵 × 𝐴𝑃𝑦 𝐴𝐵 × 𝑃𝑦 𝐷 𝐴𝐵 × (𝐴𝑃𝑦 + 𝑃𝑦 𝐷) 𝐴𝐵 × 𝐴𝐷
+
=
=
2
2
2
2

Which means ...