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Math 531, Exam 2 Information.
10/26/11, LC 310, 9:05 - 9:55.
Exam 2 will be based on:
• Sections 1G, 1H, 2A, 2B, 2C, 2D.
• The corresponding assigned homework problems
(see http://www.math.sc.edu/∼boylan/SCCourses/531Fa11/531.html)
At minimum, you need to understand how to do the homework problems.
Topic List (not necessarily comprehensive):
You will need to know: theorems, results, and definitions from class.
§1G: Polygons in circles.
Definition: An n-gon (n ≥ 3) is regular if and only if
1. it has all sides equal (it is equilateral), and
2. it has all angles equal (it is equiangular).
Fact: Every regular polygon can be inscribed in a circle. (Problem 1G.2.)
Facts on circles:
1. The ratio of circumference to diameter in every circle is a constant, π.
2. The area of a circle with radius r is πr2 .
§1H: Similarity.
Definition: 4ABC is similar to 4XY Z (4ABC ∼ 4XY Z) if and only if 3 pairs of
corresponding angles are similar:
∠A = ∠X, ∠B = ∠Y, ∠C = ∠Z.
More generally, two geometric figures P1 and P2 are similar if and only if there exists a
distance-preserving bijection f : P1 → P2 .
Note: 4ABC ∼ 4XY Z does not imply that 4ABC ∼ 4Y XZ.
Conditions equivalent to similarity: Consider 4ABC and 4XY Z. Then we have
AA: Two pairs of angles equal.
SSS: Three pairs of sides proportional:
YZ
XZ
XY
=
=
= λ > 0.
AB
BC
AC
The real constant λ is the scale factor. If you enlarge 4ABC by a factor of λ, you get
4XY Z.
SAS: Two pairs of sides proportional and the angles between them equal. For example, we
could have:
XZ
XY
=
, ∠A = ∠X.
AB
AC
Key facts: Consider 4ABC.
• Let U be on side AB, and let V be on side AC (U and V are arbitrary). Then we have
(Lemma 1.29)
AV
AU
=
.
U V k BC ⇐⇒
AB
AC
• Let U = midpoint(AB), and let V = midpoint(AC). Then we have (Cor. 1.31)
U V k BC and U V = (1/2) · BC.
Fact: Let S be a circle, and suppose that P is a point not on X.
• If P is inside S, suppose that U V and XY are chords of S through P .
• If P is outside S, suppose that U V P and XY P are secants os S.
Then we have (Thm. 1.35)
P X · P Y = P U · P V.
Fact: Suppose that 4ABC has ∠C = 90◦ . Let CP = alt(C). Then we have (Lemma 1.36)
4ACP ∼ 4ABC ∼ 4CBP.
Pythagorean Theorem: Suppose that 4ABC has ∠C = 90◦ . Then we have a2 + b2 = c2 .
Fact: Suppose that P1 and P2 are similar with scale factor λ > 0: enlarge P2 by a factor of
λ to get P1 . Then we have
Area(P1 ) = λ2 Area(P2 ).
Fact: The midpoints of the four sides of an arbitrary quadrilateral are the vertices of a
parallelogram. (Problem 1H.3).
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§2A: The circumcircle.
Definition: Let A, B, C be non-collinear points. Then the unique circle through these
points is the circumcircle of 4ABC. Its center is the circumcenter; its radius is the circumradius, R.
Theorem: The perpendicular bisectors of the sides of 4ABC meet at the circumcenter of
the triangle.
Extended law of sines: In 4ABC, we have
a
b
c
=
=
= 2R.
sin A
sin B
sin C
Fact: Let K be the area of 4ABC. Then we have (Cor. 2.5)
K=
abc
.
4R
Definition: A quadrilateral ABCD is cyclic if and only if it can be inscribed in a circle.
Fact: Quadrilateral ABCD is cyclic if and only if one pair of opposite angles is supplementary. (Problem 2A.1.)
§2B: The centroid.
Theorem: The three medians of 4ABC, AX, BY , and CZ, meet at the centroid, G, of the
triangle. Moreover, the point G is 2/3 of the way along each median from the corresponding
vertex:
AG = (2/3) · AX, BG = (2/3) · BY, CG = (2/3) · CZ.
Definition: Let 4ABC have sides with midpoints X, Y , and Z. Then 4XY Z is the
medial triangle of 4ABC.
Fact: The medial triangle of 4ABC is similar to 4ABC: 4ABC ∼ 4XY Z.
§2C: The Euler line, orthocenter, and nine-point circle.
Fact: Triangle 4ABC is equilateral if and only if its circumcenter and centroid are the
same.
Definition: Suppose that 4ABC is non-equilateral. The Euler line of 4ABC is OG.
Note: An equilateral triangle has no Euler line.
Theorem: Suppose that 4ABC is non-equilateral. Then the altitudes meet at a point H,
the orthocenter of 4ABC, that lies on the Euler line GO on the side opposite G from O.
Furthermore, we have
HG = 2GO.
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Definitions: Consider 4ABC.
1. Suppose that alt(A) = AD. Then D is the foot of alt(A).
2. If 4ABC is not a right triangle, then the altitude feet D, E, F are distinct and noncollinear. The pedal triangle of 4ABC is 4DEF .
Definition: Let 4ABC have orthocenter H. The Euler points of 4ABC are:
X = midpoint(AH),
Y = midpoint(BH),
Z = midpoint(CH).
I.e., the Euler points are the midpoints between vertices and the orthocenter.
Nine-point circle theorem: Let 4ABC be given. Then the following points lie on a common circle, the nine-point circle of 4ABC
1. The 3 altitude feet: D, E, F .
2. The 3 midpoints of the sides of 4ABC: P , Q, R.
3. The 3 Euler points: X, Y , Z.
Remarks:
1. The nine-point circle has diameters formed by joining an Euler point to the midpoint
of the opposite side: P X, QY , ZR.
2. The nine-point circle is the circumcircle of the medial and pedal triangles of 4ABC.
3. The nine points may not always be distinct.
Theorem: Consider 4ABC. The following are true (Thm. 2.14).
1. Let R be the circumradius of 4ABC. The nine-point circle of 4ABC has radius R/2.
2. Let N be the center of the nine-point circle of 4ABC. Then N lies on the Euler line
of 4ABC. Moreover, N is the midpoint of HO, where H is the orthocenter and O is
the circumcenter.
Note: The Euler line has the points H, N , G, O, and this is their relative order on the line.
Moreover, the relative distances are
HN = (3/2) · GO, N G = (1/2) · GO.
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§2D: Computations.
Law of cosines: c2 = a2 + b2 − 2ab cos C.
Heron’s formula: Let s =
4ABC is
a+b+c
be the semi-perimeter of 4ABC. Then the area of
2
p
K = s(s − a)(s − b)(s − c).
This formula expresses the area in terms of side lengths only.
Fact: Suppose that 4ABC has area K and circumradius R. Then we have (Prob. 2.19)
R=
abc
abc
.
= p
4K
4 s(s − a)(s − b)(s − c)
Stewart’s Theorem: Let 4ABC be given. Let P be an arbitrary point on side AB, and let
CP have length t. The point P divides AB into AP of length x and P B of length y (so that
x + y = c). Then we have
ct2 + xyc = xa2 + yb2 .
You can use this theorem to obtain a formula for t, the length of CP .
Facts: The triangle 4ABC is isosceles if and only if any of the following are true.
1. Two altitudes are equal. (Problem 1B. 7).
2. Two medians are equal. (Problem 2.8).
3. Two angle bisectors are equal. (Problem 2.22).
Ptolemy’s Theorem: Let ABCD be a cyclic quadrilateral with opposite pairs of side lengths
(a, c), (b, d) and diagonals of lengths x and y. Then we have
ac + bd = xy.
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