Further Geometry: Chapter Review

Vocabulary and notation

Pythagorean triple and primitive triple

A Pythagorean triple has positive integer legs and hypotenuse satisfying the Pythagorean theorem. It is primitive when the greatest common factor (GCF) of all three entries is $1$.

Coprime and opposite parity

Coprime integers have GCF $1$. Integers of opposite parity have one even member and one odd member.

Included angle and opposite pair

In triangle $ABC$, the side lengths $a,b,c$ lie across from angles $A,B,C$, respectively. An opposite pair matches a side to the angle across from it. An included angle is at the shared endpoint of two sides.

Unit semicircle and signed horizontal projection

The unit semicircle is the upper half of the radius-one circle centered at the origin. A ray turns counterclockwise from the positive horizontal direction. The horizontal coordinate of its tip at a chosen radius is the signed horizontal projection.

Supplementary angles

Two angles whose measures add to $180^\circ$.

ASA, AAS, SAS, SSS, and SSA

ASA gives two angles and the side between them. AAS gives two angles and another side. SAS gives two sides and their included angle. SSS gives all three sides. SSA gives two sides and the angle opposite one of them.

Nondegenerate triangle and semiperimeter

A nondegenerate triangle encloses positive area. Its semiperimeter $s$ is half the perimeter.

Simple polygon and lattice polygon

A simple polygon has a closed boundary whose straight edges meet only at shared endpoints of consecutive edges. A lattice polygon has integer coordinates at every vertex.

Signed area

With positive $x$ to the right and positive $y$ up, signed area $S$ is positive for a counterclockwise walk around a simple polygon and negative for a clockwise walk. $K$ denotes ordinary area.

Auxiliary circle for tangents

For a given circle centered at $O$ and an outside point $P$, the auxiliary circle has diameter $OP$ and center $M$, the midpoint of $OP$. Its intersections $T_1,T_2$ with the given circle are the contacts.

Formulas and theorems

• Euclid's formulas

  $$\begin{gathered} a=m^2-n^2\\ b=2mn\\ c=m^2+n^2 \end{gathered}$$

  Use when Use integers $m>n>0$. The output is primitive exactly when $m,n$ are coprime and of opposite parity. Here $a,b$ are legs in either size order and $c$ is the hypotenuse.

• Scaling and recovering a primitive triple

  $$\begin{gathered} k=\gcd(a,b,c)\\ \left(\frac ak,\frac bk,\frac ck\right) \end{gathered}$$

  Use when Here $a,b,c$ form a Pythagorean triple. Dividing by $k$ gives the primitive triple, and multiplying its entries by $k$ recovers the original triple. In Euclid's formulas a common factor $d$ of the parameters supplies a factor $d^2$ to each output. Coprime odd parameters give output GCF $2$.

• Recovering the integer parameters

  $$\begin{gathered} m^2=\frac{c+a}{2}\\ n^2=\frac{c-a}{2}\\ b=2mn \end{gathered}$$

  Use when Reduce to a primitive triple first. Use its odd leg as $a$, its even leg as $b$, and its hypotenuse as $c$. The positive square roots give the unique pair $m>n>0$.

• Semicircle coordinates and supplements

  $$\begin{gathered} (x,y)=(r\cos\theta,r\sin\theta)\\ \sin(180^\circ-\theta)=\sin\theta\\ \cos(180^\circ-\theta)=-\cos\theta \end{gathered}$$

  Text description

  Two rays from the origin meet the upper semicircle at reflected points. The acute ray ends to the right of the vertical axis. Its supplementary ray ends to the left at the same height. Perpendicular drops to the horizontal axis give projections of equal magnitude and opposite signs.

  Use when Turn through $0^\circ\le\theta\le180^\circ$ from the positive horizontal axis about the origin. Use $r=1$ for the unit semicircle. Reflection across the vertical axis preserves height and reverses the horizontal coordinate. For interior angles, height is positive and the horizontal projection is positive, zero, or negative for an acute, right, or obtuse turn, respectively.

• Triangle area with sine

  $$\begin{gathered} h=a\sin C\\ K=\tfrac12 ab\sin C \end{gathered}$$

  Use when Use the included angle $C$ between positive lengths $a,b$, with $0^\circ<C<180^\circ$. Height $h$ is measured to the line of base $b$, extended if needed. For fixed lengths, supplementary angles give equal areas and $90^\circ$ gives the maximum.

• Law of Sines

  $$\begin{gathered} \frac a{\sin A}=\frac b{\sin B}\\ \frac a{\sin A}=\frac c{\sin C} \end{gathered}$$

  Use when In a nondegenerate triangle, compare opposite pairs. A complete pair fixes the common ratio, which is $abc/(2K)$. Use it after completing the angles for ASA or AAS. For SSA, check each possible angle.

• SSA solution count for an acute given angle

  $$\begin{cases} 0 & a<h\\ 1 & a=h\\ 2 & h<a<b\\ 1 & a\ge b \end{cases}$$

  Text description

  A ray extends right from A. The fixed vertex C is above the ray, and its perpendicular foot H is ahead of A. The circle centered at C has radius longer than the perpendicular height but shorter than AC. It crosses the ray at B two between A and H, and at B one beyond H. The nearer crossing B two gives an obtuse angle. The farther crossing B one gives the acute inverse-sine angle.

  Use when Given acute $A$ and positive $a,b$, set $h=b\sin A$. Count where the circle centered at $C$ with radius $a$ meets the ray from $A$ beyond its start. At $a=h$ the contact gives $B=90^\circ$. At $a=b$ exclude the crossing at $A$.

• SSA solution count for a right or obtuse given angle

  $$\begin{cases} 0 & a\le b\\ 1 & a>b \end{cases}$$

  Use when Given $90^\circ\le A<180^\circ$ and positive $a,b$, the perpendicular foot is at or behind $A$. The distance from $C$ starts at $b$ at the excluded point $A$ and grows as you move forward along the ray.

• Law of Cosines

  $$\begin{gathered} c^2=a^2+b^2-2ab\cos C\\ \cos C=\frac{a^2+b^2-c^2}{2ab} \end{gathered}$$

  Use when In a nondegenerate triangle, $C$ is between $a$ and $b$ and opposite $c$. SAS gives the positive length root. Valid SSS data give one angle via inverse cosine in degrees. The signed projection $a\cos C$ makes the correction to $a^2+b^2$ negative for acute $C$, zero for right $C$, and positive for obtuse $C$. The zero correction gives the Pythagorean theorem.

• Triangle validity and Heron's formula

  $$\begin{gathered} s=\frac{a+b+c}{2}\\ K=\sqrt{s(s-a)(s-b)(s-c)} \end{gathered}$$

  Use when Use three positive side lengths in the same unit. The two shorter lengths must add to more than the longest. Equality gives a flat arrangement. For a valid triangle every difference from $s$ is positive. The formula uses three sides directly, so an angle or altitude need not be found first.

• Heron factors from the cosine numerator

  $$\begin{gathered} D=a^2+b^2-c^2\\ 16K^2=(2ab-D)(2ab+D)\\ 2ab-D=4(s-a)(s-b)\\ 2ab+D=4s(s-c) \end{gathered}$$

  Use when For a nondegenerate triangle, $D=2ab\cos C$. Both factors of $16K^2$ are positive. The expression pairs the sine area with the cosine numerator.

• Signed trapezoid for a directed edge

  $$2E_i=(x_i-x_{i+1})(h_i+h_{i+1})$$

  Use when For an edge from $P_i$ to $P_{i+1}$, take heights $h_i=y_i-k$ above a reference line $y=k$ below all vertices. Leftward edges contribute positively, rightward edges negatively, and vertical edges contribute zero. In a counterclockwise walk, pair each upper edge with its lower edge within a vertical strip. Their signed contributions leave the area between them.

• Shoelace formula

  $$\begin{gathered} S=\tfrac12\sum_{i=1}^{n}(x_i y_{i+1}-x_{i+1}y_i)\\ K=|S| \end{gathered}$$

  Text description

  The two panels use the same four positions. A is at the lower left, B at the lower right, C at the upper right, and D at the upper left. The walk A, B, C, D, A follows the four outer sides. In the walk A, C, B, D, A, the edge from A to C crosses the edge from B to D. That walk fails the simple boundary requirement.

  Use when List successive vertices $P_i=(x_i,y_i)$ of a simple polygon in boundary order, with $P_{n+1}=P_1$. A different starting vertex keeps the directed edges and preserves $S$. Reversing the walk reverses every edge and negates $S$. Coordinates need not be integers.

• Tangent contacts and the diameter circle

  For a point $T$ on the given circle, $PT$ is tangent at $T$ exactly when $\angle OTP=90^\circ$. Equivalently, $T$ lies on the circle with diameter $OP$.

  Text description

  The solid circle is centered at O. The outside point P is to its right. The dashed auxiliary circle is centered at M, halfway from O to P, and passes through both O and P. The two circles cross at T one above the diameter and T two below it. At each crossing a right-angle mark sits between the solid-circle radius and the tangent segment from P.

  Use when The given circle has center $O$ and positive radius $r$, and $OP>r$. The semicircle theorem supplies the right angle at each intersection. Its converse places every possible contact on the same auxiliary circle.

• Number of tangents through a point

  $$\begin{cases} 2 & OP>r\\ 1 & OP=r\\ 0 & OP<r \end{cases}$$

  Use when For a circle of positive radius $r$, an outside point gives two contacts. At a boundary point draw the perpendicular to $OP$ through $P$. Every line through an inside point is a secant.

Problem types, step by step

Exam traps

• Trap Using the shorter leg to recover the parameters

  Fix Reduce by the GCF first, then use the odd leg. It can be either the shorter or the longer leg.

• Trap Checking only parity or only coprimality

  Fix A primitive output requires both. Two odd inputs still give a triple, with all entries even. Shared input factors enter the outputs squared.

• Trap Turning an obtuse angle's negative projection into a negative height

  Fix Cosine gives the signed horizontal coordinate and sine gives the positive height factor for an interior angle. Extend the base line if the foot is outside, which can also happen with an acute included angle.

• Trap Confusing an included angle with an opposite pair

  Fix For area with sine or the cosine correction, use the angle at the common endpoint of the two factors. For the Law of Sines, pair each side with the angle across from it.

• Trap Stopping after the first inverse sine answer or counting a flat SSA case

  Fix Test each distinct candidate with the angle sum. Keep only positive remaining angles. Use degrees and delay rounding until after testing.

• Trap Using the magnitude of cosine in the Law of Cosines

  Fix Keep the sign. Subtracting a negative correction product makes the opposite side squared larger than the sum of the adjacent squares.

• Trap Treating the perimeter or squared area as the final quantity in Heron's formula

  Fix Use half the perimeter to form the factors. After confirming strict closure, take the positive square root and report square units.

• Trap Sorting polygon vertices or taking absolute values edge by edge

  Fix Preserve the simple boundary and include the closing edge. Sum signed differences first, then take the absolute value of the final signed area.

• Trap Joining the two contacts or using any circle that crosses twice

  Fix Check that $O$ and $P$ are opposite endpoints of the auxiliary diameter. Draw each tangent through $P$ and its own contact. The line through both contacts is a secant.