Geometry

Theorem: If two distinct lines intersect at a point, then they cannot intersect at any other point.

Given: Two distinct lines (l_1) and (l_2) intersect at point (P).

To prove: (l_1) and (l_2) do not intersect at any other point.

Proof:
Suppose, if possible, that the two lines also intersect at another point (Q), where (Q \neq P).

Then both lines (l_1) and (l_2) pass through the two distinct points (P) and (Q).

But by the fundamental axiom of Euclidean geometry:

Through two distinct points, one and only one straight line can be drawn.

Therefore, the line passing through (P) and (Q) is unique. Hence (l_1) and (l_2) must be the same line.

This contradicts the given condition that (l_1) and (l_2) are two distinct intersecting lines.

So our supposition is false. Therefore, the two lines cannot intersect at any point other than (P).

Hence, if two lines intersect at a point, they do not intersect at any other point. (\blacksquare)

Theorem: If a line in the plane of a triangle intersects the triangle, then it intersects at least two sides of the triangle.

Proof:

Let the triangle be (ABC), and let the given line be (l). The triangle is the convex region bounded by the three sides (AB), (BC), and (CA).

A line (l) divides the plane into two open half-planes. Suppose (l) intersects the triangle. Then all three vertices (A,B,C) cannot lie strictly in the same half-plane determined by (l); otherwise the whole triangle, being the convex hull of (A,B,C), would also lie in that half-plane and would not meet (l).

So there are two possible situations.

Case 1: No vertex lies on (l).

Then the vertices must lie on opposite sides of (l). Since there are three vertices and only two half-planes, two vertices lie in one half-plane and the third lies in the other. Suppose (A) lies in one half-plane and (B,C) lie in the other. Then the segments (AB) and (AC) join points on opposite sides of (l), so each of them must cross (l) at an interior point. Hence (l) intersects the two sides (AB) and (AC).

Thus (l) intersects at least two sides of the triangle.

Case 2: At least one vertex lies on (l).

If two vertices lie on (l), say (A) and (B), then (l) is the line (AB). It intersects the side (AB), and also intersects the sides (AC) and (BC) at the vertices (A) and (B) respectively. Hence (l) intersects at least two sides.

If exactly one vertex, say (A), lies on (l), then:

• If (B) and (C) lie on opposite sides of (l), then the segment (BC) crosses (l) at an interior point. So (l) intersects the side (BC). Also (A) lies on the sides (AB) and (AC). Hence (l) intersects at least two sides.
• If (B) and (C) lie on the same side of (l), then (l) meets the triangle only at the vertex (A). But (A) belongs to both sides (AB) and (AC). Hence (l) intersects the two sides (AB) and (AC).

In every possible case, the line (l) intersects at least two sides of the triangle.

[ \boxed{\text{Hence proved.}} ]

Given:
A plane $\pi$ is perpendicular to one of two parallel lines $l_1$ and $l_2$.
Let $l_1 \parallel l_2$ and $l_1 \perp \pi$ at the point $A$.

To prove:
$l_2 \perp \pi$.

Construction:
Let $l_2$ meet the plane $\pi$ at $B$. Take any line $m$ in the plane $\pi$ passing through $B$. Through $A$, draw a line $m'$ in $\pi$ parallel to $m$.

Proof:

1. Since $l_1 \perp \pi$, the line $l_1$ is perpendicular to every line in $\pi$ passing through its foot $A$.
   Therefore, [ l_1 \perp m' ]

2. Also, it is given that [ l_2 \parallel l_1 ] and by construction, [ m \parallel m' ]

3. Hence the angle between $l_2$ and $m$ is equal to the angle between $l_1$ and $m'$.
   Since $l_1 \perp m'$, we get [ \angle(l_2,m)=90^\circ ]

4. Therefore, [ l_2 \perp m ]

5. But $m$ is any arbitrary line in the plane $\pi$ passing through $B$. So $l_2$ is perpendicular to every line in $\pi$ through $B$.

6. Hence, by definition, [ l_2 \perp \pi ]

Thus, the plane $\pi$ is perpendicular to both of the parallel lines $l_1$ and $l_2$.
Hence proved.

The statement as written is false unless (O) is not on either line. For example, if (s) and (t) intersect at (O), then [ d(O,s)=d(O,t)=0, ] but no positive-radius circle centered at (O) can be tangent to either line. The correct theorem is:

[ O \text{ is the centre of a non-degenerate circle tangent to both } s \text{ and } t \iff d(O,s)=d(O,t)>0. ]

Equivalently,

[ O \text{ is equidistant from } s \text{ and } t \text{ and } O\notin s\cup t. ]

Let (d(O,s)) and (d(O,t)) denote the perpendicular distances from (O) to (s) and (t).

Key fact

A line (l) is tangent to a circle with centre (O) and radius (r>0) if and only if [ d(O,l)=r. ]

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Proof

1. Suppose (O) is the centre of a non-degenerate circle tangent to both (s) and (t)

Let the radius be (r>0).

Since the circle is tangent to (s), [ d(O,s)=r. ]

Since the circle is tangent to (t), [ d(O,t)=r. ]

Therefore, [ d(O,s)=d(O,t)=r>0. ]

So (O) is equidistant from (s) and (t), and (O) lies on neither line.

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2. Conversely, suppose

[ d(O,s)=d(O,t)=r>0. ]

Draw the circle with centre (O) and radius (r).

Since [ d(O,s)=r, ] the line (s) is tangent to this circle.

Similarly, since [ d(O,t)=r, ] the line (t) is tangent to the same circle.

Thus (O) is the centre of a circle tangent to both (s) and (t).

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Hence the corrected statement is

[ O \text{ is the centre of a non-degenerate circle tangent to both } s \text{ and } t \iff d(O,s)=d(O,t)>0. ]

[ \boxed{\text{Proved.}} ]

If one allows a degenerate circle of radius (0), then the original statement would hold only in that degenerate sense; for ordinary circles, the positivity condition is necessary.

First statement

Let the angle (BAD) be non-degenerate. Take (A) as origin and choose unit vectors (\mathbf e_1,\mathbf e_2) along the rays (\overrightarrow{AB}) and (\overrightarrow{AD}), respectively. Since the angle is non-degenerate, (\mathbf e_1,\mathbf e_2) are linearly independent.

Write [ B=b\mathbf e_1,\qquad D=d\mathbf e_2,\qquad P=p\mathbf e_1,\qquad Q=q\mathbf e_2, ] where [ b=d(A,B),\quad d=d(A,D),\quad p=d(A,P),\quad q=d(A,Q). ]

The direction vector of (BD) is [ \overrightarrow{BD}=D-B=d\mathbf e_2-b\mathbf e_1. ] The direction vector of (PQ) is [ \overrightarrow{PQ}=Q-P=q\mathbf e_2-p\mathbf e_1. ]

Case 1: (BD) and (PQ) are identical

Since (P) lies on (\overrightarrow{AB}) and the line (BD) meets (\overrightarrow{AB}) only at (B), we must have (P=B). Similarly, (Q=D). Hence [ p=b,\qquad q=d, ] so [ \frac{d(A,P)}{d(A,B)}=\frac{p}{b}=1=\frac{q}{d}=\frac{d(A,Q)}{d(A,D)}. ]

Case 2: (BD\parallel PQ)

Then their direction vectors are proportional. So for some scalar (\lambda), [ q\mathbf e_2-p\mathbf e_1=\lambda(d\mathbf e_2-b\mathbf e_1). ] Comparing coefficients of the independent vectors (\mathbf e_1,\mathbf e_2), [ -p=-\lambda b,\qquad q=\lambda d. ] Thus [ \frac{p}{b}=\lambda=\frac{q}{d}. ] Therefore [ \frac{d(A,P)}{d(A,B)}=\frac{d(A,Q)}{d(A,D)}. ]

Conversely, suppose [ \frac{d(A,P)}{d(A,B)}=\frac{d(A,Q)}{d(A,D)}=\lambda. ] Then [ p=\lambda b,\qquad q=\lambda d. ] Hence [ P=\lambda B,\qquad Q=\lambda D. ] If (\lambda=1), then (P=B) and (Q=D), so the lines (PQ) and (BD) are identical. If (\lambda\neq 1), then [ \overrightarrow{PQ}=Q-P=\lambda(D-B)=\lambda\overrightarrow{BD}, ] so (PQ\parallel BD).

Thus the lines (BD) and (PQ) are identical or parallel if and only if [ \frac{d(A,P)}{d(A,B)}=\frac{d(A,Q)}{d(A,D)}. ]

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OR

Pythagoras theorem

Let (ABC) be a right-angled triangle with the right angle at (C). Then the hypotenuse is (AB). We prove that [ AB^2=AC^2+BC^2. ]

Drop a perpendicular (CD) from (C) to the hypotenuse (AB), where (D) lies on (AB).

Now consider the triangles (ACD) and (ABC).

• (\angle CAD=\angle BAC) is common.
• (\angle ADC=90^\circ=\angle ACB), since (CD\perp AB).

Therefore, [ \triangle ACD\sim \triangle ABC. ] Hence corresponding sides are proportional: [ \frac{AC}{AB}=\frac{AD}{AC}. ] So [ AC^2=AB\cdot AD. \quad (1) ]

Similarly, consider the triangles (BCD) and (ABC).

• (\angle CBD=\angle CBA) is common.
• (\angle BDC=90^\circ=\angle BCA).

Therefore, [ \triangle BCD\sim \triangle ABC. ] Hence [ \frac{BC}{AB}=\frac{BD}{BC}. ] So [ BC^2=AB\cdot BD. \quad (2) ]

Adding (1) and (2), [ AC^2+BC^2=AB\cdot AD+AB\cdot BD ] [ =AB(AD+BD). ] But (AD+BD=AB), since (D) lies on (AB). Therefore, [ AC^2+BC^2=AB\cdot AB=AB^2. ]

Hence, [ \boxed{AB^2=AC^2+BC^2}. ]

So the square of the length of the hypotenuse of a right-angled triangle is equal to the sum of the squares of the lengths of the other two sides.

Theorem: In the four-point geometry, every distinct line has exactly one line parallel to it.

Axioms used:

1. There are exactly four points.
2. Any two distinct points determine exactly one line.
3. Every line contains exactly two points.

Proof:

Let (l) be any line in the four-point geometry.
By Axiom 3, (l) contains exactly two points. Since there are exactly four points in total, exactly two points remain outside (l). Let those two points be (P) and (Q).

By Axiom 2, there is exactly one line through (P) and (Q). Call this line (m).
Now (l) contains the other two points, while (m) contains (P) and (Q). Therefore (l) and (m) have no common point. Hence

[ m \parallel l. ]

So at least one line parallel to (l) exists.

Now suppose (n) is any line parallel to (l).
If (n) contained a point of (l), then (n) and (l) would intersect at that point, contradicting parallelism. Therefore both points of (n) must lie outside (l). But outside (l) there are only the two points (P) and (Q). Hence (n) must be the line through (P) and (Q). By Axiom 2, there is only one such line, namely (m). Thus

[ n = m. ]

So no second line parallel to (l) can exist.

Therefore every line (l) has exactly one line parallel to it. Since (l) was chosen arbitrarily, in the four-point geometry each distinct line has exactly one line parallel to it. (\square)

Example: If the four points are (A,B,C,D), then
[ AB \parallel CD,\qquad AC \parallel BD,\qquad AD \parallel BC. ] Each line has exactly one parallel line.

Definition:
A regular polyhedron (also called a Platonic solid) is a convex polyhedron in which
(i) all faces are congruent regular polygons, and
(ii) the same number of faces meet at every vertex.
Thus all faces are alike and all vertices are alike.

Proof that there are exactly five types:

Let each face be a regular (n)-gon and let (m) faces meet at each vertex. Since a polyhedron must have at least three faces at a vertex and each face has at least three sides,

[ n \ge 3,\qquad m \ge 3. ]

The interior angle of a regular (n)-gon is

[ \theta=\frac{(n-2)180^\circ}{n}. ]

At any vertex of a convex polyhedron, the sum of the face angles must be less than (360^\circ); otherwise the faces would lie flat or overlap. Therefore,

[ m\theta < 360^\circ. ]

Substituting (\theta),

[ m\cdot \frac{(n-2)180}{n}<360. ]

Dividing by (180),

[ \frac{m(n-2)}{n}<2. ]

So

[ m(n-2)<2n. ]

Rearranging,

[ mn-2m-2n<0, ]

[ (m-2)(n-2)<4. ]

Now (m-2) and (n-2) are positive integers. The only positive integer pairs whose product is less than (4) are

[ (1,1),\ (1,2),\ (2,1),\ (1,3),\ (3,1). ]

Hence the possible pairs ((m,n)) are

[ (3,3),\ (3,4),\ (4,3),\ (3,5),\ (5,3). ]

These give exactly the five regular polyhedra:

Each of these five is a regular polyhedron. Therefore, there are exactly five types of regular polyhedra.

Defect of a triangle. In a neutral (absolute) geometry, the defect of a triangle $ABC$ is $\delta(\triangle ABC)=180^\circ-(\angle A+\angle B+\angle C).$ Two triangles are equivalent if they can be cut into finitely many pieces that are pairwise congruent.

Theorem. If triangles $ABC$ and $A'B'C'$ have equal defects and $AB\cong A'B'$, then they are equivalent.

Proof.

1. Triangle to Saccheri quadrilateral. Let $M,N$ be the midpoints of $AC$ and $BC$, and drop perpendiculars $AP$, $BQ$, $CR$ from $A,B,C$ to line $MN$. Right triangles $AMP$ and $CMR$ are congruent (hypotenuses $AM=CM$, vertical angles at $M$ equal), so $AP=CR$. Likewise $\triangle BNQ\cong\triangle CNR$, so $BQ=CR$. Hence $AP=BQ$, and $PQBA$ is a Saccheri quadrilateral with base $PQ$ and summit $AB$. It is made of the pieces of $\triangle ABC$ rearranged ($\triangle AMP$ and $\triangle BNQ$ replace $\triangle CMR$ and $\triangle CNR$), so $\triangle ABC$ is equivalent to $PQBA$.

2. Summit angles. The angle sum of $PQBA$ equals that of $\triangle ABC$ (the angles at $C$, $A$, $B$ are redistributed, and the angles at $M$ and $N$ become straight angles). The base angles are right angles, and the two summit angles are equal, so each summit angle is $\theta=\tfrac12(\angle A+\angle B+\angle C)=90^\circ-\tfrac12\delta.$

3. Same data, same quadrilateral. The same construction for $\triangle A'B'C'$ gives a Saccheri quadrilateral with summit $A'B'\cong AB$ and summit angles $\theta'=90^\circ-\tfrac12\delta'$. Since $\delta=\delta'$, we get $\theta=\theta'$. Two Saccheri quadrilaterals with congruent summits and congruent summit angles are congruent. So the two quadrilaterals are congruent.

4. Conclusion. $\triangle ABC\sim_{eq}PQBA\cong P'Q'B'A'\sim_{eq}\triangle A'B'C'$, so the two triangles are equivalent. $\blacksquare$

(a) If (T) is a projectivity, then (T^{-1}) is also a projectivity.

Let (T) be a projectivity from a line (l) to a line (m) (or from a projective plane to itself). By definition, a projectivity is a one-to-one and onto map which preserves the cross ratio of four collinear points.

Since (T) is one-to-one and onto, its inverse (T^{-1}) exists. We show that (T^{-1}) also preserves cross ratio.

Take any four collinear points (A',B',C',D') in the range of (T). Since (T) is onto, there exist points (A,B,C,D) in the domain such that [ T(A)=A',\quad T(B)=B',\quad T(C)=C',\quad T(D)=D'. ] Because (T) preserves cross ratio, [ (A',B';C',D')=(T(A),T(B);T(C),T(D))=(A,B;C,D). ] But (A=T^{-1}(A'),\ B=T^{-1}(B'),\ C=T^{-1}(C'),\ D=T^{-1}(D')). Hence [ (A',B';C',D')=(T^{-1}(A'),T^{-1}(B');T^{-1}(C'),T^{-1}(D')). ] Thus (T^{-1}) preserves cross ratio. Also (T^{-1}) is one-to-one and onto. Therefore (T^{-1}) is a projectivity.

Equivalently, if (T) is represented by a nonsingular (3\times 3) matrix (A), then (T^{-1}) is represented by (A^{-1}), which is also nonsingular. Hence (T^{-1}) is a projectivity.

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(b) For any simple polyhedron, prove that

[ V-E+F=2. ]

Let (P) be a simple polyhedron with (V) vertices, (E) edges and (F) faces. Remove one face of (P) and flatten the remaining surface onto a plane. This gives a connected planar graph (G), called a Schlegel diagram. The vertices and edges of (G) are the same as those of (P). The faces of (G) in the plane are the remaining faces of (P) together with the outer face, which corresponds to the removed face. Hence [ V(G)=V,\qquad E(G)=E,\qquad F(G)=F. ] So it is enough to prove that for every connected planar graph, [ V-E+F=2. ]

Now take a spanning tree (T) of (G). A tree with (V) vertices has (V-1) edges and only one face, namely the outer face. Therefore for (T), [ V-E_T+F_T=V-(V-1)+1=2. ] Now add the remaining edges of (G) one by one in the order in which they occur in a planar embedding of (G). Each added edge lies inside one existing face and splits that face into two. Thus, when an edge is added, [ E \text{ increases by }1,\qquad F \text{ increases by }1, ] while (V) remains unchanged. Hence the value of [ V-E+F ] does not change. After all edges of (G) have been added, we obtain the original connected planar graph (G). Therefore [ V-E+F=2. ] Since (G) has the same (V,E,F) as the polyhedron (P), we get [ \boxed{V-E+F=2}. ] This is Euler’s polyhedron formula.

(a) Let the base be $b$. Then the height is $h=3b-12$, and area $=bh=288$: $b(3b-12)=288\;\Rightarrow\;3b^2-12b-288=0\;\Rightarrow\;b^2-4b-96=0\;\Rightarrow\;(b-12)(b+8)=0.$ A length is positive, so $b=12$ units and $h=3(12)-12=24$ units. Check: $12\times24=288$.

(b) Claim: $V=\tfrac13\,(\text{area of base})\times(\text{height})$.

Triangular pyramid. Take a triangular prism $ABC\text{-}DEF$ with base $ABC$, top $DEF$ and height $h$. Its volume is (base area)$\times h$. Cut it by the planes $DBC$ and $DBF$ into three tetrahedra: $D\text{-}ABC$, $D\text{-}BCF$ and $D\text{-}BEF$.

• $D\text{-}BEF$ is the same solid as $B\text{-}DEF$. Its base $DEF$ is congruent to $ABC$ and its height is $h$, so its volume equals that of $D\text{-}ABC$ (base $ABC$, height $h$).
• $D\text{-}BCF$ and $D\text{-}BEF$ have the same apex $D$ and bases $BCF$, $BEF$ that are congruent halves of the parallelogram $BCFE$, so their volumes are equal.

So all three tetrahedra have equal volume, and each is one third of the prism: $V=\tfrac13\,(\text{base area})\times h.$

General pyramid. Divide the base polygon into triangles by diagonals from one vertex. The pyramid splits into triangular pyramids with the same apex and height $h$. Adding their volumes, $V=\tfrac13h\,(\text{sum of triangle areas})=\tfrac13\,(\text{base area})\times h.\qquad\blacksquare$