Coordinate & Solid Geometry

1.Distance, midpoints, slopes, and lines

Coordinates turn a picture into arithmetic. Every point of the plane is a pair $(x,y)$ measured from two perpendicular axes, so every geometric question about points can be restated as a question about those pairs. The most basic bridge is the distance formula. If $P=(x_1,y_1)$ and $Q=(x_2,y_2)$, draw the horizontal and vertical segments that complete $PQ$ to a right triangle. Its legs have lengths $|x_2-x_1|$ and $|y_2-y_1|$, so the Pythagorean theorem gives the length of the hypotenuse $PQ$.

Stated precisely, $PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. The absolute values disappear because they are squared, so the order of the two endpoints never matters. The formula needs no special case for tilted segments or for segments running through any quadrant: the signs of the legs are irrelevant once they are squared. This is why one formula handles every position of the segment, and why it is the first tool to reach for whenever a figure is placed in coordinates.

The midpoint $M$ of $PQ$ is the point whose coordinates are the averages, $M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$. To justify this, note that $x_M-x_1=\frac{x_2-x_1}{2}=x_2-x_M$, so $M$ sits exactly halfway in the horizontal direction, and the same computation works vertically. A useful consequence: if you know one endpoint and the midpoint, the other endpoint is $2M-P$, computed coordinate by coordinate.

Slope measures steepness: $m=\frac{y_2-y_1}{x_2-x_1}$ when $x_2\ne x_1$. Any two points on the same line give the same slope, because the right triangles formed by the rise and the run are similar. A line can be written in slope-intercept form $y=mx+b$, in point-slope form $y-y_1=m(x-x_1)$, or in general form $Ax+By+C=0$. The general form is the safe choice when the line might be vertical: a vertical line has no slope, but it has the equation $x=c$.

Perpendicularity is the most useful relation among slopes, and it is really a dot product fact. Direction vectors of the two lines are $(1,m_1)$ and $(1,m_2)$, and the lines are perpendicular exactly when these vectors have zero dot product, which is $1+m_1m_2=0$. So $m_1m_2=-1$. The same idea works without slopes: the line $Ax+By=C$ has normal vector $(A,B)$, and two lines are perpendicular exactly when their normal vectors are perpendicular.

Contest problems are often easier after a smart choice of axes. If the figure has a right angle, put that vertex at the origin with the legs on the axes. If it has a symmetry axis, make that axis a coordinate axis. Then the distance and slope formulas become short, and the answer usually falls out after one substitution. Choosing coordinates well is a skill that pays off in every later section of this topic.

• d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}
• M=\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)
• m=\frac{y_2-y_1}{x_2-x_1},\qquad y-y_1=m(x-x_1)
• \ell_1\perp\ell_2 \iff m_1m_2=-1 \text{ (when neither line is vertical)}

• Dividing by $x_2-x_1$ when the segment is vertical. A vertical line has no slope; write its equation as $x=c$ instead.
• Dropping the square on one term, or taking the square root of only part of the sum. The distance is the square root of the whole sum of squares.
• Treating $\tfrac12$ as the perpendicular slope of $2$. The perpendicular slope is the negative reciprocal, $-\tfrac12$, and the sign matters.

2.Shoelace formula and areas of polygons

Heights and bases are awkward once a triangle is tilted, so coordinates need their own area tool. The shoelace formula gives the area of any simple polygon from its vertices alone. Label the vertices $(x_1,y_1),\dots,(x_n,y_n)$ in order around the boundary, and set $(x_{n+1},y_{n+1})=(x_1,y_1)$. Then $\text{Area}=\tfrac12\left|\sum_{i=1}^{n}(x_i y_{i+1}-x_{i+1}y_i)\right|$.

Why does this work? Take one edge $P_iP_{i+1}$ and drop perpendiculars from its endpoints to the $x$-axis. The edge, the two perpendiculars and the axis enclose a trapezoid whose signed area is $\tfrac12(x_{i+1}-x_i)(y_i+y_{i+1})$. The sign is positive when the edge runs to the right and negative when it runs to the left. Summing over all edges, the signed trapezoid areas add up to the signed area enclosed by the boundary, and the absolute value removes the dependence on orientation.

Now expand the trapezoid term: $(x_{i+1}-x_i)(y_i+y_{i+1})=x_{i+1}y_i+x_{i+1}y_{i+1}-x_iy_i-x_iy_{i+1}$. Sum over the cycle. The terms $x_{i+1}y_{i+1}$ and $x_iy_i$ each appear once with a plus sign and once with a minus sign, so they cancel. What remains is $\sum(x_{i+1}y_i-x_iy_{i+1})$, which is exactly the negative of $\sum(x_iy_{i+1}-x_{i+1}y_i)$. Multiplying by $\tfrac12$ gives the formula.

The triangle case is worth memorizing in its determinant form. For $A=(x_1,y_1)$, $B=(x_2,y_2)$, $C=(x_3,y_3)$, the area is $\tfrac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|$. If one vertex is the origin, this collapses to $\tfrac12|x_1y_2-x_2y_1|$, which is half the magnitude of a 2D cross product. The same quantity appears in the vector form of the area of a parallelogram, so the shoelace formula is a sum of parallelogram areas.

The formula gives quick comparisons as well as computations. Two triangles with the same base line and equal heights have determinants of equal absolute value, and this is visible at once from the coordinates. For lattice polygons (vertices with integer coordinates) every term is an integer, so the area is a multiple of $\tfrac12$. This observation is the first step of Pick's theorem, which you may meet in later problems.

Practical rule: list the vertices in order around the polygon, write the first vertex again at the end, multiply each $x$ by the next $y$ and subtract, then halve the absolute value of the total. The most common error is to list the vertices out of order, for example jumping across a diagonal. That traces a different (often self-intersecting) polygon, and the number you get is the area of the wrong figure.

• \text{Area}=\tfrac12\left|\sum_{i=1}^{n}(x_iy_{i+1}-x_{i+1}y_i)\right|
• \text{Area}(OAB)=\tfrac12|x_1y_2-x_2y_1|
• \text{Area}(ABC)=\tfrac12\left|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\right|

• Forgetting to close the polygon by repeating the first vertex. The final product $x_ny_1-x_1y_n$ is then missing from the sum.
• Listing the vertices out of order, which traces a different polygon and gives the area of the wrong figure.
• Omitting the absolute value and reporting a negative area for a clockwise listing.

3.Circles, tangents, and chords

A circle with center $(a,b)$ and radius $r$ is the set of points at distance $r$ from the center, so its equation is $(x-a)^2+(y-b)^2=r^2$. This is just the distance formula set equal to $r$. When the equation is given in expanded form $x^2+y^2+Dx+Ey+F=0$, complete the square in each variable: $\left(x+\tfrac D2\right)^2+\left(y+\tfrac E2\right)^2=\tfrac{D^2+E^2}{4}-F$. The center is then $\left(-\tfrac D2,-\tfrac E2\right)$ and the radius is read off directly. The condition $\tfrac{D^2+E^2}{4}-F>0$ is exactly what makes the set a genuine circle, rather than a single point or the empty set.

Tangency is a perpendicularity statement. A line touches a circle at a point $P$ exactly when it is perpendicular to the radius $OP$. In vector language, a point $X$ lies on the tangent at $P$ when $(X-O)\cdot(P-O)=r^2$. Indeed $X-P$ is perpendicular to $P-O$, so $(X-P)\cdot(P-O)=0$, which rearranges to $(X-O)\cdot(P-O)=(P-O)\cdot(P-O)=r^2$. For the circle $x^2+y^2=r^2$, the tangent at $(x_0,y_0)$ is therefore $x_0x+y_0y=r^2$.

For a line $\ell$ and a circle with center $O$, let $d$ be the distance from $O$ to $\ell$. If $d>r$ the line misses the circle; if $d=r$ it is tangent; if $d<r$ it cuts a chord $AB$. The foot $H$ of the perpendicular from $O$ bisects the chord, since $OA=OB=r$. Pythagoras in the right triangle $OHA$ gives $HA^2=r^2-d^2$, so the chord has length $2\sqrt{r^2-d^2}$. This is much faster than solving a quadratic and is the standard tool for chord problems.

The distance from a point $(x_0,y_0)$ to a line $Ax+By+C=0$ is $\dfrac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$. To derive it, pick any point $Q=(x_1,y_1)$ on the line, so $Ax_1+By_1+C=0$. The distance is the length of the projection of $\vec{QP}$ onto the unit normal $\frac{(A,B)}{\sqrt{A^2+B^2}}$, and the dot product $(A,B)\cdot(P-Q)$ equals $Ax_0+By_0+C$ after the $Ax_1+By_1$ term is replaced using the line's equation.

Power of a point: from an external point $P$ with center $O$, the tangent segment to a circle of radius $r$ has length $t$ with $t^2=|PO|^2-r^2$. The proof is Pythagoras in the right triangle formed by $P$, $O$ and a point of tangency $T$, since $OT\perp PT$. This is the cleanest route to a tangent length, and it works without ever finding $T$.

Two circles meet along a radical line. Subtracting their equations, the squared terms cancel, and the result is linear, so the common chord lies on a straight line. This is a useful reduction: to intersect two circles, first subtract the equations to get a line, then intersect that line with one of the circles. Many problems that look like a system of two quadratics become a single quadratic this way.

• (x-a)^2+(y-b)^2=r^2 \iff x^2+y^2+Dx+Ey+F=0,\ \text{center }\left(-\tfrac D2,-\tfrac E2\right),\ r^2=\tfrac{D^2+E^2}{4}-F
• \text{Tangent at }(x_0,y_0)\text{ on }x^2+y^2=r^2:\quad x_0x+y_0y=r^2
• \text{Chord length}=2\sqrt{r^2-d^2}
• t^2=|PO|^2-r^2\ \text{for a tangent segment from }P

• Reading the center as $(D,E)$ rather than $\left(-\tfrac D2,-\tfrac E2\right)$. The sign and the halving are the two usual slips.
• Forgetting that the constant on the right side is $r^2$, not $r$, so the radius is the square root of that constant.
• Solving a tangency condition with a quadratic and forgetting that the discriminant must be zero. A tangency condition is an equation with specific solutions, not an inequality.

4.Ellipses and parabolas: the basics

Conics are most useful in contests through their equations and defining properties. A parabola is the set of points equidistant from a fixed point, the focus, and a fixed line, the directrix. Take focus $F=(0,p)$ and directrix $y=-p$. A point $(x,y)$ lies on the parabola when $x^2+(y-p)^2=(y+p)^2$. Expanding, $x^2-2py+p^2=2py+p^2$, so $x^2=4py$. The factor $4p$ is the thing to remember: the focus sits $p$ units from the vertex, and the latus rectum (the chord through the focus perpendicular to the axis) has length $4p$.

An ellipse is the set of points whose distances to two foci sum to a constant. Let the foci be $(\pm c,0)$ and let $2a$ be the constant sum. Testing the point $(0,b)$ on the minor axis gives $2\sqrt{b^2+c^2}=2a$, so $a^2=b^2+c^2$. Squaring out the general condition and using this relation gives the standard form $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $c^2=a^2-b^2$. The major axis has length $2a$, the minor axis has length $2b$, and each focus is $c$ from the center.

Affine scaling is the key to ellipse problems. The map $(u,v)\mapsto(au,bv)$ sends the unit circle $u^2+v^2=1$ onto the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, and it multiplies every area by $ab$. Since the unit circle has area $\pi$, the ellipse has area $\pi ab$. The same map preserves midpoints, parallel lines and tangency, so an ellipse question can be pulled back to a circle question, solved there, and scaled back at the end.

Tangent lines to conics come from a discriminant condition. For the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, substitute $y=mx+k$ and multiply by $a^2b^2$ to get $(b^2+a^2m^2)x^2+2a^2mkx+a^2(k^2-b^2)=0$. The line is tangent exactly when the discriminant vanishes, and that simplifies to $k^2=a^2m^2+b^2$. This formula is worth memorizing, because it turns a tangency problem into a single equation in $m$ and $k$.

The parabola has the same kind of tangent test. Substituting $y=mx+k$ into $x^2=4py$ gives $x^2-4pmx-4pk=0$, whose discriminant vanishes exactly when $k=-pm^2$. For the circle $x^2+y^2=r^2$ the tangency test is $|k|/\sqrt{1+m^2}=r$, which is the distance formula in disguise. Practicing all three tests side by side makes the pattern clear: tangency is always a discriminant equal to zero.

On the AMC, conic facts are usually used lightly: identify the center, read off $a$, $b$, $c$ or the focus, and compute a distance or an area. Watch for two classic traps. For an ellipse, $c^2=a^2-b^2$ (the relation $a^2+b^2$ belongs to hyperbolas). For $x^2=4py$ the focus is $(0,p)$, not $(0,\tfrac p4)$ and not $(0,4p)$.

• x^2=4py \text{ has focus }(0,p)\text{ and directrix }y=-p
• \frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad c^2=a^2-b^2,\qquad \text{Area}=\pi ab
• \text{Sum of distances from a point of the ellipse to the foci}=2a
• y=mx+k\text{ tangent to the ellipse}\iff k^2=a^2m^2+b^2

• Using $c^2=a^2+b^2$, the hyperbola relation, for an ellipse. The ellipse has $c^2=a^2-b^2$ with $a$ the larger semi-axis.
• Placing the focus of $x^2=4py$ at $(0,4p)$ or $(0,\tfrac p4)$. The focus is $p$ from the vertex.
• Computing an ellipse area as $\pi(a+b)$. The scaling factor is the product $ab$.

5.Reflections, rotations, and transformations

A rigid motion preserves distances, and every such motion of the plane is a composition of translations, rotations and reflections. Reflections are the most used in contests, because they swap the two sides of a line without changing any distance. Reflecting across the $x$-axis sends $(x,y)\mapsto(x,-y)$; across the $y$-axis, $(x,y)\mapsto(-x,y)$; across $y=x$, $(x,y)\mapsto(y,x)$; across $y=-x$, $(x,y)\mapsto(-y,-x)$. A rotation by $90^\circ$ counterclockwise about the origin is $(x,y)\mapsto(-y,x)$.

To reflect a point across a general line $\ell:Ax+By+C=0$, move from $P=(x_0,y_0)$ along the normal direction $(A,B)$ to the foot of the perpendicular, then continue the same distance. The image is $P'=P-\dfrac{2(Ax_0+By_0+C)}{A^2+B^2}(A,B)$. Two checks confirm the formula: a point on $\ell$ is fixed, since its value $Ax_0+By_0+C$ is zero, and the midpoint of $PP'$ lies on $\ell$ by construction.

Why do reflections help with distances? Suppose $A$ and $B$ lie on the same side of a line $\ell$, and a point $P$ moves along $\ell$. Reflect $B$ across $\ell$ to get $B'$. Because $P$ is fixed by the reflection, $PB=PB'$ for every $P$ on $\ell$. So $PA+PB=PA+PB'\ge AB'$ by the triangle inequality, with equality exactly when $P$ lies on the segment $AB'$, where that segment crosses $\ell$. This is the light-ray argument: the angle of incidence equals the angle of reflection.

Composition is order-sensitive. A translation followed by a reflection is not the same map as the reflection followed by the translation, because the reflection line moves relative to the translated figure. A reliable habit is to write each map as a formula, apply them in the order described in the problem, and substitute the output of one into the next. Do not combine two maps mentally unless you have checked the composition on a test point.

Transformations also control areas. Rigid motions preserve area. A dilation by factor $k$ about any center multiplies area by $k^2$. More generally, the linear map $(x,y)\mapsto(ax+by,\,cx+dy)$ multiplies every area by $|ad-bc|$, the absolute value of the determinant. This is the same determinant that appears in the shoelace formula and in the ellipse scaling argument, so these are two faces of one idea.

Symmetry is the cheapest transformation of all. If a figure is symmetric across a line, reflecting one half generates the other, and the total area is twice the area of a half. Before coordinatizing a problem, look for such symmetry, and make the axis of symmetry a coordinate axis; this often removes half of the unknowns at once.

• \text{Across }y=x:\ (x,y)\mapsto(y,x);\qquad\text{across }y=-x:\ (x,y)\mapsto(-y,-x)
• P'=P-\dfrac{2(Ax_0+By_0+C)}{A^2+B^2}(A,B)\ \text{for the line }Ax+By+C=0
• |PA|+|PB|\ge|AB'|\ \text{for }P\text{ on }\ell,\ \text{where }B'\text{ is the reflection of }B
• (x,y)\mapsto(ax+by,cx+dy)\ \text{multiplies areas by }|ad-bc|

• Reflecting when the two points are already on opposite sides of the line. Then the segment joining them already crosses the line, and the minimum is simply their distance.
• Composing the maps in the wrong order, for instance translating first when the problem says to reflect first.
• Using $(x,y)\mapsto(x,-y)$ for reflection across $y=x$. Reflection across $y=x$ swaps the two coordinates.

6.Loci: sets of points with a metric condition

A locus is the set of all points satisfying a geometric condition, and in coordinates it is the solution set of an equation. The most important loci come from comparing distances. For fixed points $A=(a_1,a_2)$ and $B=(b_1,b_2)$, the condition $|PA|=|PB|$ means $|PA|^2=|PB|^2$. Expanding both sides, the terms $x^2$ and $y^2$ cancel, leaving a linear equation. So the locus is a line, and it is the perpendicular bisector of $AB$: it passes through the midpoint and is perpendicular to $AB$.

The same cancellation works for any condition of the form $|PA|^2-|PB|^2=k$. Expanding gives $-2(a_1-b_1)x-2(a_2-b_2)y+(|A|^2-|B|^2)=k$, which is again a line perpendicular to $AB$. So a difference of squared distances produces a line, and this gives a clean method for proving that several lines meet at one point: impose two such conditions and intersect the resulting lines.

The ratio condition $|PA|=k|PB|$ with $k\ne1$ gives a circle, the Apollonius circle. Squaring, $(x-a_1)^2+(y-a_2)^2=k^2\big((x-b_1)^2+(y-b_2)^2\big)$. The coefficient of $x^2+y^2$ is $1-k^2\ne0$. Dividing by it puts the equation in the form $x^2+y^2+Dx+Ey+F=0$, which is a circle. By symmetry its center lies on the line $AB$, and the circle is the complete locus of points with the given ratio.

The angle condition is a dot product condition, and it has a beautiful identity. For a variable point $X$ and fixed points $A$, $B$ with midpoint $M$, one has $(X-A)\cdot(X-B)=|X-M|^2-\tfrac14|A-B|^2$. To see this, write $X-M=u$ and $A-M=w$, so $B-M=-w$, and then $(X-A)\cdot(X-B)=(u-w)\cdot(u+w)=|u|^2-|w|^2$. Therefore the points with $(X-A)\cdot(X-B)=0$, which are the points seeing $AB$ at a right angle, form the circle with diameter $AB$. This is Thales' theorem written as an identity.

Once a locus is a line or a circle, extremal questions become geometry. The nearest point of a circle with center $O$ and radius $r$ to an outside point $P$ lies on segment $OP$, at distance $|OP|-r$ from $P$. The farthest point is at distance $|OP|+r$. Many contest questions ask for the least or greatest distance from a fixed point to a locus, and they reduce to exactly this computation.

Recognizing the locus is the skill. Equal distances give lines. Fixed distance ratios give circles. Right angles give circles on a diameter. A fixed sum or difference of distances to two points gives an ellipse or a hyperbola, which is the subject of the conic section. A problem that mentions a constant sum of distances is usually steering you toward an ellipse, so identify the foci first.

• |PA|=|PB| \iff P\text{ lies on the perpendicular bisector of }AB
• |PA|=k\,|PB|\ (k\ne1)\ \text{is a circle (Apollonius)}
• (X-A)\cdot(X-B)=|X-M|^2-\tfrac14|AB|^2,\quad M=\text{midpoint of }AB
• |PA|^2-|PB|^2=\text{constant}\ \text{is a line perpendicular to }AB

• Squaring only one side of $|PA|=|PB|$, which drops a term and gives a wrong line.
• Calling every ratio locus a circle. The case $k=1$ gives a line, not a circle.
• Assuming a constant sum of distances gives a circle. It gives an ellipse with foci $A$ and $B$; it is a circle only in the degenerate case $A=B$.

7.Solids: volumes, surface areas, and Cavalieri

Solids on the AMC are mostly prisms, pyramids, cones, cylinders and spheres, and the questions ask for a volume, a surface area, or a ratio between two solids. Almost every volume formula in this family follows from one principle, Cavalieri's principle. If two solids sit between the same pair of parallel planes, and every plane parallel to them cuts the two solids in regions of equal area, then the solids have equal volume. Cavalieri lets you compare volumes by comparing slices, without any direct computation.

Prisms and cylinders have volume $Bh$, where $B$ is the base area and $h$ the height. Pyramids and cones have volume $\tfrac13Bh$, and the factor $\tfrac13$ is the main thing to remember. One way to see it: slice a pyramid with base area $B$ and height $h$ by planes parallel to the base. At height $t$ the slice is similar to the base with ratio $1-\tfrac th$, so its area is $B\left(1-\tfrac th\right)^2$. Adding up these slices gives $\int_0^h B(1-\tfrac th)^2\,dt=\tfrac13Bh$. A calculus-free route to the same factor is the classical dissection of a triangular prism into three tetrahedra of equal volume.

Cones are the same argument with circles. A cone with base radius $r$ and height $h$ has base area $\pi r^2$, so $V=\tfrac13\pi r^2h$. Its lateral surface area is $\pi r\ell$, where $\ell=\sqrt{r^2+h^2}$ is the slant height. A common error is to use $\ell$ in the volume formula. The volume always uses the perpendicular height $h$, while the slant height enters only surface area.

Spheres follow from Archimedes' comparison. Let the hemisphere be $x^2+y^2+z^2\le r^2$ with $z\ge0$. At height $z$ its cross-section is a disk of area $\pi(r^2-z^2)$. Inside the cylinder $x^2+y^2\le r^2$, $0\le z\le r$, remove the cone $x^2+y^2\le z^2$ with apex at the origin. At height $z$ what remains is a ring of area $\pi r^2-\pi z^2=\pi(r^2-z^2)$. The slices agree, so by Cavalieri the hemisphere has volume $\pi r^3-\tfrac13\pi r^3=\tfrac23\pi r^3$, and the sphere has volume $\tfrac43\pi r^3$. Differentiating this volume with respect to $r$ gives the surface area $4\pi r^2$.

Similarity sets the scaling rules. If a solid is scaled by a factor $k$ in every direction, its areas multiply by $k^2$ and its volumes by $k^3$. This explains many contest ratios. A plane parallel to the base of a pyramid, located one-third of the way from the apex to the base, cuts off a pyramid whose volume is $\tfrac1{27}$ of the whole. The remaining piece, called a frustum, has the rest of the volume.

A practical plan: identify the solid, find its base and its perpendicular height (not the slant height), and then compute. For a cone or sphere, look for the axial triangle or the circle through the axis, because two-dimensional similarity usually supplies the missing radius or height.

• V_{\text{prism}}=Bh,\qquad V_{\text{pyramid}}=\tfrac13Bh,\qquad V_{\text{cone}}=\tfrac13\pi r^2h
• V_{\text{sphere}}=\tfrac43\pi r^3,\qquad S_{\text{sphere}}=4\pi r^2
• S_{\text{cone, lateral}}=\pi r\ell,\qquad \ell=\sqrt{r^2+h^2}
• \text{Scaling every length by }k:\ \text{areas}\times k^2,\ \text{volumes}\times k^3

• Using the slant height $\ell$ in the cone volume formula instead of the perpendicular height $h$.
• Dropping the factor $\tfrac13$ for pyramids and cones, or copying the $\tfrac12$ from the triangle area formula.
• Confusing the scaling of area ($k^2$) with the scaling of volume ($k^3$) when a solid is enlarged.

8.Cross-sections and inscribed solids

A cross-section is the intersection of a solid with a plane, and most contest questions about solids are really questions about cross-sections, because a section reduces a three-dimensional situation to a two-dimensional one. A plane at distance $d$ from the center of a sphere of radius $R$ cuts a circle of radius $\sqrt{R^2-d^2}$. This is Pythagoras in the right triangle formed by the center, the foot of the perpendicular to the plane, and any point of the circle, so the section has area $\pi(R^2-d^2)$. Note that the radius is not $R-d$.

A plane parallel to the base of a cone or pyramid cuts a similar copy of the base. If the plane is at distance $t$ from the apex and the height is $h$, the section is scaled by $t/h$. For a cone, the sections are circles of radius $\tfrac{t}{h}r$. An inscribed sphere, which touches the base and the lateral surface, is found in the axial cross-section: a triangle whose incircle is the great circle of the sphere. Its radius is the inradius of that triangle.

Sections through a cube are polygons with vertices on its edges. A plane through three vertices gives a triangle, and a plane through two opposite edges gives a rectangle. A plane through the midpoints of six edges gives a regular hexagon. Concretely, take the cube $0\le x,y,z\le2$ and the plane $x+y+z=3$. It contains the six edge midpoints $(2,1,0)$, $(2,0,1)$, $(1,2,0)$, $(0,2,1)$, $(1,0,2)$, $(0,1,2)$, none of the vertices (whose coordinate sums are even numbers other than $3$), and it is perpendicular to the main diagonal. Adjacent midpoints are $\sqrt2$ apart, so the section is a regular hexagon of side $\sqrt2$.

Inscribed and circumscribed solids are the other half of the topic. A sphere inscribed in a cube of edge $s$ has diameter $s$, because it touches opposite faces. A sphere circumscribed about a box with edges $a,b,c$ has diameter equal to the space diagonal $\sqrt{a^2+b^2+c^2}$, which follows from applying Pythagoras twice: the diagonal of the base is $\sqrt{a^2+b^2}$, and it is perpendicular to the vertical edge of length $c$.

For a regular tetrahedron of edge $a$, the incenter is the centroid. The altitude has length $H=a\sqrt{\tfrac23}$, since the circumradius of an equilateral triangle of side $a$ is $\tfrac{a}{\sqrt3}$ and $H^2=a^2-\tfrac{a^2}{3}=\tfrac{2a^2}{3}$. The centroid divides the altitude in the ratio $3:1$ from the apex, so the inradius is $\tfrac H4=\tfrac{a\sqrt6}{12}$. For $a=12$ this is $\sqrt6$.

For a sphere inscribed in a cone or pyramid, draw the axial cross-section and find the incircle of the resulting triangle. The sphere is tangent to the base and to the lateral surface, and each tangency appears in the axial plane as a tangency of the circle to a side. Once the radius is known, the volume follows from $\tfrac43\pi r^3$. This two-step method, section first and then formula, handles almost every inscribed-sphere question.

• \text{Plane at distance }d\text{ from the center of a sphere of radius }R:\ \text{section radius }\sqrt{R^2-d^2}
• \text{Cube of edge }s:\ \text{inscribed sphere radius }\tfrac s2
• \text{Box }a\times b\times c:\ \text{circumscribed sphere diameter }\sqrt{a^2+b^2+c^2}
• \text{Regular tetrahedron of edge }a:\ r_{\text{in}}=\dfrac{a\sqrt6}{12}

• Using the edge length as the sphere radius. The inscribed sphere of a cube of edge $s$ has radius $\tfrac s2$.
• Writing the section radius as $R-d$. The correct value is $\sqrt{R^2-d^2}$, from Pythagoras.
• Naming the section shape without counting the edges the plane crosses. A plane can cut a cube in a triangle, rectangle, pentagon or hexagon.

9.Vectors in 3D and distance to a plane

Three-dimensional coordinates extend the distance formula by one term: $PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$. The Pythagorean theorem is applied twice, once in the base plane to get the horizontal diagonal and once in the vertical plane through that diagonal. Vectors are the natural language here. The displacement from $P$ to $Q$ is $\vec{PQ}=(x_2-x_1,\,y_2-y_1,\,z_2-z_1)$, and its length is its norm, $|\vec{PQ}|=\sqrt{v_1^2+v_2^2+v_3^2}$.

The dot product $u\cdot v=u_1v_1+u_2v_2+u_3v_3$ measures alignment, because $u\cdot v=|u||v|\cos\theta$ where $\theta$ is the angle between the vectors. This follows from the law of cosines in the triangle with sides $u$, $v$ and $u-v$: $|u-v|^2=|u|^2+|v|^2-2\,u\cdot v$, and expanding the left side in coordinates gives the coordinate formula. Two vectors are perpendicular exactly when their dot product is zero, which is the most useful test in three-dimensional problems.

The cross product $u\times v=(u_2v_3-u_3v_2,\ u_3v_1-u_1v_3,\ u_1v_2-u_2v_1)$ is perpendicular to both $u$ and $v$, and its length is $|u\times v|=|u||v|\sin\theta$. That length is the area of the parallelogram spanned by $u$ and $v$, so the area of triangle $PQR$ equals $\tfrac12|\vec{PQ}\times\vec{PR}|$. This is the three-dimensional version of the shoelace idea: an area is a determinant-like quantity in coordinates.

A plane can be written $ax+by+cz+d=0$, and the vector $n=(a,b,c)$ is normal to it: if $P$ and $Q$ both lie on the plane then $n\cdot\vec{PQ}=0$. To find the distance from a point $P_0$ to the plane, take any point $Q$ on the plane and project $\vec{QP_0}$ onto the unit normal. The projection length is $\dfrac{|n\cdot\vec{QP_0}|}{|n|}$. Since $n\cdot Q=-d$, this equals $\dfrac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}$.

The distance from a point $P$ to a line through $A$ with direction $u$ is $\dfrac{|u\times(P-A)|}{|u|}$. The reason is that $|u\times(P-A)|=|u|\cdot|P-A|\sin\theta$, and dividing by the base length $|u|$ leaves exactly the height of the point above the line, which is the distance.

In a three-dimensional problem, place the figure with the most symmetric vertex at the origin and its edges along the axes. Write the coordinates of every relevant point, and then use the distance, dot product and cross product as needed. Avoid working with angles directly: coordinates give exact answers and are less error-prone than trigonometry.

• |\vec{PQ}|=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}
• u\cdot v=u_1v_1+u_2v_2+u_3v_3=|u||v|\cos\theta
• |u\times v|=|u||v|\sin\theta=\text{area of the parallelogram spanned by }u,v
• \operatorname{dist}(P_0,\ ax+by+cz+d=0)=\dfrac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}

• Omitting the square root in the denominator, or using $a^2+b^2+c^2$ where its square root is needed.
• Mishandling the constant term: in $ax+by+cz=k$ the constant is $d=-k$, and it enters the numerator with its sign.
• Confusing the dot product with the cross product. The dot product is a number; the cross product is a vector whose length is a parallelogram area.

Example 1: Midpoint, then distance to the origin

Let $A=(-3,1)$ and $B=(5,7)$, and let $M$ be the midpoint of segment $AB$. What is the distance from the origin to $M$?

Example 2: Quadrilateral by the shoelace formula

The quadrilateral with vertices $(0,0)$, $(4,0)$, $(5,3)$, $(1,4)$, listed in order around its boundary, has what area?

Example 3: Plane section of a sphere

A sphere has radius $5$. A plane cuts it in a circle of area $16\pi$. How far is the plane from the center of the sphere?

Example 4: Sphere inscribed in a cone

A right circular cone has base radius $3$ and height $4$. A sphere is inscribed in the cone, touching the base and the lateral surface. What is the volume of the sphere divided by $\pi$?

Example 5: Chord of a circle

The line $3x+4y+4=0$ meets the circle $(x-1)^2+(y-2)^2=25$ at two points. What is the distance between those two points?

Example 6: Ellipse minus an inner circle

Find the area of the region inside the ellipse $\dfrac{x^2}{4}+y^2=1$ and outside the circle $x^2+y^2=1$.