How to "coordinate bash" a geometry problem?

Since you wanted a couple of examples, here are two problems where "coordinate bashing" can be used. Note: Both of these problems were pulled from the 2004 Harvard-MIT Math Tournament.

Example 1: $\Delta ABC$ has side lengths $AB = 8$, $AC = 15$, and $BC = 17$. If $D$, $E$, and $F$ are the circumcenter, centroid, and incenter respectively of $\Delta ABC$, find the area of $\Delta DEF$.

Since $\Delta ABC$ is an $8$-$15$-$17$ right triangle, we can place it on the coordinate plane. Let $A = (0,0)$, $B = (8,0)$, and $C = (0,15)$. (Check for yourself that $AB = 8$, $AC = 15$, $BC = 17$.)

The circumcenter of a triangle is the center of the circle which passes through the vertices of a triangle. In a right triangle, the circumcenter is the midpoint of the hypotenuse. The hypotenuse $BC$ has endpoints $B = (8,0)$ and $C = (0,15)$, the midpoint is $D = (\tfrac{8+0}{2},\tfrac{0+15}{2}) = (4,\tfrac{15}{2})$.

The centroid of a triangle is the intersection of the medians of the triangle. This sounds hard to do with coordinates, but as it turns out, the centroid of a triangle is simply the average of the coordinates of the three vertices. So, $E = (\tfrac{0+8+0}{3},\tfrac{0+0+15}{3}) = (\tfrac{8}{3},5)$.

The incenter of a triangle is the center of the inscribed circle. The area of $\Delta ABC$ is $K := \tfrac{1}{2} \cdot 8 \cdot 15 = 60$, and the semi-perimeter is $s := \tfrac{1}{2}(8+15+17) = 20$. Therefore, the radius of the inscribed circle is $r = \frac{K}{s} = \frac{60}{20} = 3$. So, the distance from $F$ to each of the three sides is $3$ units. Since $AB$ lies on the $x$-axis ($y = 0$), $F$ must lie on one of the lines $y = \pm 3$. Since $AC$ lies on the $y$-axis ($x = 0$), $F$ must lie on one of the lines $x = \pm 3$. Since $F$ is inside $\Delta ABC$, $F = (3,3)$.

Now that we have the coordinates of $D$, $E$, and $F$, we can simply use the Shoelace formula to find that the area of $\Delta DEF$ is $\tfrac{1}{2}\left|4 \cdot 5 + \tfrac{8}{3} \cdot 3 + 3 \cdot \tfrac{15}{2} - \tfrac{8}{3} \cdot \tfrac{15}{2} - 3 \cdot 5 - 4 \cdot 3 \right| = \tfrac{7}{4}$.

Example 2: A tetrahedron has all its faces triangles with sides $13$, $14$, $15$. What is its volume?

The volume of a tetrahedron is $\tfrac{1}{3}Bh$ where $B$ is the area of the base and $h$ is the height. If you know that a $13$-$14$-$15$ triangle is a $5$-$12$-$13$ right triangle glued to a $9$-$12$-$15$ right triangle, then it is easy to get that the area of each face is $\tfrac{1}{2} \cdot 14 \cdot 12 = 84$. But how to find the height? There is a non-coordinate solution on page 14 of their solutions, but here is a coordinate bash solution:

Let $A,B,C,D$ be the vertices of the tetrahedron with $AB = CD = 13$, $AC = BD = 14$, and $BC = AD = 15$. Orient the tetrahedron such that $A = (-5,0,0)$, $B = (0,12,0)$, $C = (9,0,0)$, and $D = (x,y,z)$. (Check for yourself that $AB = 13$, $AC = 14$, $BC = 15$.) Since the base $\Delta ABC$ lies in the $xy$-plane, the height from $D$ to $\Delta ABC$ is simply $z$.

To solve for $z$, we will need to use the facts that $CD = 13$, $BD = 14$, and $AD = 15$, along with the distance formula: $$CD^2 = (x-9)^2+y^2+z^2 = 13^2$$ $$BD^2 = x^2+(y-12)^2+z^2 = 14^2$$ $$AB^2 = (x+5)^2+y^2+z^2 = 15^2$$ Subtracting the first equation from the third gives $28x-56 = 56$, and so, $x = 4$. Substitute this back in to get $y^2+z^2 = 144$ and $(y-12)^2+z^2 = 180$. Taking the difference between these equations yields $24y-144=-36$, and so, $y = \tfrac{9}{2}$. Substituting this back in and solving for $z$ gives $z = \pm \tfrac{3\sqrt{55}}{2}$. So the height of the tetrahedron is $h = |z| = \tfrac{3\sqrt{55}}{2}$, and thus, the volume is $\tfrac{1}{3}Bh = \tfrac{1}{3} \cdot 84 \cdot \tfrac{3\sqrt{55}}{2} = 42\sqrt{55}$.

Basically, in order to prove a geometric theorem, you just put the situation on a Euclidean plane, and do the algebra. For example, here's a proof of the law of cosines:

Say we have a triangle with vertices at $A=(0,0), B=(0, b),$ and $C=(c_1, c_2)$, where $b,c_1 > 0$. Note that this case is general, since you can always rotate/translate a triangle so that it lies on the right half of the plane with any vertex at the origin, and a side lying on the positive $y$-axis. The angle at $A$ is just $\theta = \pi/2 - \tan^{-1}(c_2/c_1)$. Then $\cos(\theta) = \frac{c_2/c_1}{\sqrt{1+(c_2/c_1)^2}}$. Then we have $$\overline{AB}^2+\overline{AC}^2-2\overline{AB}\overline{AC}\cos(\theta) = $$ $$b^2+c_1^2+c_2^2-2b\sqrt{c_1^2+c_2^2}\cdot\frac{c_2}{\sqrt{c_1^2+c_2^2}} = $$ $$(c_2-b)^2+c_1^2 = \overline{BC}^2$$

Notation: Let $\triangle (PQR)$ denote the area of triangle PQR. Let $|ST|$ denote the length of the line segment $ST.$

Let $A,B,C$ be the vertices of a triangle. Let $C'$ be a point on $AB$ and $B'$ be a point on $AC,$ with $C'$ strictly between $A,B,$ and $B'$ strictly between $A,C.$

Let $0<x<1$. Let $C''$ lie on $C'C$ and let $B''$ lie on $B'B$ such that $$|C'C''|=x|C'C| \quad \text { and }\quad |B'B''|=(1-x)|B'B|.$$ Theorem: The ratio of $\triangle (AB''C'')$ to the area of the quadrilateral $BC'B'C$ is $x(1-x).$

Proof: Choose orthogonal co-ordinate axes with $A=(0,0).$ Let $p\times q$ denote the scalar outer product. That is, $(a,b)\times (c,d)=ad-bc.$ We use the following tools:

(1). The area of any triangle $APQ$ is $\frac {1}{2}|P\times Q|.$

(2). The basic rules for the outer product: (i).$ u\times v=-(v\times u) .$ (ii). $u\times u=0.$ (iii). For real $r,s$ and vectors $u,v$ we have $(ru)\times v=u\times rv=r(u\times v),$ $(rs)(u\times v)=r((su)\times v)=(rsu\times v),$ and $(r+s)(u\times v)=(ru)\times v +(su)\times v.$

Now let $B'=y C$ and $C'= z B$ with $y,z \in (0,1).$ Then $\triangle (AB'C')=\frac {1}{2}yz |B\times C| =yz \triangle (ABC).$ The area of quadrilateral $BC'B'C$ is therefore $$\triangle (ABC)-\triangle (AB'C')=(1-yz)\triangle (ABC).$$

We have $B''=xB'+(1-x)B$ and $C''=(1-x)C'+xC.$ Plug the values $B'=yC$ and $C'=zB$ into these and compute $\triangle (AB''C'')=\frac {1}{2}|B''\times C''|.$ The calculation effortlessly simplifies to $$\frac {1}{2}x(1-x)(1-yz)|B\times C|=x(1-x)\cdot (1-yz)\triangle (ABC).$$

For a proof without co-ordinates, prove that the line thru $C''$ parallel to $AB,$ and the line thru $B''$ parallel to $AC$ both meet the segment $BC$ in a common point $D.$ Subdivide $ABC$ into 6 triangles,one of which is $AB''C"$ and one is $B''C''D,$ compute the areas of each of the 5 triangles other than AB''C'' and subtact their total from $\triangle (ABC)$ to find $\triangle (AB''C'').$ A complication is that, reading the vertices of triangle $AB''C''$ counter-clockwise from $A,$ they may be in that order, or in the order $A,C'',B''. $