How do you perform "coordinate bashing"? I read through this which gave some useful information, but I still don't know how to perform it. What I know about coordinate bashing:

  • Coordinate bashing is assigning geometrical figures points on the coordinate plane.
  • It involves using formulas such as the distance formula, slope formula, shoelace theorem, point to line distance, etc.

Can someone provide an illustrated example or two?

  • 1
    $\begingroup$ it obviously depends on the problem, sometimes you can also put the vertices into the complex plane, for example when you want to check if a quad is cyclic or something. $\endgroup$
    – Asinomas
    Dec 28 '16 at 6:17
  • 1
    $\begingroup$ It seems to me like all they're saying is give coordinates (possibly generic) to the shapes involved. If my understanding is correct, you've probably done this before and just have never heard it by that name. $\endgroup$
    – Dair
    Dec 28 '16 at 6:50
  • 3
    $\begingroup$ Note that "bashing" is not a technical term. It is just a rather informal catch-all word for many different techniques where all they have in common is that they are based on symbolic reasoning about coordinates rather than on diagrams and visual intuition. It is a slightly pejorative term, suggesting that even though a coordinate-bashing proof produces a correct result, it would have been nice to have an argument that produces more understanding of why it is so. The neutral term for the same techniques is analytic geometry. $\endgroup$ Dec 28 '16 at 8:56
  • $\begingroup$ I hate that term. More often than not, it takes insight to be able to choose coordinates and arrangements to give an illuminating answer. $\endgroup$ Dec 29 '16 at 4:16

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 $\tfrac{1}{3}Bh = \tfrac{1}{3} \cdot 84 \cdot \tfrac{3\sqrt{55}}{2} = 42\sqrt{55}$.

  • $\begingroup$ Excellent, thanks! $\endgroup$
    – suomynonA
    Dec 29 '16 at 4:06

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 $$\overline{AB}^2+\overline{AC}^2-2\overline{AB}\overline{AC}\cos(\theta) = $$ $$b^2+c_1^2+c_2^2-2b\sqrt{c_1^2+c_2^2}\cdot\frac{c_2}{\sqrt{c_1^2+c_2^2}} = $$ $$(c_2-b)^2+c_1^2 = \overline{BC}^2$$

  • 2
    $\begingroup$ Some questions don't respond well to co-ordinates and some others become surprisingly simple. A very early computer program for manipulating and simplifying elementary algebraic expressions was tested (successfully) by having it prove the existence of the incenters of triangles, using co-ordinates only, and no trigonometry. Don't try it by hand unless you're not busy for a couple of months. $\endgroup$ Dec 29 '16 at 7:32

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''. $


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