Slopefessor Fr1day – Page 41 – High School Geometry

Median of a triangle

In a triangle $ABC$ , the median from vertex $A$ is a line segment from $A$ to the midpoint of side $BC$ (red line $AM$ in the diagram below).

Similarly, the median from vertex $B$ is the line segment from $B$ to the midpoint of $AC$ , and the median from vertex $C$ is a line segment from $C$ to the midpoint of $AB$ . The three medians of a triangle always intersect at a point called the centroid.

Since a median is a line segment, we can find its equation, its length, and its midpoint. We can also find the angle the median makes with the opposite side.

Equation of a median

Recall that the equation of a straight line can be given in the form $y=mx+b$ , where $m$ is the slope and $b$ is the $y$ -intercept. So, supposing we want to find the equation of the median $AM$ in the diagram above, then:

we first determine the coordinates of point $M$ using the midpoint formula (since $M$ is the midpoint of the opposite side $BC$ by definition);
$\boxed{\textrm{\textbf{MIDPOINT FORMULA:=}}\Big(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\Big)}$
we find the slope of $AM$ using the slope formula (call this slope $m$ );
$\boxed{\textrm{\textbf{SLOPE FORMULA:=}}\frac{y_2-y_1}{x_2-x_1}}$
we substitute the slope $m$ and the coordinates of $M$ (or $A$ ) into the straight line equation $y=mx+b$ to find $b$ ;
we substitute $b$ back into $y=mx+b$ , which will then be the desired equation.

So long as you’re comfortable with finding equations of lines, you won’t have any issues with equations of medians.

Example 1

$\triangle PQR$ has vertices at $P(-4,0),~Q(6,0),~R(0,4)$ . Find the equation of the median from $R$ .

The median from vertex $R$ is the red line segment $RN$ shown above. Following the steps we outlined previously, we first find the midpoint of $PQ$ ; it is:

$\Big(\frac{-4+6}{2},\frac{0+0}{2}\Big)=(1,0).$

Let’s label this midpoint as $N(1,0)$ . Next, we find the slope of $RN$ ; it is:

$\frac{0-4}{1-0}=\frac{-4}{1}=-4=m,~\textrm{say}.$

We can now write $y=-4x+b$ , and then determine the value of $b$ using the coordinates of $N$ or $R$ . Let’s use $R(0,4)$ :

$\begin{equation*} \begin{split} y&=-4x+b\\ 4&=-4(0)+b,\quad\textrm{since}~x=0,y=4~\textrm{at}~R(0,4)\\ 4&=0+b\\ 4&=b. \end{split} \end{equation}$

Finally, in $y=-4x+b$ , we replace the $b$ by $4$ . Therefore, the desired equation is $y=-4x+4$ . (Notice that we could have saved a little time by observing that the point $R(0,4)$ is on the $y$ -axis, and so $b=4$ follows immediately.)

Example 2

$\triangle PQR$ has vertices at $P(-4,0),~Q(6,0),~R(0,4)$ . Find the equation of the median from $Q$ .

We first find the midpoint of $PR$ ; it is:

$\Big(\frac{0+-4}{2},\frac{4+0}{2}\Big)=(-2,2).$

Let’s label it as $M(-2,2)$ . Next, we find the slope of $QM$ ; it is:

$\frac{2-0}{-2-6}=\frac{2}{-8}=-\frac{1}{4}=m,~\textrm{say}.$

We can then write $y=-\frac{1}{4}x+b$ , and use the coordinates of $Q$ (or $M$ ) to determine the value of $b$ . Let’s use $M(-2,2)$ :

$\begin{equation*} \begin{split} y&=mx+b\\ 2&=-\frac{1}{4}(-2)+b\\ 2&=\frac{2}{4}+b\\ 2&=\frac{1}{2}+b\\ b&=2-\frac{1}{2}\\ b&=\frac{4}{2}-\frac{1}{2}\\ b&=\frac{3}{2} \end{split} \end{equation}$

So, the desired equation is $y=-\frac{1}{4}x+\frac{3}{2}$ .

Example 3

$\triangle PQR$ has vertices at $P(-4,0),~Q(6,0),~R(0,4)$ . Find the equation of the median from $P$ .

We first find the midpoint of $RQ$ ; it is:

$\Big(\frac{0+6}{2},\frac{4+0}{2}\Big)=(3,2).$

Let’s label it as $L(3,2)$ . Next, we find the slope of $PL$ ; it is:

$\frac{2-0}{3-(-4)}=\frac{2}{7}=m,~\textrm{say}.$

We can now write $y=\frac{2}{7}x+b$ , and use the coordinates of $P$ (or $L$ ) to determine $b$ . Let’s use $L(3,2)$ :

$\begin{equation*} \begin{split} y&=mx+b\\ 2&=\frac{2}{7}(3)+b\\ 2&=\frac{6}{7}+b\\ b&=2-\frac{6}{7}\\ b&=\frac{14}{7}-\frac{6}{7}\\ b&=\frac{8}{7} \end{split} \end{equation}$

So, the desired equation is $y=\frac{2}{7}x+\frac{8}{7}$ . (This can be written in other forms if one wishes.)

Example 4

$\triangle ABC$ has vertices located at points $A(-2,4),~B(4,2),~C(r,s)$ . The equation of the median through $B$ is $y=-\frac{1}{7}x+\frac{18}{7}$ , while the median through $C$ has equation $y=-x+4$ . Determine the values of $r$ and $s$ .

Since we have equations for two medians, it is possible to determine the coordinates of $C$ uniquely. Let’s draw a diagram first (we do not know precisely where the point $C$ will be, so we just insert it somewhere).

Since $C(r,s)$ lies on the median $y=-x+4$ , its coordinates satisfy this equation, so

(1) $\begin{equation*} s=-r+4 \end{equation*}$

Also, the midpoint of $AC$ , namely

$\Big(\frac{r+-2}{2},\frac{s+4}{2}\Big)=\Big(\frac{r-2}{2},\frac{s+4}{2}\Big)$

lies on the median through $B$ (equation: $y=-\frac{1}{7}x+\frac{18}{7}$ ), so we obtain our second equation:

(2) $\begin{equation*} \frac{s+4}{2}=-\frac{1}{7}\Big(\frac{r-2}{2}\Big)+\frac{18}{7} \end{equation*}$

The lowest common denominator in equation (2) is $14$ . We clear the fractions by multiplying both sides of the equation by $14$ :

$14\times\Big(\frac{s+4}{2}\Big)=14\times\Big(-\frac{1}{7}\Big(\frac{r-2}{2}\Big)+\frac{18}{7}\Big)$

That is,

$7s+28=-r+2+36,\quad\implies r=10-7s.$

Substitute $10-7s$ for $r$ in equation (1):

$\begin{equation*} \begin{split} s&=-r+4\\ s&=-(10-7s)+4\\ s&=-10+7s+4\\ s-7s&=-10+4\\ -6s&=-6\\ s&=1 \end{split} \end{equation}$

Since $r=10-7s$ , we obtain $r=10-7\times 1=10-7=3$ . So $r=3$ and $s=1$ .

Example 5

$\triangle RST$ has vertices located at $R(-2,2),~S(4,6),~T(a,b)$ . If the equation of the median through $T$ is $y=4x$ , determine possible values for $a$ and $b$ . How many possibilities are there?

Imagine that we’ve represented the points in the diagram below:

Since the median $TM$ lies on the line $y=4x$ , the coordinates of point $T(a,b)$ must satisfy this equation, and so

$b=4a.$

That’s all we’ve got from the given information.

It follows that the solution is not unique. Any ordered pair $(a,b)$ that satisfies $b=4a$ — except $(1,4)$ — will do. For example, $(0,0),(2,8),(-1,-4)$ are acceptable ordered pairs for $(a,b)$ . Thus, for the second part of the question, there are infinite possibilities.
It is essential to exclude the point $M(1,4)$ from our list of possible solutions. Since $M(1,4)$ is collinear with $R(-2,2)$ and $S(4,6)$ , we cannot have a triangle $RST$ if $T$ is chosen as the point $T(1,4)$ .

Length of a median

We can make use of the distance formula to calculate the length of a median.

$\boxed{\textrm{\textbf{DISTANCE FORMULA:=}}\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$

Example 6

$\triangle ABC$ has vertices located at $A(0,0),~B(10,0),~C(0,4)$ . Find the length of the median through $A$ .

We first find the midpoint of $BC$ ; it is:

$\Big(\frac{10+0}{2},\frac{0+4}{2}\Big)=(5,2).$

Let’s label this midpoint as $M(5,2)$ . The required median is line segment $AM$ . Using the distance formula, the length of $AM$ is:

$\sqrt{(5-0)^2+(2-0)^2}=\sqrt{29}.$

Example 7

$\triangle ABC$ has vertices located at $A(0,0),~B(10,0),~C(0,4)$ . Find the length of the median through $B$ .

Since we need the median through $B$ , we’ll first find the midpoint of $AC$ ; it is:

$\Big(\frac{0+0}{2},\frac{0+4}{2}\Big)=(0,2).$

Let’s label it as $N$ . Then, by the distance formula, the length of $BN$ is:

$\sqrt{(0-10)^2+(2-0)^2}=\sqrt{104}=\sqrt{4\times 26}=2\sqrt{26}.$

The length of the median through $B$ is $2\sqrt{26}\approx 10.20$ units.

Example 8

The vertices of a triangle $TUV$ are located at $T(-1,2),~U(5,4),~V(a,b)$ . If the length of the median through $V$ is $5$ units, determine possible values for $a$ and $b$ .

Since we’re dealing with the median through $V$ , we’ll first find the midpoint of $T$ and $U$ ; it is:

$\Big(\frac{-1+5}{2},\frac{2+4}{2}\Big)=(2,3).$

Let’s label this midpoint as $L(2,3)$ . The median is then $VL$ . Since its length is $5$ units, we have, by the distance formula, that:

$\sqrt{(a-2)^2+(b-3)^2}=5.$

Square both sides: $(a-2)^2+(b-3)^2=25$ . We’ve solved this type of question before during our post on the length of a line segment. In the present case, we can take $a=-2$ and $b=0$ . There are several other possibilities.

Centroid of a triangle

We’ll apply our knowledge of linear systems to find the centroid of a triangle — the point of intersection of the three medians.

Example 9

Find the coordinates of the centroid of $\triangle ABC$ , where the vertices are located at $A(-2,0),~B(4,-2),~C(2,2)$ .

The easiest approach to this question is to use the centroid formula (see the exercises at the end), but we don’t recommend this.

Another approach is to get a grid paper and a well-sharpened pencil. Then plot the points, draw the medians, and determine where they intersect. But there’s a limitation with this approach: if the point of intersection of the medians is not comprised of integer coordinates, then reading them off from the grid may be difficult.

What we’ll do here is to go through the derivation of the equations of the medians $AM$ and $CL$ as shown in the diagram above, then solve the resulting system of equations to find where they intersect.

Median $AM$ (orange-coloured line above) is (a section of) the $x$ -axis, so its equation is simple, short, straightforward:

(3) $\begin{equation*} y=0 \end{equation*}$

Next, we find the equation of $CL$ (green line above). Notice that $L$ is the point $L(1,-1)$ . Let’s use another form of the equation of a straight line:

$\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}$

and then substitute the coordinates of $C(2,2)$ as $(x_1,y_1)$ and $L(1,-1)$ as $(x_2,y_2)$ . This way we obtain:

$\begin{equation*} \begin{split} \frac{y-y_1}{x-x_1}&=\frac{y_2-y_1}{x_2-x_1}\\ \frac{y-2}{x-2}&=\frac{-1-2}{1-2}\\ \frac{y-2}{x-2}&=\frac{-3}{-1}\\ \frac{y-2}{x-2}&=3\\ y-2&=3(x-2)\\ y&=3x-4 \end{split} \end{equation}$

So, the equation of median $CL$ is:

(4) $\begin{equation*} y=3x-4 \end{equation*}$

We now solve the linear system given by equations (3) and (4). From equation (3), $y=0$ . Substitute $y=0$ in equation (4):

$\begin{equation*} \begin{split} y&=3x-4\\ 0&=3x-4\\ 4&=3x\\ \frac{4}{3}&=x \end{split} \end{equation}$

Thus, the point of intersection of medians $AM$ and $CL$ is $(\frac{4}{3},0)$ . Put differently, the centroid is $(\frac{4}{3},0)$ . (It is also essential to ensure that median $BN$ also passes through this point, but we’ll leave that part.)

Midpoint of midpoint of midpoint

That English doesn’t sound right. Right? Well, you might be right. But let’s see what that might mean.

Example 10

Prove that the midpoint of a median always coincides with the midpoint of the midpoints of the other two sides.

Let’s use a diagram to clarify things. Consider $\triangle ABC$ pictured below:

Here, $P$ is the midpoint of $AB$ ; $Q$ is the midpoint of $AC$ ; and $R$ is the midpoint of $BC$ . We want to show that the midpoint of $PQ$ and the midpoint of $AR$ coincide. This is very simple.

Let’s find the coordinates of $P,Q,R$ :

$\begin{equation*} \begin{split} P&:=\Big(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\Big);\\ Q&:=\Big(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2}\Big);\\ R&:=\Big(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2}\Big). \end{split} \end{equation}$

The midpoint of $PQ$ is:

$\Big(\frac{\frac{x_1+x_2}{2}+\frac{x_1+x_3}{2}}{2},\frac{\frac{y_1+y_2}{2}+\frac{y_1+y_3}{2}}{2}\Big)=\Big(\frac{2x_1+x_2+x_3}{4},\frac{2y_1+y_2+y_3}{4}\Big).$

The midpoint of $AR$ is:

$\Big(\frac{x_1+\frac{x_2+x_3}{2}}{2},\frac{y_1+\frac{y_2+y_3}{2}}{2}\Big)=\Big(\frac{2x_1+x_2+x_3}{4},\frac{2y_1+y_2+y_3}{4}\Big).$

So we see that these two midpoints coincide. Congrats! You’ve proved a geometric property.

Takeaway

In a triangle, a median is a line segment connecting a vertex to the midpoint of the opposite side. A triangle has three medians, and the medians intersect at a point called the centroid. Finding the equation of a median is as simple as finding the equation of a regular straight line.

Tasks

$\triangle ABC$ has vertices located at $A(0,0),~B(1,4),~C(6,0)$ . Through which of the vertices should a median be drawn, so that the median is parallel to the line $2x+y+3=0$ ?
Find the centroid of the triangle whose vertices are $(-2,0),~(2,0),(0,2)$ .
The vertices of $\triangle ABC$ are $A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3)$ . Prove that the centroid of $\triangle ABC$ is located at $\Big(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\Big)$ .
[A simple way to obtain this result is to use the fact that the centroid divides the median in the ratio $2:1$ . But what if one is unaware of this fact? Then one has to resort to finding the equations of two medians, and solving the resulting linear system. We highly encourage anyone who has the time to spare, tenacity of spirit, and the tools and stationery to try the latter approach. It is an abundantly rewarding exercise. You won’t get many opportunities to solve a math problem where each line occupies the entire length of a $1$ m chalkboard. Do the drill and feel the thrill. Then thank us when you’re free.]
Determine an appropriate choice of coordinates for the vertices $A,B,C$ of $\triangle ABC$ for which its centroid will be the origin $(0,0)$ . Is the answer unique?
Prove that the three medians in an equilateral triangle are equal in length. (It happens that medians and altitudes coincide for equilateral triangles, and so they have lengths of the form $a\sqrt{3}$ , where $a$ is any non-zero real number.)
Prove that if the three medians in a triangle are equal in length, then the triangle is equilateral. (In view of the preceding exercise, it means that we can characterize equilateral triangles using medians.)
Prove that two of the three medians in an isosceles triangle are equal in length.
Prove that if two medians in a triangle are equal in length, then the triangle is isosceles. (In view of the preceding exercise, it follows that we can characterize isosceles triangles using medians.)
Find the angle the median from $A$ makes with side $BC$ , if the triangle $ABC$ has vertices at $A(-1,-1),~B(1,1),~C(3,-3)$ .
In $\triangle ABC$ , it is desired that the median from vertex $A$ and the median from vertex $B$ meet at $90^{\circ}$ . Determine an appropriate choice of coordinates for the vertices $A,B,C$ that will make this happen.

Thanksgiving theorems, part I

Today is thanksgiving day in our jurisdiction; and, being that it falls on the $14$ th — a number that is “HTMLy” (read as extremely) significant to us — we’ve chosen to mark the occasion with math. While this choice may seem unconventional, it doesn’t rule out the fact that our genuine gratitude to The Generous Giver is unconditional.

Now to the math itself. Consider the prime factorizations of $14$ and $15$ :

$14=2\times 7,\quad 15=3\times 5$

and an enumeration of the first four prime numbers:

$2,3,5,7.$

Clearly, we have that $2\times 7 < 3\times 5$ ; this was what motivated the theorems in this post. By the way, our use of the word theorem is a deliberate exaggeration, but since we’re not dealing with advanced math, we hope that we’ll be offered exoneration, despite our overt exaggeration.

Theorem 1

Let $x,y,z,t$ be an arithmetic sequence. Then $yz>xt$ .

If $x,y,z,t$ are as given, we can suppose that the common difference of the arithmetic sequence is $d$ , and then write $y=x+d,z=x+2d,t=x+3d$ . Consider $yz-xt$ :

$\begin{equation*} \begin{split} yz-xt&=(x+d)(x+2d)-(x)(x+3d)\\ &=(x^2+3xd+2d^2)-(x^2+3xd)\\ &=2d^2\\ &>0 \end{split} \end{equation}$

Since the difference $yz-xt>0$ , we conclude that $yz>xt$ .

Theorem 2

Let $x,y,z,t$ be integers that form an arithmetic sequence. Then $yz-xt$ can never be a perfect square.

If $x,y,z,t$ are integers that form an arithmetic sequence, then the common difference $d$ must be an integer as well (note that we exclude the case when $d=0$ ). From our previous calculation, we saw that $yz-xt=2d^2$ . Since $2d^2$ is not a perfect square when $d$ is an integer (except for $d=0$ ), so is $yz-xt$ .

Theorem 3

Let $x,y,z,t$ be consecutive, positive even numbers. Then $yz-xt=8$ .

This should have been better described as a corollary (to the first “theorem”), but we don’t want to introduce another jargon here. We’ll ask you to prove it in the exercises at the end; for now, let’s see a few instances:

Consider the first four positive even numbers: $2,4,6,8$ . We have that $4\times 6 - 2\times 8=24-16=8$ ;
Consider the numbers $10,12,14,16$ . We have that $12\times 14-10\times 16=168-160=8$ ;
Consider the numbers $16, 18,20,22$ . We have that $18\times 20-16\times 22=360-352=8$ ;
And so on. You feel that?

Theorem 4

Let $x,y,z,t$ be consecutive, positive odd numbers. Then $yz-xt=8$ .

Actually, the “positive” requirement in the two preceding theorems can be dropped, what’s essential is that none of the numbers should be zero. Few examples below:

Consider the first four positive odd numbers: $1,3,5,7$ . We have that $3\times 5-1\times 7=15-7=8$ ;
Consider the numbers $11,13,15,17$ . We have $13\times 15-11\times 17=195-187=8$ ;
Consider the numbers $61,63,65,67$ . We have $63\times 65-61\times 67=4095-4087=8$ .

Theorem 5

Let $x,y,z,t$ be consecutive integers, none of which is zero. Then, $yz-xt=2$ .

Consecutive integers form an arithmetic sequence with a common difference of $d=1$ . So, in view of our first theorem, we have that $yz-xt=2d^2=2(1^2)=2$ .

The next result is well-known, and is in a class of its own:

Theorem 6

Let $x,y,z,t$ be four consecutive terms of a Fibonacci sequence. Then $yz-xt=\pm 1$ .

Usually, the terms in a Fibonacci sequence are denoted by $F_{n},F_{n+1},F_{n+2},\cdots$ ; we’ve used $x,y,z,t$ for convenience. Also, in order to keep things simple, we’ve omitted the proof of the above result, but we invite our readers to try it out — this can be accomplished, for example, by using the fact that the $n$ th term is given by

$F_{n}=\frac{1}{\sqrt{5}}\Big(\frac{1+\sqrt{5}}{2}\Big)^n-\frac{1}{\sqrt{5}}\Big(\frac{1-\sqrt{5}}{2}\Big)^n$

and then expanding and simplifying $F_{n+1}F_{n+2}-F_{n}F_{n+3}$ .

Theorem 7

Let $x,y,z,t$ be a geometric sequence. Then $yz-xt=0$ .

Let $x,y,z,t$ be as given. Suppose that the common ratio of the geometric sequence is $r$ , then we can write $y=xr,z=xr^2,t=xr^3$ . Now:

$\begin{equation*} \begin{split} yz-xt&=(xr)(xr^2)-(x)(xr^3)\\ &=x^2r^3-x^2r^3\\ &=0, \end{split} \end{equation}$

which is what we want.

Linear systems with special coefficients

Do our “theorems” have any where we can apply them? Well, somewhat — think linear systems. If we have a linear system in two variables $x$ and $y$ , like so:

(1) $\begin{equation*} ax+by=k \end{equation*}$

(2) $\begin{equation*} cx+dy=l, \end{equation*}$

then the solution can be given by (Cramer’s rule ):

$x=\frac{dk-bl}{ad-bc},\quad y=\frac{al-ck}{ad-bc}.$

As can be seen from the above explicit formula, the quantity $ad-bc$ is akin to what we’ve been playing around with. That makes us to ask: “What if the coefficients in a linear system are so arranged that they’re consecutive terms of any of the number patterns we’ve considered in this post”? This is certainly an obscure possibility, but we’ve included it out of curiosity.

Theorem 8

Let $a,b,c,d$ be consecutive integers, none of which is zero. Then the linear system

(3) $\begin{equation*} ax+by=k \end{equation*}$

(4) $\begin{equation*} cx+dy=l \end{equation*}$

has integer solutions, if $k$ and $l$ are even numbers.

The proof relies on the fact that the quantity $ad-bc$ (called the determinant) appears in the denominator when the above linear system is solved. Recall that the solution is:

$x=\frac{dk-bl}{ad-bc},\quad y=\frac{al-ck}{ad-bc}.$

Since $a,b,c,d$ are consecutive non-zero integers, we know, from Theorem 5, that $bc-ad=2$ , or $ad-bc=-2$ . Thus the solution may be re-written as:

$x=\frac{dk-bl}{-2},\quad y=\frac{al-ck}{-2}.$

Since both $k$ and $l$ are even numbers, we know that the numerators above, namely $dk-bl$ and $al-ck$ , are both even. Therefore, the expressions for $x$ and $y$ are divisible by $2$ , resulting in integer quotients.

Theorem 9

Let $a,b,c,d$ be consecutive terms of a Fibonacci sequence. Then the linear system

(5) $\begin{equation*} ax+by=k \end{equation*}$

(6) $\begin{equation*} cx+dy=l \end{equation*}$

has integer solutions if $k$ and $l$ are integers.

Notice that we have $bc-ad=\pm 1$ if $a,b,c,d$ are consecutive terms of a Fibonacci sequence (Theorem 6 above), and so the general solution

$x=\frac{dk-bl}{ad-bc},\quad y=\frac{al-ck}{ad-bc}$

simplifies to $x=\pm(dk-bl),~y=\pm(al-ck)$ , which are integers.

Theorem 10

Let $a,b,c,d$ be a geometric sequence with common ratio $r$ . Then the linear system

(7) $\begin{equation*} ax+by=k \end{equation*}$

(8) $\begin{equation*} cx+dy=l \end{equation*}$

has no solution so long as $k\neq \frac{l}{r^2}$ (and $k$ and $l$ are not both zero).

A proof of our present theorem relies on the fact that $ad-bc=0$ when $a,b,c,d$ form a geometric sequence (Theorem 7), together with the fact that the solution to the linear system is:

$x=\frac{dk-bl}{ad-bc},\quad y=\frac{al-ck}{ad-bc}.$

Since division by zero is not allowed, we’ll be unable to determine $x$ and $y$ from the above expressions, thus no solution. But wait. How about the assumption that $k$ and $l$ are not both zero? If they are both zero, then we’ll have a trivial solution: $x=0,y=0$ . What of the assumption that $k\neq \frac{l}{r^2}$ ? Suppose that $k=\frac{l}{r^2}$ . Then, we can divide equation (8) by $r^2$ , obtaining

(9) $\begin{equation*}\frac{c}{r^2}x+\frac{d}{r^2}y=\frac{l}{r^2}.\end{equation*}$

However, since $a,b,c,d$ form a geometric sequence with common ratio $r$ , it follows that $a=\frac{c}{r^2}$ and $b=\frac{d}{r^2}$ . This would then mean that equation (9) is the same as:

(10) $\begin{equation*} ax+by=k, \end{equation*}$

which is exactly equation (7). This is saying that if $k=\frac{l}{r^2}$ , then equations (7) and (8) are the same, meaning that there’s an infinite number of solutions to the original linear system. So we want $k\neq \frac{l}{r^2}$ for our theorem to hold.

Takeaway

If four numbers $x,y,z,t$ are arranged in ascending order, then the quantity $yz-xt$ always takes on constant values, under certain circumstances. In the case of even/odd numbers, it is $8$ ; in the case of Fibonacci numbers, it is $\pm 1$ ; in the case of geometric sequences, it is $0$ .

Tasks

Construct an example to show that $x< y< z< t$ does not always imply that $yz>xt$ .
Without requiring that $x,y,z,t$ form a geometric sequence, prove that it is still possible to have $yz-xt=0$ , where $x,y,z,t$ are non-zero integers that are not all equal and have been sorted in ascending order.
Let $x< y< z< t$ be numbers. Under what conditions, other than the numbers forming an arithmetic sequence, can we always have $yz>xt$ ?
Let $x,y,z,t$ be consecutive positive even numbers. Prove that $yz-xt=8$ .
Let $x,y,z,t$ be consecutive positive odd numbers. Prove that $yz-xt=8$ .

Thanksgiving

In the spirit of today’s occasion, the poster finds it appropriate to appreciate God for a certain Thursday, June 14, 2018.

Thursday, June 14, 2018. Beautiful, beautiful day. Thursday, June 14, 2018. Beautiful, beautiful day. The poster can’t refrain from singing this refrain again and again. He’s got chance today; he just can’t refrain; he just chants that refrain.

No doubt we should keep our personal things to ourselves, but the poster couldn’t help it this time, especially considering the coincidence with today’s date, today’s event, and “that Thursday”. The magnitude of what happened makes the poster to offer tribute and gratitude to The One whose gracious attribute makes Him to distribute good things generously to all men.

So, what happened exactly? Refer to a strange word the poster used at the beginning of this post, namely “HTMLy”. Read it as “each-team-ly” (imagine how a 2 year old might pronounce it $\cdots$ ). “HTMLy” was coined to capture two words: HTML and extremely. “That Thursday” is extremely significant to the poster because that was the day the poster was introduced to $\cdots$ you guessed it $\cdots$ HTML. That singular and spectacular Thursday transformed the poster’s way of thinking, the poster’s way of teaching, and — of course — the poster’s way of “thanking”.

Did you get the point the poster is trying to make? Make it a point of duty to be grateful, even for things that are little.