Geometric sequence on a staircase

In case you’ve used a staircase before, you’ve “experienced” a special case of a geometric sequence as a consequence.

For a case study, consider the diagram below:

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Join A to D, D to G, and G to A to form \triangle ADG:

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We’ll prove that the slopes of the sides of \triangle ADG form a geometric sequence, if AC=EG.

Example

A portion of a staircase is shown below:

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Verify that the slopes of the sides of \triangle ADG form a geometric sequence.

Easy-peasy.

The slopes of sides DG,GA,AD are \frac{1}{2},1,2, respectively. They form a geometric sequence with a common ratio 2.

(The fact that the slope of side GA is 1 implies that it makes an angle of 45^{\circ} with the horizontal. In real life, however, the angle is usually around 30^{\circ}.)

Example

A portion of a staircase is shown below:

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Verify that the slopes of the sides of \triangle ADG form a geometric sequence.

Easy-peasy as before.

The slopes of sides DG,GA,AD are \frac{11}{40},\frac{11}{20},\frac{11}{10}, respectively. They form a geometric sequence with a common ratio of 2.

(It shouldn’t come as a surprise that the common ratio is 2 this time too. Indeed, it is almost always the case in a staircase, because the “rise” and “run” stay constant throughout the entire span of the staircase. For this reason, the geometric sequence obtained on a staircase is a special case.)

Basic background

From Example 3 to Example 6 below, we lay the foundation for our staircase story. Don’t worry if it seems to “escalate” quite quickly; remember we’re on a staircase \cdots.

PROVE that if the slopes of the sides of a triangle form a geometric sequence with a positive common ratio, then the triangle contains an obtuse angle.

Actually, any triangle in which the three sides have positive slopes contains an obtuse angle (for now we’re content with slopes in geometric sequence).

Let \triangle ABC be such that sides AB,BC,CA have slopes a,ar,ar^2, respectively. If the common ratio r is positive, then the angle at A is obtuse. To see this, let the lengths of sides AB,BC,CA be l_1,l_2,l_3, respectively. Then, as we’ll show in our next post, the following pseudo-pythagorean property holds:

    \[l_1^2+l_3^2=\frac{1+r^2}{(1+r)^2}l_2^2.\]

Thus, we can use the cosine law to compute the cosine of the angle at A:

    \begin{equation*} \begin{split} \cos A&=\frac{l_1^2+l_3^2-l_2^2}{2\times l_1\times l_3}\\ &=\frac{\left(\frac{1+r^2}{(1+r)^2}l_2^2\right)-l_2^2}{2\times l_1\times l_3}\\ &=\frac{\left(1+r^2-(1+r)^2\right)l_2^2}{2\times(1+r)^2\times l_1\times l_3}\\ &=\frac{-2r\times l_2^2}{2\times (1+r)^2\times l_1\times l_3}\\ &=\frac{-r}{(1+r)^2}\frac{l_2^2}{l_1l_3}\\ \implies\cos A& < 0\\ \therefore\angle A&~~\textrm{is OBTUSE} \end{split} \end{equation*}

The obtuse angle at A is well-oriented in such a way that a special 90^{\circ} can be cut out from it.

In \triangle ABC, let a,ar,ar^2~(r> 0) be the slopes of sides AB,BC,CA, respectively. Then a point X divides BC in the ratio r:1 if and only if it shares the same x-coordinate as vertex A.

Thus, AX will be a vertical line. Take note of this.

The problem deals with coordinates, so let’s begin by specifying the coordinates of the vertices of the parent triangle ABC. Assume A(x_1,y_1),~B(x_2,y_2),~C(x_3,y_3) are the vertices. Since the slopes of sides AB,BC,CA are a,ar,ar^2, there are relationships among the coordinates:

(1)   \begin{equation*} \begin{split} x_1&=x_3+\frac{1}{ar(r+1)}(y_2-y_3)\\ y_1&=\frac{r}{r+1}y_2+\frac{1}{r+1}y_3\\ x_2&=x_3+\frac{1}{ar}(y_2-y_3) \end{split} \end{equation*}

Now, suppose a point X divides side BC in the ratio r:1, with r> 0. Using the formula for internal division of a line segment, we can write the x-coordinate of point X in terms of the x-coordinates of B(x_2,y_2) and C(x_3,y_3) as follows:

    \begin{equation*} \begin{split} x&=\frac{rx_3+1x_2}{r+1}\\ &=\frac{rx_3+1\left(x_3+\frac{1}{ar}(y_2-y_3)\right)}{r+1}\\ &=\frac{(r+1)x_3+\frac{1}{ar}(y_3-y_3)}{r+1}\\ &=x_3+\frac{1}{ar(r+1)}(y_2-y_3)\\ &=x_1 \end{split} \end{equation*}

Therefore, point X shares the same x-coordinate as vertex A(x_1,y_1).

Conversely, suppose that point X shares the same x-coordinate as vertex A(x_1,y_1). We proceed to find the y-coordinate of X. Since the slope of BC is ar and it goes through (x_3,y_3), its equation is y-y_3=ar(x-x_3). Using the fact that x=x_1=x_3+\frac{1}{ar(r+1)}(y_2-y_3) at point X, we have:

    \begin{equation*} \begin{split} y-y_3&=ar\left(x_3+\frac{1}{ar(r+1)}(y_3-y_3)-x_3\right)\\ y&=ar\left(\frac{1}{ar(r+1)}(y_3-y_3)\right)+y_3\\ &=\frac{1}{r+1}(y_2-y_3)+\left(\frac{r+1}{r+1}\right)y_3\\ \therefore y&=\frac{y_2+ry_3}{r+1} \end{split} \end{equation*}

Let’s manipulate x_1=x_3+\frac{1}{ar(r+1)}(y_2-y_3) slightly:

    \begin{equation*} \begin{split} x_1&=\left(\frac{r+1}{r+1}\right)x_3+\frac{1}{ar(r+1)}(y_2-y_3)\\ &=\frac{rx_3+x_3+\frac{1}{ar}(y_2-y_3)}{r+1}\\ &=\frac{rx_3+x_2}{r+1} \end{split} \end{equation*}

Thus, X is the point \left(\frac{x_2+rx_3}{r+1},\frac{y_2+ry_3}{r+1}\right). By examining its coordinates, we see that it actually divides side BC in the ratio r:1.

In \triangle ABC, let a,ar,ar^2~(r> 0) be the slopes of sides AB,BC,CA, respectively. Then a point Y divides BC in the ratio 1:r if and only if it shares the same y-coordinate as vertex A.

Thus, AY will be a horizontal line. Together with the previous example, the parent \triangle ABC contains a special right triangle AXY in which both legs AX and AY are parallel to the coordinate axes. Further, BX=CY. The presence of this right triangle is key to our staircase story.

Let’s find the y coordinate of point Y which divides BC in the ratio 1:r. Using equation (1) and the formula for internal division of a line segment:

    \begin{equation*} \begin{split} y&=\frac{ry_2+y_3}{r+1}\\ &=y_1\\ \end{split} \end{equation*}

Easier than thought. Conversely, suppose that Y shares the same y-coordinate as vertex A(x_1,y_1), so that y=y_1=\frac{ry_2+y_3}{r+1}. We’ll find its x-coordinate, using the fact that it lies on side BC whose equation is y-y_3=ar(x-x_3):

    \begin{equation*} \begin{split} \left(\frac{ry_2+y_3}{r+1}\right)-y_3&=ar(x-x_3)\\ \left(\frac{ry_2+y_3}{r+1}\right)-\left(\frac{r+1}{r+1}\right)y_3&=ar(x-x_3)\\ \frac{r}{r+1}(y_2-y_3)&=ar(x-x_3)\\ \frac{1}{a(r+1)}(y_2-y_3)&=x-x_3\\ x&=x_3+\frac{1}{a(r+1)}(y_2-y_3)\\ &=\left(\frac{r+1}{r+1}\right)x_3+\frac{1}{a(r+1)}(y_2-y_3)\\ &=\frac{x_3+rx_3+\frac{1}{a}(y_2-y_3)}{r+1}\\ &=\frac{x_3+r\left(x_3+\frac{1}{ar}(y_2-y_3)\right)}{r+1}\\ \therefore x&=\frac{x_3+rx_2}{r+1} \end{split} \end{equation*}

These calculations show that Y is the point \left(\frac{rx_2+x_3}{r+1},\frac{ry_2+y_3}{r+1}\right), and it divides side BC in the ratio 1:r if we inspect its coordinates.

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In the above diagram, point C divides AG in the ratio r:1 while point E divides AG in the ratio 1:r.

(Main goal)

Let \triangle ABC be a triangle. Then the following two statements are equivalent:

  1. the slopes of the sides form a geometric sequence with positive common ratio;
  2. there are internal points X and Y on BC such that AX is vertical, AY is horizontal, and BX=CY.

The implication (1)\implies(2) follows from the three previous examples. For (2)\implies (1), consider the diagram below:

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We’ll show that

(2)   \begin{equation*}\left(\frac{f-b}{e-a}\right)\times\left(\frac{b-d}{a-c}\right)=\left(\frac{f-d}{e-c}\right)^2. \end{equation*}

Since BX=CY, we have:

    \[(a-c)^2+(v-d)^2=(f-b)^2+(e-u)^2.\]

Because B,X,Y,C are co-linear, we have \frac{f-b}{e-u}=\frac{v-d}{a-c}, and so f-b=(e-u)\left(\frac{v-d}{a-c}\right). Using this in the previous equation, we have:

    \begin{equation*} \begin{split} (a-c)^2+(v-d)^2&=\left[(e-u)\left(\frac{v-d}{a-c}\right)\right]^2+(e-u)^2\\ (a-c)^2+(v-d)^2&=(e-u)^2\left[\frac{(v-d)^2+(a-c)^2}{(a-c)^2}\right]\\ \therefore (a-c)^2&=(e-u)^2 \end{split} \end{equation*}

The latter means that either a-c=e-u or a-c=-(e-u). However, in our diagram we have a> c and e> u, thus ruling out the possibility of a-c=-(e-u). We therefore settle for a-c=e-u.

Returning to (2) and using some equality relations among the slopes of the co-linear points B,X,Y,C, we have:

    \begin{equation*} \begin{split} \left(\frac{f-d}{e-c}\right)^2&=\left(\frac{b-v}{u-a}\right)^2\\ \left(\frac{f-b}{e-a}\right)\left(\frac{b-d}{a-c}\right)&=\left[(e-u)\frac{b-v}{u-a}\right].\frac{1}{e-a}\left(\frac{b-d}{a-c}\right)\\ &=\left(\frac{b-v}{u-a}\right)\left(\frac{b-d}{e-a}\right)\\ &=\left(\frac{b-v}{u-a}\right)\left(\frac{b-d}{u-c}\right)\\ &=\left(\frac{b-v}{u-a}\right)\left(\frac{b-v}{u-a}\right) \end{split} \end{equation*}

This proves that the slopes of the sides of \triangle ABC form a geometric progression. The common ratio is positive in view of the diagram.

In \triangle ABC, let a,ar,ar^2~(r> 0) be the slopes of sides AB,BC,CA, respectively. For any positive integer n\neq 1,r,\frac{1}{r}, PROVE that there are points X and Y on BC such that the slopes of the sides of \triangle AXY form a geometric sequence \left(\frac{n-r}{nr-1}\right)ar,ar,\left(\frac{nr-1}{n-r}\right)ar.

Note that both Example 4 and Example 5 result from the special case n=r, but the associated slopes do not form a geometric sequence. Also, n need not be a positive integer; any positive number works as well.

We need to construct points X and Y first.

Let X be the point which divides side BC in the ratio n:1. (The restriction that n\neq 1 is essential to exclude the midpoint of BC.) Using equation (1) and the formula for internal division of a line segment, we can determine the coordinates of X as follows:

    \begin{equation*} \begin{split} X_x&=\frac{nx_3+x_2}{n+1}\\ &=\frac{nx_3+x_3+\frac{1}{ar}(y_2-y_3)}{n+1}\\ &=\frac{(n+1)x_3+\frac{1}{ar}(y_2-y_3)}{n+1}\\ &=x_3+\frac{1(y_2-y_3)}{ar(n+1)}\\ X_y&=\frac{ny_3+y_2}{n+1} \end{split} \end{equation*}

So, X is the point \left(x_3+\frac{(y_2-y_3)}{ar(n+1)},\frac{ny_3+y_2}{n+1}\right). Next, let’s construct point Y by dividing side BC in the ratio 1:n.

    \begin{equation*} \begin{split} Y_x&=\frac{x_3+nx_2}{n+1}\\ &=\frac{x_3+nx_3+\frac{n}{ar}(y_2-y_3)}{n+1}\\ &=x_3+\frac{n}{ar(n+1)}(y_2-y_3)\\ Y_y&=\frac{y_3+ny_2}{n+1} \end{split} \end{equation*}

Thus, Y is the point \left(x_3+\frac{n(y_2-y_3)}{ar(n+1)},\frac{y_3+ny_2}{n+1}\right). Together with equation (1), the vertices of \triangle AXY are:

    \begin{equation*} \begin{split} A&\left(x_3+\frac{(y_2-y_3)}{ar(r+1)},\frac{ry_2+y_3}{r+1}\right)\\ X&\left(x_3+\frac{(y_2-y_3)}{ar(n+1)},\frac{ny_3+y_2}{n+1}\right)\\ Y&\left(x_3+\frac{n(y_2-y_3)}{ar(n+1)},\frac{y_3+ny_2}{n+1}\right) \end{split} \end{equation*}

From these, we obtain \left(\frac{nr-1}{n-r}\right)ar,~ar,~\left(\frac{n-r}{nr-1}\right)ar as the slopes of sides AX,XY,YA, respectively. Note that n=0 yields a,ar,ar^2 as the slopes of AX,XY,YA — same slopes as those of the parent triangle ABC.

The next example shows that if we exclude three points from BC, namely, the midpoint of BC, the point on BC that shares the same x-coordinate as A, and the point on BC that shares the same y-coordinate as A , then we obtain what may be regarded as a stronger version of Example 6.

For any \triangle ABC, the following statements are equivalent:

  1. the slopes of sides AB,BC,CA form a geometric sequence with positive common ratio;
  2. for any internal point X on BC, there is an internal point Y on BC such that BX=YC and the slopes of \triangle AXY form a geometric sequence.

Bear in mind the three excluded points on BC. Basically, these are the points where the slope of AX will be undefined, the slope of AY will be zero, and where X=Y. Naturally, these were excluded to ensure that we have proper slopes.

To show (1)\implies (2), let X be the point which divides BC in the ratio n:1, where n\neq 1 is a positive integer. Choose Y as the point which divides BC in the ratio 1:n. Then BX=CY. By Example 7, the slopes of the sides of \triangle AXY form a geometric progression.

To show (2)\implies (1), take X=B. Then Y has to be taken as C, so that the equality BX=YC is satisfied (both sides being zero).

PROVE that the two infinite geometric progressions

(3)   \begin{equation*} \cdots,-\frac{1}{4},-\frac{1}{2},-1,-2,-4,\cdots \end{equation*}

(4)   \begin{equation*} \cdots,\frac{1}{4},\frac{1}{2},1,2,4,\cdots \end{equation*}

can be obtained from the same triangle.

You already know how to generate (4); if not, see here. So we focus on obtaining (3), following a similar procedure and using the same triangle that yields (4).

Begin with a triangle whose slopes are 1,2,4 — like the one shown below:

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where A(1,4), B(0,0), C(3,6) are the vertices.

For an integer n, set X_{n}:=\left(\frac{2^{n-1}+2}{2^{n-1}+1},\frac{2^{n}+4}{2^{n-1}+1}\right). Then X_n is an internal point on BC; further, X_{n}\rightarrow (1,2) as n\rightarrow\infty and X_{n}\rightarrow (2,4) as n\rightarrow -\infty. These limits ensure that X_{n} always resides within side BC.

Next, let’s compute the slope of cevian AX_{n} for each n:

    \begin{equation*} \begin{split} \textrm{slope of}~AX_{n}&=\frac{4-\left(\frac{2^{n}+4}{2^{n-1}+1}\right)}{1-\left(\frac{2^{n-1}+2}{2^{n-1}+1}\right)}\\ &=\frac{4\left(2^{n-1}+1\right)-\left(2^{n}+4\right)}{2^{n-1}+1-\left(2^{n-1}+2\right)}\\ &=\frac{2(2^n)-2^n}{-1}\\ &=-2^n \end{split} \end{equation*}

Letting n=\cdots, -2,-1,0,1,2,\cdots gives rise to the infinite geometric sequence in (3).

That special right triangle

We conclude by isolating a unique property of the special right triangle that appears in staircases.

PROVE that a triangle is a right triangle with legs parallel to the coordinate axes if and only if the slopes of its three medians form a geometric sequence with common ratio -2 .

First suppose that we have a right triangle with legs parallel to the coordinate axes. This direction of the proof was done in our previous post, but let’s go through it briefly. Fix the vertices at A(0,a), B(0,0), and C(c,0); a little modification will be needed for the general case.

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Now calculate the slopes of medians CN,BM,AL:

    \begin{equation*} \begin{split} \textrm{slope of median}~ CN&=-\frac{a}{2c}\\ \textrm{slope of median}~ BM&=\frac{a}{c}\\ \textrm{slope of median}~ AL&=-\frac{2a}{c} \end{split} \end{equation*}

These form a geometric progression with common ratio of -2.

Conversely, suppose that we have \triangle ABC with vertices A(x_1,y_1), B(x_2,y_2), and C(x_3,y_3) in such a way that the slopes of the three medians form a geometric sequence with common ratio -2.

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If we denote the slopes of the medians from vertices A,B,C by m_A,m_B,m_C respectively, then:

    \begin{equation*} \begin{split} m_A&=\frac{y_2+y_3-2y_1}{x_2+x_3-2x_1}\\ m_B&=\frac{y_1+y_3-2y_2}{x_1+x_3-2x_2}\\ m_C&=\frac{y_1+y_2-2y_3}{x_1+x_2-2x_3} \end{split} \end{equation*}

Suppose that the geometric sequence of median slopes is m_C,m_B,m_A, then m_B=-2m_C, and m_A=-2m_B:

    \begin{equation*} \begin{split} \frac{y_1+y_3-2y_2}{x_1+x_3-2x_2}&=-2\left(\frac{y_1+y_2-2y_3}{x_1+x_2-2x_3}\right)\\ \frac{y_2+y_3-2y_1}{x_2+x_3-2x_1}&=-2\left(\frac{y_1+y_3-2y_2}{x_1+x_3-2x_2}\right)\\ \end{split} \end{equation*}

Clear fractions from the first equation above:

(5)   \begin{equation*} \begin{split} (x_1+x_2-2x_3)(y_1+y_3-2y_2)+2(x_1+x_3-2x_2)(y_1+y_2-2y_3)&=0\\ x_1(y_1-y_3)+x_2(-y_1-2y_2+3y_3)-2x_3(y_3-y_2)&=0 \end{split} \end{equation*}

Now clear fractions from the second equation:

(6)   \begin{equation*} \begin{split} (x_1+x_3-2x_2)(y_2+y_3-2y_1)+2(x_2+x_3-2x_1)(y_1+y_3-2y_2)&=0\\ x_1(2y_1-3y_2+y_3)+2x_2(-y_1+y_2)+x_3(y_2-y_3)&=0 \end{split} \end{equation*}

It turns out that y_1 can be eliminated if we perform the operation 2\times(5)-(6):

    \begin{equation*} \begin{split} x_1(3y_2-3y_3)+x_2(-6y_2+6y_3)+x_3(-3y_3+3y_2)&=0\\ \implies (x_1+x_3-2x_2)(y_2-y_3)&=0\\ &\vdots\cdots\vdots\\ x_1+x_3-2x_2&=0~\textrm{or}~y_2-y_3=0 \end{split} \end{equation*}

Since the slopes of the three medians form a geometric sequence, we can’t have x_1+x_3-2x_2=0. Otherwise this would result in an undefined slope for the median from vertex B, namely m_B=\frac{y_1+y_3-2y_2}{x_1+x_3-2x_2}, and a geometric sequence cannot contain such a term. So the only option is to take y_2-y_3=0, whence y_2=y_3. Our triangle now has one side parallel to the x-axis.

Put y_2=y_3 in equation (5):

    \begin{equation*} \begin{split} x_1(y_1-y_3)+x_2(-y_1-2y_2+3y_3)-2x_3(y_3-y_2)&=0\\ \implies x_1(y_1-y_2)+x_2(-y_1-2y_2+3y_2)-2x_3(y_2-y_2)&=0\\ \implies (x_1-x_2)(y_1-y_2)&=0\\ \implies x_1-x_2&=0\quad\textrm{or}~y_1-y_2=0 \end{split} \end{equation*}

Since we already had y_2=y_3, it’s impossible to have y_1=y_2 in addition, otherwise this would lead to y_1=y_2=y_3, which is not allowed for a triangle. So we take x_1-x_2=0, whence x_1=x_2. Our triangle now has a side that is parallel to the y-axis.

Therefore, \triangle ABC has side AB parallel to the y-axis, and side BC parallel to the x-axis.

Takeaway

We’ve used staircases in various cases and in various places — homes, airports, train stations, offices, etc. As we ascend our staircase, we “run” forward and “rise” higher, so our slopes are positive. In turn, the geometric sequence we describe has a positive common ratio — in most cases, 2.

Embellished with nice properties, triangles with slopes in geometric progression are established as objects to cherish and relish. And, they not only appeal in theory, they appear in real life too.

Tasks

  1. (Three out) Given \triangle ABC with vertices A(1,9), B(0,0), C(4,12), the slopes of the sides form a geometric sequence with common ratio 3. Find two points X and Y on BC for which the slopes of the sides of \triangle AXY do not form a geometric sequence.
    (The third point is the midpoint of BC, namely (2,6). Apart from these three points, given any other point X' within BC, there is a corresponding point Y' within BC for which the slopes of the sides of \triangle AX'Y' form a geometric sequence.)
  2. Consider the diagram below:

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    in which C and D have the same x-coordinates, D and E share the same y-coordinates, and AC=EG.

    • If also CE=EG, PROVE that the slopes of the sides of \triangle ADG form a geometric sequence with a common ratio of 2;
    • Give an example to show that if C and E do not trisect AG, then the common ratio is no longer 2.
  3. PROVE that if the slopes of the sides of a triangle are all positive (and non-zero), then the triangle contains an obtuse angle.
    (Similarly, if the slopes are all negative, there’ll be an obtuse angle.)
  4. In \triangle ABC, let a,ar,ar^2 be the slopes of sides AB,BC,CA, respectively. If r> 0, PROVE that side BC is the longest side.
  5. In \triangle ABC, let a,ar,ar^2 be the slopes of sides AB,BC,CA, respectively. If r< 0, PROVE that side BC is NOT always the longest side.
  6. PROVE that the two infinite geometric sequences \cdots,-\frac{1}{9},-\frac{1}{3},-1,-3,-9,\cdots and \cdots,\frac{1}{9},\frac{1}{3},1,3,9,\cdots can be obtained from the same triangle, using slopes.
  7. (Close enough) Let ABC be a triangle in which sides AB,BC,CA have slopes a,ar,ar^2, with a\neq 0 and r> 0. Let X and Y be points on BC which divide BC in the ratios r:1 and 1:r. PROVE that the area of \triangle AXY is \left(\frac{r-1}{r+1}\right) times the area of \triangle ABC.
    (In the limit r\rightarrow\infty, we have area of \triangle AXY \approx area of \triangle ABC.)
  8. Find coordinates for the vertices A,B,C of a non-right triangle ABC which contains a right triangle AXY such that the area of \triangle AXY is \frac{1}{3} times the area of \triangle ABC.
  9. Let \triangle ABC be such that the slopes of sides AB,BC,CA form the geometric sequence a,ar,ar^2, with r> 0.
    • Let X be the point which divides BC in the ratio r:1 and let X' be the midpoint of X and BC. PROVE that the slope of cevian AX' is -3 times the slope of BC.
    • Let Y be the point which divides BC in the ratio 1:r and let Y' be the midpoint of BC and Y. PROVE that the slope of cevian AY' is -\frac{1}{3} times the slope of BC.
    • Conclude that every triangle with slopes a,ar,ar^2 (where r> 0) contains a sub-triangle with slopes -\frac{1}{3}ar,ar,-3ar, a geometric sequence with common ratio -3.
  10. (Right inside) Let a,ar,ar^2 (r> 0) be the slopes of sides AB,BC,CA in \triangle ABC. PROVE that there are points X and Y on BC for which XY^2=AX^2+AY^2.
  11. Let a,ar,ar^2 (r> 0) be the slopes of sides AB,BC,CA in \triangle ABC. PROVE that there two points R and S on BC for which the slopes of \triangle ARS form an arithmetic sequence with a common difference of 2ar, namely: -3ar,-ar,ar.
  12. Let a,ar,ar^2 (r> 0) be the slopes of sides AB,BC,CA in \triangle ABC. PROVE that there two points U and T on BC for which the slopes of \triangle ATU form an arithmetic sequence with a common difference of ar, namely: -ar,0,ar.
  13. Let A(0,0), B(0,b), C(c,0) be a right triangle in the first quadrant. Choose any vertex, say A, and draw two cevians AP and AQ to side BC.
    • For a positive integer n, let P and Q divide side BC in the ratios n:1 and 1:n, respectively. PROVE that \triangle APQ and \triangle ABC have the same centroid
    • Can this be extended to all triangles?
  14. (Almost right) Let a,ar,ar^2,~r> 0, be the slopes of sides AB,BC,CA in \triangle ABC. PROVE that there are points X and Y on BC such that AB^2+AC^2=BC.XY.
    (Compare this with exercise 10 above.)

Thanksgiving

Today is a special day in this poster’s life. As such, he gives praise to the Owner of life.

Mind always goes back to that marked Thursday, June 14, 2018. In turn, it CAN`T hold back from giving thanks.

What happened was very basic (as simple as HTML), but its effect became “drastic” (actually, dramatic, dynamic), and so one’s gratitude is made public.

To the reader:

Hopefully this poster’s gratitude to the Heavenly Father doesn’t ruffle your feathers. If it does, please (don’t bother to) read further.

Wish:

May you find that “spark” you need to start what will lead you to succeed.