Step 1: Ratio of GH and DE
Similar Right-Angle Triangles
The triangles in the Crossing Ladders Problem are all right
triangles. A right triangle means a triangle whose base forms a 90 degree
angle from its height. Since the problem assumes that the
"walls" and "ruler" are 90 degrees from the
"floor", then the triangles which concern us are all right
Several of the triangles in the Crossing Ladders Problem are
similar. As taught in basic trigonometry, two triangles are similar if
they are equally proportional to each other. This means, for example, that
a triangle's height is proportional to the height of a similar triangle, as are
the triangle's base and hypotenuse.
In the Crossing Ladders Problem, we will concentrate on two sets
of three similar triangles, which is a total of six different triangles. Note
also that each of these four triangles is a right triangle.
Triangles AOB and OED
The first set
of similar triangles includes the two triangles shown in the following diagram, in which the blue
triangle is similar to the red triangle, which defines the triangles AOB and OED.
Because the triangles AOB (blue) and OED (red) are similar, we
may establish the ratios between the sides of these two triangles. First,
we first note the ratio in the following diagram between the blue line (AO) and
the red line (EO), which we can express as (AO/EO).
ratio (AO/EO), as shown in the diagram above, is the same ratio, as shown in the
diagram below, between the blue line (OB) and the red
line (DE), which we can express as (OB/DE):
The ratios (AO/EO) and (OB/DE), diagrammed above, are the same
ratio, as shown in the diagram below, between
the blue line (AB) and the red line (DO), which we can express as (AB/DO):
In summary, the triangles AOB (red) and OED (blue) are similar,
so the ratios of their sides are equal, as follows:
(AO/EO) = (BO/DE) = (AB/DO)
Triangles COB and OGH
The second set of similar triangles COB (blue) and OGH (red) is shown in the following
diagram where, again, the triangles are similar.
Again, remember that similar triangles, as shown in the above
diagrams, have proportional bases, heights, and hypotenuses. Therefore, as
we did above, we construct the ratios of the sides of the triangles COB and OGH.
that the triangles COB (blue) and OGH (red) have the following ratios of their
(CO/OG) = (OB/GH) = (CB/OH)
We can construct a solution to the Crossing Ladders Problem by
solving first either for line (AB) or for line (BC) -- using either line to
start the solution will lead to the complete solution. In this case, we elected to begin with line (AB).
From the ratios
above, we know that line (AB) is similar to line (DO). Thus:
considering the first set of triangles, AOB and OED, we know that the line AB
equals the line DO multiplied by the ratio described above, (OB/DE).
Therefore, our first basic equation is:
AB = OD x (OB / DE)
||You should stop here to
be sure to understand this formula. Notice that |AB| equals |OD|
multiplied by the ratio (|OB| / |ED|) which was derived above.
Now we will make the same calculation with the other set of
triangles, COB and OGH. However, we're not looking for the equation for
the line CB but for the line AB, so these calculations are more difficult that
those above. The analysis
is difficult but straightforward.
that the line CB equals the line OH multiplied by the ratio described above, (OB/GH).
CB = OH * (OB / GH)
Now we have a problem: can we find a way to transform the
equation above so that is solves for the line AB rather than for the line
CB. Look at the diagram above. We start by transposing the equation
above by multiplying both sides by (GH):
GH * CB = OH * OB
Now divide both sides by OB:
GH * CB / OB = OH
If we reverse sides and rearrange the terms, we have:
OH = CB * (GH / OB)
Now we are close to our goal of getting an equation whose terms
include only (AB), (CB), (GH) and (OB). First, notice that line (CB) is
exactly the same as line (OD). We can thus substitute:
OH = OD * (GH / OB)
Next, we notice that line (OH) is exactly the same as line
(AB). After we make this substitution, we have the second basic equation:
AB = OD * (GH / OB)
From this point onward, only algebra is needed to determine the
length of the line AC. Equations (1) and (2) can be rearranged, as
OD x (OB / DE) = OD x (GH / OB)
This simplifies, first by dropping OD:
(OB / DE) = (GH / OB)
Next, multiply by OB:
OB2 / DE = GH
and reverse the terms:
GH = OB2 / DE
We know that the line OB equals 5, so:
GH = 52 / DE
This completes Step 1, and we now have the ratio between GH
GH = 25 /
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