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Let $a$ and $b$ be any two (distinct, positive) numbers. We can interpret the arithmetic mean (AM) and the harmonic mean (HM) of these numbers as line segments parallel to the bases of a trapezoid of lengths $a$ and $b$: specifically, the AM segment passes through the midpoints of the legs; the HM segment passes through the intersection of the diagonals.

How does one show, geometrically, that the corresponding parallel line segment representing the geometric mean (GM) of $a$ and $b$ lies between the line segments representing the other two means?

Moreover, how do we show that the HM segment is closer to the trapezoid's shorter base, while the AM segment is closer to the longer base? That is, how can we use this trapezoid interpretation to establish the inequality "$\text{HM} < \text{GM} <\text{AM}$"?

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  • $\begingroup$ You might want to specify $a\neq b$ so that the spacing between those "line segments parallel to the bases" is uniquely determined. $\endgroup$
    – hardmath
    Jul 25, 2016 at 21:46
  • $\begingroup$ cut-the-knot.org/pythagoras/GeometricMean.shtml $\endgroup$ Jul 25, 2016 at 22:35
  • $\begingroup$ @Jack D'Aurizio That is an interesting link, but it is not a response to my post. I am looking for a "line segment" in a trapezoid equal in length to the geometric mean of the bases. $\endgroup$
    – user74973
    Jul 27, 2016 at 14:55

2 Answers 2

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I'll assume that the lengths $a$ and $b$ are unequal, for otherwise the three means are all the same. For convenience in describing the constructions, assume without loss of generality that $a < b$.

The difficulty with this problem in my opinion is that while the constructions of the arithmetic and harmonic means have simple constructions that directly use the form of the trapezoid itself (namely, the midpoints of the lateral sides and the intersection of the diagonals, respectively), the geometric mean does not.

There is a relatively simple construction of the geometric mean of the two bases of a trapezoid on the page linked here. In first attempting to answer this question I followed a very similar construction (although with different labels for the points), but I choose now to use a variation of the construction that makes use of some steps that we would otherwise have to add later in order to show that the geometric mean is between the other two means. My construction, however, will still use the same basic idea as the other construction, which is to first find a segment of length $\sqrt{ab}$ (the geometric mean) somewhere on the plane, then find an equal segment that is parallel to $\overline{AB}$ and $\overline{CD}$ and has endpoints on $\overline{BC}$ and $\overline{AD}$.

In order to find a segment of length $\sqrt{ab}$, the first step is to construct a segment of length $a$ adjacent to segment $\overline{CD}$. One way to do this is to construct the midpoint $M$ of side $\overline{AD}$ and extend the lines $\overleftrightarrow {BM}$ and $\overleftrightarrow {CD}$ until they intersect at $E$. The point $D$ now divides the segment $\overline{CE}$ into segments of lengths $CD=b$ and $DE=a$. We construct a semicircle using $\overline{CE}$ as its diameter, and extend a line perpendicular to $\overline{CE}$ through $D$ until it intersects the semicircle at $F$. Then $DF = \sqrt{ab}$. We have then already constructed the geometric mean of $a$ and $b$, although not in the location in the plane where we ultimately want to construct it.

To construct an equal-length segment at the desired location, we can use a compass to construct a point $G$ on segment $\overline{CD}$ such that $DG = DF$. (We know there is such a point between $C$ and $D$ because $a < b$ and therefore $DF = \sqrt{ab} < b = CD$.) Next, we find the intersection of segment $\overline{AG}$ with diagonal $\overline{BD}$ at $K$. Finally, we construct the line through $K$ parallel to $\overline{AB}$ and $\overline{CD}$ and find its points of intersection $P$ and $Q$ with the sides of the trapezoid $\overline{AD}$ and $\overline{BC}$, respectively.

figure constructing three means of the bases of a trapezoid

Now triangles $\triangle ABK$ and $\triangle GDK$ are similar, and their bases $AB$ and $DG$ (hence also their corresponding altitudes, and the altitudes of trapezoids $ABQP$ and $CDPQ$) are in the ratio $a:\sqrt{ab}$. Therefore $AB:PQ = PQ:CD = a:\sqrt{ab}$, and $PQ = \sqrt{ab}$; we have constructed the geometric mean of $a$ and $b$ at the desired location.

For the next step (showing that $\frac{2ab}{a+b} < \sqrt{ab} < \frac{a+b}{2}$), we construct point $H$ on segment $\overline{CD}$ such that $DH = AB$. (We can do this by drawing a compass arc from $E$ around center $D$ until the arc intersects $\overline{CD}$ at $H$.) Since $a < \sqrt{ab} < b$, we have $DH < DG < CD$, and therefore $G$ is between $C$ and $H$.

Now let $J$ be the intersection of segment $\overline{AC}$ with diagonal $\overline{BD}$, and construct segment $\overline{MN}$ through $J$ with endpoint $N$ on side $\overline{BC}$ of the trapezoid. Then $\overline{MN}$ is parallel to $\overline{AB}$ and $\overline{CD}$ (one reason is that $M$ and $J$ bisect $AD$ and $AH$), and $MN$ is the arithmetic mean of $a$ and $b$.

Next, let $L$ be the intersection of diagonal $overline{AC}$ with diagonal $overline{BD}$, and construct segment $\overline{RS}$ through $J$ parallel to $\overline{AB}$ and $\overline{CD}$ so that the endpoints $R$ and $S$ are on the trapezoid's sides $\overline{AD}$ and $\overline{BC}$, respectively. Then $RS$ is the harmonic mean of $a$ and $b$.

Now the points $J$, $K$, and $L$ all lie along the diagonal $\overline{BD}$, and they are on the line segments from $A$ to (respectively) $H$, $G$, and $C$. But $H$, $G$, and $C$ occur in that order along $\overline{CD}$, and therefore $J$, $K$, and $L$ also occur in that order along $\overline{BD}$, and the segment $\overline{PQ}$ through $K$, representing the geometric mean, lies between the segments representing the harmonic and arithmetic means.

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  • $\begingroup$ There is one small edit to make. At the beginning of your construction, you need to specify that $a < b$. You say that $G$ is on the same side of $D$ as $C$ and that $|DG| = |DF|$. $\endgroup$
    – user74973
    Jul 30, 2016 at 16:17
  • $\begingroup$ @user74973 I considered making the assumption $a<b$ earlier than I did, but I noticed there are two ways that "$C$ and $G$ are on the same side of $D$" might be true: if the points lie along a single line in the sequence $DGC$ or if they lie on the line in the sequence $DCG$. The construction works in either case as long as $DG = \sqrt{ab}$. On the other hand, if I made the assumption $a<b$ at the beginning rather than later, I could just say $G$ is between $C$ and $D$. I'm considering it. $\endgroup$
    – David K
    Jul 30, 2016 at 22:12
  • $\begingroup$ @user74973 In the end I decided that it is indeed simpler to just assume $a<b$ initially, so I made that suggested edit. But I also changed a number of other details of the construction in a way that I think is better suited to showing the desired result. $\endgroup$
    – David K
    Jul 31, 2016 at 3:57
  • $\begingroup$ @David K I appreciate you taking the time to provide such an excellent explanation. Apparently, several other members of this web site do too. $\endgroup$
    – user74973
    Aug 6, 2016 at 15:18
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In a trapezoid with bases of lengths "a" and "b," the geometric mean, "G," is the length of the segment that is parallel to the bases and that also divides the trapezoid into two similar trapezoids.

Source: http://jwilson.coe.uga.edu/emt668/emat6680.2000/umberger/EMAT6690smu/Essay3smu/Essay3smu.html

So if you take that for granted, it must be between the two bases so that it can divide the trapezoid.

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    $\begingroup$ I think the Question asks for something stronger. Having identified the "heights" of line segments (parallel to the bases) representing the arithmetic mean and the harmonic mean, the problem is to show the one representing the geometric mean has a height intermediate to those two segments. $\endgroup$
    – hardmath
    Jul 25, 2016 at 21:48
  • $\begingroup$ Oh, I see. I misinterpreted "first two line segments" as "a" and "b". $\endgroup$
    – kcborys
    Jul 25, 2016 at 21:51
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    $\begingroup$ Sorry, Captain, the crucial statement in the web page by Shannon you link to is that the geometric mean line "also divides the trapezoid into two similar trapezoids." Yet any line parallel to the bases divides the trapezoid into two similar trapezoids. The construction he liniks to is correct, but the statement made is meaningless. So this does not answer the question. $\endgroup$ Jul 25, 2016 at 22:24
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    $\begingroup$ @MarkFischler: "Yet any line parallel to the bases divides the trapezoid into two similar trapezoids." This is not true. Certainly, any parallel line divides the trapezoid into sub-trapezoids whose corresponding angles match, but rarely does the line create sub-trapezoids whose edge-lengths are in the proportions that make them similar. $\endgroup$
    – Blue
    Jul 25, 2016 at 22:47
  • $\begingroup$ @Captain I have looked at the construction of the line segment as described in the web site you referenced. That is clever. For now, I can grant that the line segment representing the harmonic mean of the two bases in the given trapezoid is closer to the smaller base than the line segment representing the average (arithmetic mean) of the two bases. I can grant that there is only one line segment parallel to the bases that divides the given trapezoid into two smaller, similar trapezoids. What is the argument that such a line segment is between the other two line segments? $\endgroup$
    – user74973
    Jul 29, 2016 at 21:26

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