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I stumbled upon an old problem on the site Gifted Mathematics. In the inscribed hexagon $ABCDEF$ of area $54$, we know that AB=CD=EF=5, BC=DE=2, AF=11. It is required to find the value of BE.

Applying Bramaguptha's formula to trapezoids ABEF and BCDE (and feeding it to Wolfram Alpha...), it is possible to find the answer, but the formulas are very ugly. Is there a smarter way to do it?

enter image description here

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  • $\begingroup$ It was a typo. It's a plain 2. $\endgroup$
    – mau
    Dec 19, 2017 at 16:30
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    $\begingroup$ It turns out that $\angle AOF=3\angle AOB$ (where $O$ is the circle center). And, conversely, from that relation one can find the solution. I wonder if that might be of help. $\endgroup$ Dec 19, 2017 at 23:18

2 Answers 2

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Ptolemy's theorem helps.

Let $|\overline{BE}|=x$.

Applying the theorem to the quadrilateral $CDEB$, we get $$|\overline{CD}|\cdot|\overline{BE}|+|\overline{DE}|\cdot|\overline{CB}|=|\overline{BD}|\cdot|\overline{CE}|$$ Since $|\overline{BD}|=|\overline{CE}|$, we get $$|\overline{BD}|=|\overline{CE}|=\sqrt{5x+4}$$ Since $|\overline{DF}|=|\overline{CE}|$, we get $$|\overline{DF}|=\sqrt{5x+4}\tag1$$

Applying the theorem to the quadrilateral $CDEF$, we get $$|\overline{CD}|\cdot|\overline{EF}|+|\overline{DE}|\cdot|\overline{CF}|=|\overline{CE}|\cdot|\overline{DF}|$$ so $$|\overline{CF}|=\frac{5x-21}{2}$$ Since $|\overline{AD}|=|\overline{CF}|$, we get $$|\overline{AD}|=\frac{5x-21}{2}\tag2$$ where we have to have $\frac{5x-21}{2}\gt 0,$ i.e. $$x\gt\frac{21}{5}\tag3$$

Applying the theorem to the quadrilateral $CDFA$, we get $$|\overline{CD}|\cdot|\overline{AF}|+|\overline{AC}|\cdot|\overline{DF}|=|\overline{AD}|\cdot|\overline{CF}|$$ Since $|\overline{AC}|=|\overline{DF}|$, we get $$55+|\overline{DF}|^2=|\overline{AD}|^2\tag4$$

From $(1)(2)(4)$, we have $$55+(5x+4)=\left(\frac{5x-21}{2}\right)^2$$ i.e. $$(x-1)(5x-41)=0$$ It follows from $(3)$ that $$|\overline{BE}|=x=\frac{41}{5}$$

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    $\begingroup$ Neat (+1). Also verifies that the given area value was both redundant and useless. $\endgroup$
    – dxiv
    Dec 20, 2017 at 6:07
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I'll leave a couple of observations below. Neither qualifies as "smarter", but may help someone find what must be the trick solution.

1.   The problem is overdetermined, in the sense that the figure can be solved without knowing the area. For example, let $r$ be the radius, and $2a,2b$ the center angles subtended by $AB,BC\,$, then:

$$ 2r\sin(a)=5 \\ 2r \sin(b) = 2 \\ 2r \sin(3a+2b)=11 $$

The system can technically be solved for $r, a, b\,$, and the rest follows.

2.   Let $BE=x\,$, then adding the areas of the two trapezoids gives:

$$ \frac{1}{2}\left(x+11\right)\sqrt{5^2 - \frac{1}{4}(11-x)^2} + \frac{1}{2}\left(x+5\right)\sqrt{2^2 - \frac{1}{4}(x-5)^2}= 54 $$

After not so pretty calculations, the above reduces to a quartic equation:

$$ \begin{align} 0 &\;=\; 1825 x^4 + 7820 x^3 - 244362 x^2 - 1594900 x + 16946161 \\ &\;=\; (5 x - 41)^2 \,(73 x^2 + 1510 x + 10081) \end{align} $$

The unique real root is obviously the answer, but what's more interesting is that it's a double root. This would seem to suggest that the chosen numbers represent some sort of "limit case", and once it is recognized what is so special about it, that would likely point to the better solution.

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