The pentagram revisited

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The pentagram revisited

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Source: problem 14 (G4) of QEDMO 1; created by myself

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#1 h Nov 7, 2005, 8:14 PM • 2 Y

Y by Adventure10, Mango247

In the following, the abbreviation $g \cap h$ will mean the point of intersection of two lines $g$ and $h$ .

Let $ABCDE$ be a convex pentagon. Let $A^{\prime}=BD\cap CE$ , $B^{\prime}=CE\cap DA$ , $C^{\prime}=DA\cap EB$ , $D^{\prime}=EB\cap AC$ and $E^{\prime}=AC\cap BD$ . Furthermore, let $A^{\prime\prime}=AA^{\prime}\cap EB$ , $B^{\prime\prime}=BB^{\prime}\cap AC$ , $C^{\prime\prime}=CC^{\prime}\cap BD$ , $D^{\prime\prime}=DD^{\prime}\cap CE$ and $E^{\prime\prime}=EE^{\prime}\cap DA$ .

Prove that:
$\frac{EA^{\prime\prime}}{A^{\prime\prime}B}\cdot\frac{AB^{\prime\prime}}{B^{\prime\prime}C}\cdot\frac{BC^{\prime\prime}}{C^{\prime\prime}D}\cdot\frac{CD^{\prime\prime}}{D^{\prime\prime}E}\cdot\frac{DE^{\prime\prime}}{E^{\prime\prime}A}=1.$

Darij

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#2 h Nov 8, 2005, 6:05 PM • 1 Y

Y by Adventure10

my solution is very ugly.
I used more than eight times the menelau's theorem

anyone have one simple solution?

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#3 h Nov 9, 2005, 1:35 AM • 2 Y

Y by Adventure10, Mango247

Take a look in there for additional identities.

http://mathworld.wolfram.com/HoehnsTheorem.html

Sorry Darij, but they beat you to it.

Was this problem the thing that inspired you for a construction problem you posted

some time ago?

Thank you.

M.T.

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#4 h Nov 9, 2005, 12:45 PM • 2 Y

Y by Adventure10, Mango247

armpist wrote:

Take a look in there for additional identities.

http://mathworld.wolfram.com/HoehnsTheorem.html

Thanks for the link! In fact, these identities are what inspired me to create this problem, I just forgot their name (Hoehn's identities). As for the construction problem you are talking about, in case you meant http://www.mathlinks.ro/Forum/viewtopic.php?t=54608 : no, it was not inspired by any pentagram identities, and I actually don't see how to apply them in the problem. (It was inspired by the problem posted at http://www.mathlinks.ro/Forum/viewtopic.php?t=38828 , but, unlike the latter one, is solvable).

As for the solution, yes, the problem is really tricky if one tries to solve it by Menelaos/Ceva only (such a solution is possible, but if I would have written it down, it would probably take 4-5 pages).

Here is my "official" solution of the problem:

After the sine law in triangle $EA^{\prime \prime }A^{\prime }$ , we have

$\frac{EA^{\prime \prime }}{A^{\prime }A^{\prime \prime }}=\frac{\sin\measuredangle EA^{\prime }A^{\prime \prime }}{\sin \measuredangle A^{\prime}EA^{\prime \prime }}=\frac{\sin \measuredangle EA^{\prime }A}{\sin\measuredangle BEC}$ .

After the sine law in triangle $AA^{\prime }E$ , we have

$\frac{\sin \measuredangle EA^{\prime }A}{\sin \measuredangle AEA^{\prime }}=\frac{EA}{AA^{\prime }}$ , thus $\sin \measuredangle EA^{\prime }A=\sin \measuredangle AEA^{\prime }\cdot \frac{EA}{AA^{\prime }}=\sin \measuredangle AEC\cdot \frac{EA}{AA^{\prime }}$ .

Hence,

$\frac{EA^{\prime \prime }}{A^{\prime }A^{\prime \prime }}=\frac{\sin\measuredangle EA^{\prime }A}{\sin \measuredangle BEC}=\frac{\sin\measuredangle AEC\cdot \frac{EA}{AA^{\prime }}}{\sin \measuredangle BEC}=\frac{EA\cdot \sin \measuredangle AEC}{AA^{\prime }\cdot \sin \measuredangle BEC}$ .

Similarly,

$\frac{BA^{\prime \prime }}{A^{\prime }A^{\prime \prime }}=\frac{AB\cdot \sin\measuredangle DBA}{AA^{\prime }\cdot \sin \measuredangle DBE}$ .

Thus,

$\frac{EA^{\prime \prime }}{BA^{\prime \prime }} =\frac{EA^{\prime \prime }}{A^{\prime }A^{\prime \prime }}: \frac{BA^{\prime \prime }}{A^{\prime}A^{\prime \prime }}=\frac{EA\cdot \sin \measuredangle AEC}{AA^{\prime}\cdot \sin \measuredangle BEC}: \frac{AB\cdot \sin \measuredangle DBA}{AA^{\prime }\cdot \sin \measuredangle DBE}$
$=\frac{EA}{AB}\cdot \frac{\sin \measuredangle AEC}{\sin \measuredangle DBA}\cdot \frac{\sin \measuredangle DBE}{\sin \measuredangle BEC}$ .

Analogously,

$\frac{AB^{\prime \prime }}{CB^{\prime \prime }} =\frac{AB}{BC}\cdot \frac{\sin \measuredangle BAD}{\sin \measuredangle ECB}\cdot \frac{\sin\measuredangle ECA}{\sin \measuredangle CAD}$ ;
$\frac{BC^{\prime \prime }}{DC^{\prime \prime }} =\frac{BC}{CD}\cdot \frac{\sin \measuredangle CBE}{\sin \measuredangle ADC}\cdot \frac{\sin\measuredangle ADB}{\sin \measuredangle DBE}$ ;
$\frac{CD^{\prime \prime }}{ED^{\prime \prime }} =\frac{CD}{DE}\cdot \frac{\sin \measuredangle DCA}{\sin \measuredangle BED}\cdot \frac{\sin\measuredangle BEC}{\sin \measuredangle ECA}$ ;
$\frac{DE^{\prime \prime }}{AE^{\prime \prime }} =\frac{DE}{EA}\cdot \frac{\sin \measuredangle EDB}{\sin \measuredangle CAE}\cdot \frac{\sin\measuredangle CAD}{\sin \measuredangle ADB}$ .

Therefore,

$\frac{EA^{\prime \prime }}{BA^{\prime \prime }}\cdot \frac{AB^{\prime\prime }}{CB^{\prime \prime }}\cdot \frac{BC^{\prime \prime }}{DC^{\prime\prime }}\cdot \frac{CD^{\prime \prime }}{ED^{\prime \prime }}\cdot \frac{DE^{\prime \prime }}{AE^{\prime \prime }}$
$=\left( \frac{EA}{AB}\cdot \frac{\sin \measuredangle AEC}{\sin\measuredangle DBA}\cdot \frac{\sin \measuredangle DBE}{\sin \measuredangle BEC}\right) \cdot \left( \frac{AB}{BC}\cdot \frac{\sin \measuredangle BAD}{\sin \measuredangle ECB}\cdot \frac{\sin \measuredangle ECA}{\sin \measuredangle CAD}\right)$
$\cdot \left( \frac{BC}{CD}\cdot \frac{\sin \measuredangle CBE}{\sin \measuredangle ADC}\cdot \frac{\sin \measuredangle ADB}{\sin \measuredangle DBE}\right) \cdot \left( \frac{CD}{DE}\cdot \frac{\sin \measuredangle DCA}{\sin \measuredangle BED}\cdot \frac{\sin \measuredangle BEC}{\sin \measuredangle ECA}\right)$
$\cdot \left( \frac{DE}{EA}\cdot \frac{\sin \measuredangle EDB}{\sin \measuredangle CAE}\cdot \frac{\sin \measuredangle CAD}{\sin \measuredangle ADB}\right)$ .

Most of the fractions in this product cancel each other; what remains is:

$\frac{EA^{\prime \prime }}{BA^{\prime \prime }}\cdot \frac{AB^{\prime\prime }}{CB^{\prime \prime }}\cdot \frac{BC^{\prime \prime }}{DC^{\prime\prime }}\cdot \frac{CD^{\prime \prime }}{ED^{\prime \prime }}\cdot \frac{DE^{\prime \prime }}{AE^{\prime \prime }}$
$=\frac{\sin \measuredangle AEC}{\sin \measuredangle DBA}\cdot \frac{\sin \measuredangle BAD}{\sin \measuredangle ECB}\cdot \frac{\sin \measuredangle CBE}{\sin \measuredangle ADC}\cdot \frac{\sin \measuredangle DCA}{\sin \measuredangle BED}\cdot \frac{\sin \measuredangle EDB}{\sin \measuredangle CAE}$
$=\frac{\sin \measuredangle AEC}{\sin \measuredangle CAE}\cdot \frac{\sin\measuredangle BAD}{\sin \measuredangle DBA}\cdot \frac{\sin \measuredangle CBE}{\sin \measuredangle ECB}\cdot \frac{\sin \measuredangle DCA}{\sin \measuredangle ADC}\cdot \frac{\sin \measuredangle EDB}{\sin \measuredangle BED}$ .

Now, $\frac{\sin \measuredangle AEC}{\sin \measuredangle CAE}=\frac{AC}{CE}$ by the sine law in triangle $ACE$ , and similarly $\frac{\sin \measuredangle BAD}{\sin \measuredangle DBA}=\frac{BD}{DA}$ , $\frac{\sin \measuredangle CBE}{\sin \measuredangle ECB}=\frac{CE}{EB}$ , $\frac{\sin \measuredangle DCA}{\sin \measuredangle ADC}=\frac{DA}{AC}$ and $\frac{\sin \measuredangle EDB}{\sin \measuredangle BED}=\frac{EB}{BD}$ . Thus,

$\frac{EA^{\prime \prime }}{BA^{\prime \prime }}\cdot \frac{AB^{\prime \prime}}{CB^{\prime \prime }}\cdot \frac{BC^{\prime \prime }}{DC^{\prime \prime }}\cdot \frac{CD^{\prime \prime }}{ED^{\prime \prime }}\cdot \frac{DE^{\prime\prime }}{AE^{\prime \prime }}=\frac{AC}{CE}\cdot \frac{BD}{DA}\cdot \frac{CE}{EB}\cdot \frac{DA}{AC}\cdot \frac{EB}{BD}=1$ ,

and the problem is solved.

Darij

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