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USAMO 1972 #5 August 4, 2009

Posted by lumixedia in Problem-solving.
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USAMO 1972 #5. A given convex pentagon {ABCDE} has the property that the area of each of the five triangles {ABC}, {BCD}, {CDE}, {DEA}, {EAB} is unity. Show that every non-congruent pentagon with the above property has the same area, and that, furthermore, there are an infinite number of such non-congruent pentagons.

Solution. Let {P} be the intersection of {BD} and {EC}. Since {\triangle EDC} and {\triangle BCD} have the same area and the same base, they must also have the same height, so we know that {BE} is parallel to {CD}. Similarly, {EC} is parallel to {AB} and {BD} is parallel to {AE}, so {ABPE} is a parallelogram and {[PEB]=[AEB]=1}. Now let {[PED]=[PBC]=y}, so that {[PDC]=1-y}. We have

\displaystyle \frac{PD}{PB}=\frac{[PED]}{[PEB]}=\frac{y}{1}

but also

\displaystyle \frac{PD}{PB}=\frac{[PCD]}{[PCB]}=\frac{1-y}{y}

so we know {y/1=(1-y)/y}, or {y^2+y-1=0}. Solving the quadratic gives

\displaystyle y=\frac{\sqrt{5}-1}{2}

and

\displaystyle [ABCDE]=[AEB]+[PEB]+[PED]+[PBC]+[PCD]

\displaystyle =1+1+y+y+1-y=3+y=\frac{\sqrt{5}+5}{2}

so {ABCDE} has only one possible area, as desired.

We can create an infinite number of noncongruent pentagons with this property as follows. First we construct an arbitrary triangle {PCD} with area {1-y} and extend {DP} to {B} and {CP} to {E} so that {[DPE]=[CPB]=y}. Then we let {A} be the intersection of the line through {B} parallel to {EC} and the line through {E} parallel to {BD}.* Then we have {[EAB]=[CDE]=[BCD]=1} and {[ABCDE]=3+y}. To show that {ABCDE} has the desired property, we just need to prove that {[DEA]=[ABC]=1}. [Edit: see comments for much better ways to proceed from here.] Let {F} be the intersection of {AD} and {EC}. We can calculate

\displaystyle \frac{FP}{PE}=\frac{FP}{AB}=\frac{DP}{DB}=\frac{[DCP]}{[DCB]}=1-y

and

\displaystyle \frac{PE}{PC}=\frac{[PDE]}{[PDC]}=\frac{y}{1-y}

so

\displaystyle \frac{FP}{PC}=\frac{FP}{PE}\cdot\frac{PE}{PC}=(1-y)\cdot\frac{y}{1-y}=y

from which

\displaystyle [DFP]=\frac{FP}{PC}[DPC]=y(1-y).

Since {EC} and {AB} are parallel, we have {\triangle DFP\sim\triangle DAB}. So we can write

\displaystyle [DAB]=[DFP]\cdot\left(\frac{DB}{DP}\right)^2

\displaystyle=y(1-y)\cdot\frac{1}{(1-y)^2}=\frac{y}{1-y}=1+y

where the last equality comes from {y=1-y^2=(1-y)(1+y)}. We conclude that

\displaystyle [DEA]=[ABCDE]-[DAB]-[BCD]

\displaystyle=3+y-(1+y)-1=1

and, by symmetry, {[ABC]=1}. So {ABCDE} has the desired property.

*Note: the official solution stops here and proclaims that {ABCDE} satisfies the desired property. This means either that the solution writers completely forgot to check that {[DEA]=[ABC]=1}, or that there is a far more obvious reason why this is true than the proof I found. Help.

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Comments»

1. Tirasan Khandhawit - August 4, 2009

Seemingly, it is a tradition for textbooks to leave some exercises to the reader.

From the construction, it is clear to me that [DCB] = [DCE] =1 and [DEA] = [EAB] = [ABC] (from parallel lines). So we only need to show that [DCB] = [ABC] which is equivalent to AD // BC.

By above argument, we have

\displaystyle \frac{PF}{PC} = y = \frac{1-y}{y} = \frac{PD}{PB}.

Thus \triangle PDF \sim \triangle PBC and AD // BC.

In the real contest, one should also justify that ABCDE is convex. This might be trivial but we have to make sure that the construction is valid.

2. lumixedia - August 4, 2009

Oh…I didn’t see that we can get [DEA]=[EAB]=[ABC] from the parallel lines. Yeah, that pretty much makes the stuff I added unnecessary since we have [EAB]=[EPB]=(PB/PD)[DPE]=1 by construction. Oops. Thanks for the pointer!

\angle AED=\angle AEB+\angle BED=\angle PDC+\angle BED
<\angle EDC+\angle BED=180^{\circ}

seems like it should be sufficient to show that ABCDE is convex, I think. Does that look right?


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