## USAMO 1972 #5 August 4, 2009

Posted by lumixedia in Problem-solving.
Tags: , , , ,

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.

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?