AIME 1987 · 第 6 题

Solution

Solution 1

Since $XY = WZ$, $PQ = PQ$ and the areas of the trapezoids $PQZW$ and $PQYX$ are the same, then the heights of the trapezoids are the same. Thus both trapezoids have area $\frac{1}{2} \cdot \frac{19}{2}(XY + PQ) = \frac{19}{4}(XY + 87)$. This number is also equal to one quarter the area of the entire rectangle, which is $\frac{19\cdot AB}{4}$, so we have $AB = XY + 87$.

In addition, we see that the perimeter of the rectangle is $2AB + 38 = XA + AD + DW + WZ + ZC + CB + BY + YX = 4XY$, so $AB + 19 = 2XY$.

Solving these two equations gives $AB = \boxed{193}$.

Solution 2

Let $YB=a$, $CZ=b$, $AX=c$, and $WD=d$. First we drop a perpendicular from $Q$ to a point $R$ on $BC$ so $QR=h$. Since $XY = WZ$ and $PQ = PQ$ and the areas of the trapezoids $PQZW$ and $PQYX$ are the same, the heights of the trapezoids are both $\frac{19}{2}$.From here, we have that $[BYQZC]=\frac{a+h}{2}*19/2+\frac{b+h}{2}*19/2=19/2* \frac{a+b+2h}{2}$. We are told that this area is equal to $[PXYQ]=\frac{19}{2}* \frac{XY+87}{2}=\frac{19}{2}* \frac{a+b+106}{2}$. Setting these equal to each other and solving gives $h=53$. In the same way, we find that the perpendicular from $P$ to $AD$ is $53$. So $AB=53*2+87=\boxed{193}$

Solution 3

Since $XY = YB + BC + CZ = ZW = WD + DA + AX$. Let $a = AX + DW = BY + CZ$. Since $2AB - 2a = XY = WZ$, then $XY = AB - a$.Let $S$ be the midpoint of $DA$, and $T$ be the midpoint of $CB$. Since the area of $PQWZ$ and $PQYX$ are the same, then their heights are the same, and so $PQ$ is equidistant from $AB$ and $CD$. This means that $PS$ is perpendicular to $DA$, and $QT$ is perpendicular to $BC$. Therefore, $PSCW$, $PSAX$, $QZCT$, and $QYTB$ are all trapezoids, and $QT = (AB - 87)/$2. This implies that

$$((a + 2((AB - 87)/2)/2) \cdot 19 = (((AB - a) + 87)/2) \cdot 19$$ $$(a + AB - 87) = AB - a + 87$$ $$2a = 174$$ $$a = 87$$ Since $a + CB = XY$, $XY = 19 + 87 = 106$, and $AB = 106 + 87 = \boxed{193}$.