First, we determine the lengths of the sides of the new triangle, in terms of $x$.

We drop perpendiculars from $X, Y$ and $Z$ to points $P, Q$ and $R$, respectively, on the line $A E$. Since each of the four triangles is isosceles, then $B P=P C=C Q=Q D=D R=R E=\frac{1}{2} x$.

Consider triangle $A R Z$, which is right-angled at $R$. Since

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-218.jpg?height=345&width=564&top_left_y=1069&top_left_x=1279)
$A Z=A E=4 x$, then by the Pythagorean Theorem,

$$
\begin{aligned}
A R^{2}+R Z^{2} & =A Z^{2} \\
R Z^{2} & =(4 x)^{2}-\left(\frac{7}{2} x\right)^{2} \\
R Z^{2} & =\frac{15}{4} x^{2}
\end{aligned}
$$

so the square of the height of each of the four isosceles triangles is $\frac{15}{4} x^{2}$.

Thus, $A Y^{2}=A Q^{2}+Q Y^{2}=\left(\frac{5}{2} x\right)^{2}+\frac{15}{4} x^{2}=10 x^{2}$, so $A Y=\sqrt{10} x$, and

$A X^{2}=A P^{2}+P X^{2}=\left(\frac{3}{2} x\right)^{2}+\frac{15}{4} x^{2}=6 x^{2}$, so $A X=\sqrt{6} x$.

Thus, the new triangle has side lengths $\sqrt{6} x, \sqrt{10} x$ and $4 x$. Since

$(\sqrt{6} x)^{2}+(\sqrt{10} x)^{2}=(4 x)^{2}$, then this new triangle is right-angled, with hypotenuse $4 x$, and so has area $\frac{1}{2}(\sqrt{6} x)(\sqrt{10} x)=\frac{1}{2} \sqrt{60} x^{2}=\sqrt{15} x^{2}$.

We would like the area to be less than 2004, so $\sqrt{15} x^{2}<2004$ or $x<\sqrt{\frac{2004}{\sqrt{15}}} \approx 22.747$. Therefore, the largest integer value of $x$ that works is 22 .
Four identical isosceles triangles $A W B, B X C, C Y D$, and $D Z E$ are arranged, as shown, with points $A, B, C, D$, and $E$ lying on the same straight line. A new triangle is formed with sides the same lengths as $A X, A Y$ and $A Z$. If $A Z=A E$, the largest integer value of $x$ such that the area of this new triangle is less than 2004 is
(A) 18
(B) 19
(C) 20
(D) 21
(E) 22

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-119.jpg?height=346&width=571&top_left_y=1714&top_left_x=1308)