Let $a,b,c,d,e$ be non-negative real numbers such that $a+b+c+d+e>0$. What is the least real number $t$ such that $a+c=tb$, $b+d=tc$, $c+e=td$?

$ \textbf{(A)}\ \frac{\sqrt 2}2
\qquad\textbf{(B)}\ 1
\qquad\textbf{(C)}\ \sqrt 2
\qquad\textbf{(D)}\ \frac32
\qquad\textbf{(E)}\ 2
$
1. We start with the given equations:
   \[
   a + c = tb, \quad b + d = tc, \quad c + e = td
   \]
   and the condition \(a + b + c + d + e > 0\).

2. We need to find the least real number \(t\) such that these equations hold. Let's analyze the equations step by step.

3. First, observe that if \(c = 0\), then from the first equation \(a + 0 = tb\), we get \(a = tb\). Similarly, from the second equation \(b + d = t \cdot 0\), we get \(b + d = 0\), which implies \(b = 0\) and \(d = 0\). From the third equation \(0 + e = td\), we get \(e = 0\). This would imply \(a + b + c + d + e = 0\), which contradicts the given condition \(a + b + c + d + e > 0\). Therefore, \(c > 0\).

4. Now, let's manipulate the given equations to find a relationship involving \(t\). From the first equation, solve for \(a\):
   \[
   a = tb - c
   \]
   From the second equation, solve for \(d\):
   \[
   d = tc - b
   \]
   From the third equation, solve for \(e\):
   \[
   e = td - c
   \]

5. Substitute \(d = tc - b\) into \(e = td - c\):
   \[
   e = t(tc - b) - c = t^2c - tb - c
   \]

6. Now, consider the sum \(a + c + c + e\):
   \[
   a + c + c + e = (tb - c) + c + c + (t^2c - tb - c) = t^2c
   \]
   Since \(a + c + c + e = t^2c\), we have:
   \[
   t^2c \geq 2c
   \]
   Dividing both sides by \(c\) (since \(c > 0\)):
   \[
   t^2 \geq 2 \implies t \geq \sqrt{2}
   \]

7. To check if \(t = \sqrt{2}\) is possible, substitute \(t = \sqrt{2}\) into the original equations:
   \[
   a + c = \sqrt{2}b, \quad b + d = \sqrt{2}c, \quad c + e = \sqrt{2}d
   \]
   Let \(a = 0\), \(e = 0\), and \(c = b\sqrt{2} = d\sqrt{2}\). This satisfies all the given conditions:
   \[
   a + c = 0 + c = \sqrt{2}b \implies c = \sqrt{2}b
   \]
   \[
   b + d = b + d = \sqrt{2}c \implies d = \sqrt{2}c - b = \sqrt{2}(\sqrt{2}b) - b = b
   \]
   \[
   c + e = c + 0 = \sqrt{2}d \implies c = \sqrt{2}d \implies c = \sqrt{2}b
   \]
   Thus, \(t = \sqrt{2}\) is indeed possible.

The final answer is \(\boxed{\sqrt{2}}\).