After the mind-boggling Dissecting Square puzzle, here is yet another geometrical puzzle. But a lot easier.

In the figure $AE = 111$ and other lengths are unknown. What is the value of $AB^2 + BC^2 + CD^2 + DE^2?$


Most elegant solution will be accepted. Cheers.
  • 1
    $\begingroup$ The top voted answer in the no-math-textbook meta question explicitly allows for math problems that have a clever or elegant, puzzlingly-type solution. Since this puzzle is very easily solved by basic geometry, I’m very interested in seeing some creative solutions, especially the one OP had in mind. Voting to reopen. $\endgroup$ – Bass Jan 7 '18 at 11:15
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    $\begingroup$ @Bass, I agree. $\endgroup$ – Seyed Jan 7 '18 at 12:02
  • $\begingroup$ I disagree. The "obvious" solution is the simplest one, and the chance of a more rational chain of thought is exceedingly low. In other words, this is not a complicated enough "puzzle" to have the freedom of multiple instructive solutions. $\endgroup$ – greenturtle3141 Jan 8 '18 at 4:58

By pythagorean principle,

$AB^2 + BC^2 + CD^2 + DE^2$
$= AC^2 + CD^2 + DE^2$
$= AD^2 + DE^2$
$= AE^2$
$= 111^2 = 12321$


Answer is

12321, the square of 111.

Logic: Applying the Pythagorean theorem on the right-angled triangles, we get to know that
$AB^2 + BC^2 = AC^2$
$AC^2 + CD^2 = AD^2$
$AD^2 + DE^2 = AE^2$

Substituting the right hand side of the above set of equations (one after the other) in the the sum $AB^2 + BC^2 + CD^2 + DE^2$, and using associativity of addition, we get: $AB^2 + BC^2 + CD^2 + DE^2 = (((AB^2 + BC^2) + CD^2) + DE^2) = (AC^2 + CD^2) + DE^2 = AD^2 + DE^2 = AE^2$
Finally by substituting the value of the length of AE, 111, we get the answer 12321.


Solving with pythagorean theorem:

\begin{align} 111^2=& & & & &AD^2&+&DE^2\\ =& & &AC^2&+&CD^2&+&DE^2\\ =&AB^2&+&BC^2&+&CD^2&+&DE^2 \end{align}


For any two points, you can replace the line segment between them with a path consisting of any two connected perpendicular line segments, and the sum of squares of the segments will not change. (Pythagoras' theorem.)

The diagram shows three such replacements, starting with $AE \xrightarrow{} AD+DE$, followed by $AD \xrightarrow{} AC+CD$ and $AC \xrightarrow{} AB+BC$.

Since $AE^2=12321$, and $AB^2+BC^2+CD^2+DE^2 $ is the resulting sum of squares after the three replacements, $AB^2+BC^2+CD^2+DE^2$ must equal $12321$.


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