Generalizing a puzzle from Mind Your Decisions, here's something that I found to be rather neat.

Geometry Puzzle

Suppose that AB$=c$, AC$=b$, and BC$=a$. What's the perimeter of $\triangle$CDE?

Clue: The coveted tick will go to the most attractive, visual proofs/arguments that don't use any additional variables! Judges decision etc.

  • 7
    $\begingroup$ Do you mean $BC=a$ ? $\endgroup$
    – hexomino
    Feb 14, 2019 at 15:57

2 Answers 2


It is probably not something OP expects (with a lot of additional variables) but I would like to publish my idea anyway:

Proposed solution

The clue is to "build" |AD| and |BE| segments from fragments of |AB| and |DE|

(we can also use a known property of quadrilateral circumscribing circle which says that |AD| + |BE| = |AB| + |DE|)

We can derive that:

a + b = (|AD| + |DC|) + (|CE| + |EB|) = c + |DE| + |DC| + |EC|

so by subtracting c from both sides:

|DE| + |DC| + |EC| = a + b - c


Since $AB=BC=c$, and $AC=b$ we know this is an isosceles triangle (which makes the diagram a poor representation of the question). This means that the point where the inscribed circle touches $AC$ (the side not equal to the others) is the midway mark. Lets call this point $M$.

Since no guidance was given on $D$ and $E$, one can move $D$ all the way towards $M$. To maintain the tangent, $E$ then nears $C$. As $D$ approaches $M$, and $E$ approaches $C$, the triangle approaches an isosceles triangle whose two long sides are both $DC$ and whose short side is $0$. Thus, the perimeter of $CDE$ is $CD+CD=2CD$. Since $AC=c$ and $CD=CM$ is half of $AC$, we know $CD=\frac{c}{2}$.

Thus, The perimeter of $CDE=2CD=2\frac{c}{2}=c$.

  • $\begingroup$ This was probably correct, but the OP had actually made a typo. AB != BC, and hence, it is not an isosceles triangle. $\endgroup$
    – Aryaman
    Feb 16, 2019 at 7:00

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