2
$\begingroup$

$1.$ Point A and B lie on the sides of a square, segment AB divides the square into two polygons each of which has an inscribed circle. One of the circles has radius 6 cm while the other one is larger. What is the difference, in cm, between the side length of the square and twice the length of segment AB ?

What must a polygon meet in order for it to have a inscribed circle? I am having trouble in just finding the right figure to work with in this problem, I think I'll have no problem finding the answer after the figure, but complete solutions are welcome

$\endgroup$
  • 2
    $\begingroup$ the only configuration I can come up look likes this. The answer do not depends on the slope of the line and is $12$. $\endgroup$ – achille hui Jul 23 '18 at 14:40
2
$\begingroup$

Refer to the diagrams below. We focus on the bottom right quarter of the square.

enter image description here

The square has a side length $$2CM=CM+CN$$.

Now, $$=\begin{align*}\\&\implies CM+CN-2AB\\&\implies CM+CN-(AP+BP)-AB\\&\implies CM+CN-(AN+BM)-AB\\&\implies(CM-BM)+(CN-AN)-AB\\&\implies BC+AC-AB\\&\implies BC+AC-(AQ+BQ)\\&\implies BC+AC-(AE+BD)\\&\implies CD+CE=12\end{align*}$$ Note that we applied equal tangent segments repeatedly.

$\endgroup$
  • $\begingroup$ +1 very clever solution. $\endgroup$ – achille hui Jul 24 '18 at 8:43
0
$\begingroup$

Let the incircles have centers $R$ and $S$ and respective radii $r$ and $s$ (with $r < s$). Note that $s$ is half the side of the original square. Let $T$ be the the original square's vertex that completes right triangle $\triangle ABT$, and define $c:= |AB|$.

enter image description here

Zooming in ...

enter image description here

The four pairs of congruent right triangles allow us to write (naming squares by their diagonals) $$\begin{align} |\square ST| - |\square RT| &= 2\,|\triangle ABS| + 2\,|\triangle ABR| \tag{1} \\[6pt] s^2 - r^2 &= 2\cdot \frac12 sc + 2\cdot\frac12 rs \tag{2} \\[6pt] (s+r)(s-r) &= c(s+r) \tag{3} \\[6pt] s-r &= c \tag{4} \end{align}$$

The relation in the original question is merely $(4)$, "doubled". $\square$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.