Tangent line to quarter circle inscribed in square ABCD intersects point X. Where to start?

Question

Disclaimer: The answer must be an integer, as all competition problems were designed to yield integral answers.

Hello! Yesterday I underwent a math tournament. There was one problem that was rather difficult in my eyes, and that is the question I am bringing up today.

Let ABCD be a square with side length 24, and suppose M is the midpoint of line AB. Construct a circle centered at C through B and D, and let the tangent line through M to this circle (other than AB) intersect AD again at X. What is the length of AX?

Of course I have drawn ABCD and the circle centered at C as described above (using GeoGebra for my first time :) ). In the image please disregard the values on the X and Y axes (as they clearly do not match up with the length of 24, as seen in the problem).

How would I go about starting this problem? Do we try to find the length of MX and use the Pythagorean Theorem to find AX?

-Many thanks

Answer 1

Note that because $AB$ and $AD$ are also tangents to the circle with center $C$, we observe that $$MB = ME, \quad XE = XD.$$ Since we know $MA = MB = 12$, and $XA + XD = 24$, this immediately suggests we let $XA = z$, and by the Pythagorean theorem, $$z^2 + 12^2 = XA^2 + MA^2 = MX^2 = (ME + XE)^2 = (12 + 24-z)^2 = (36-z)^2.$$ Solving this equation for $z$ easily yields $$72z = 36^2 - 12^2 = (36+12)(36-12) = (48)(24),$$ hence $z = 16.$

Answer 2

Produce CE to cut the y-axis at K.

$\angle BCE =2 \times \tan^{-1} (\dfrac {1}{2}) = 53.xxx^0$. This means $\triangle BCK$ is an 8 times 3-4-5 triangle in disguise.

[Added: An alternate approach in getting the above result is:-

$\triangle KME \sim \triangle KCB \rightarrow \dfrac {KM}{KC} = \dfrac{ME}{CB} = \dfrac {EK}{BK}$

$∴\dfrac {y + 12}{t + 24} = \dfrac{12}{24} = \dfrac {t}{y + 24}$

$(1): \dfrac {y + 12}{t + 24} = \dfrac{1}{2}$

$(2): \dfrac {t}{y + 24} = \dfrac{1}{2}$

$∴ y = 8$]

By exterior angle of cyclic quadrilateral (BCEM), $\triangle AMX$ is an 4 times 3-4-5 triangle in disguise.

Therefore, AX = 4(4).