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

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

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 $z = 16.$

• How can we determine that MB = ME and XE = XD? I do not see how the observation supports that. – Mario Ishac May 18 '16 at 18:01
• @MarDev The two tangent segments drawn from any given point outside of a circle are of equal length. In other words, draw a circle. Pick a point outside of the circle. Draw the two external tangent lines to the circle that pass through the point. The distances from the points of tangency on the circle to the given point are equal. – heropup May 18 '16 at 18:12
• Yes, but the tangent segments are not originating from the same point in the diagram above. X D and A are all different points. – Mario Ishac May 18 '16 at 21:31
• @MarDev $MB = ME$ because $\overline{MB}$ and $\overline{ME}$ are tangent from the same point $M$ to the points of tangency $B$ and $E$, respectively. $XE = XD$ because $\overline{XE}$ and $\overline{XD}$ are tangent from the same point $X$ to the points of tangency $E$ and $D$. – heropup May 18 '16 at 21:39
• Ah, now I see. Thank you for your help, marking this answer as accepted. Sorry that it didn't come so obvious to me :P – Mario Ishac May 18 '16 at 22:23

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).

• How does CE cut the Y axis? My mistake that I forgot to include the use of no calculator. Your solution relies on some calculator work, How would I approach this problem without a calculator? – Mario Ishac May 16 '16 at 22:49
• @MarDev Let KA = y and KE = t. From the fact that $\triangle KME \sim \triangle KBC$, setup the corresponding ratios. Two simultaneous equations in y and t are obtained and hence y can be found. – Mick May 17 '16 at 5:25
• @MarDev Correction:- $\triangle KME \sim \triangle KCB$ – Mick May 17 '16 at 5:35
• Still don't understand where point K is being formed, could you elaborate on that? – Mario Ishac May 17 '16 at 18:14
• @MarDev I have modified my post with more details in the added. – Mick May 18 '16 at 5:31