Common tangent to diameter circles

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Common tangent to diameter circles

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Source: Spain MO 2025 P2

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#1 h Mar 31, 2025, 1:06 PM • 1 Y

Y by AlexCenteno2007

The cyclic quadrilateral $ABCD$ , inscribed in the circle $\Gamma$ , satisfies $AB=BC$ and $CD=DA$ , and $E$ is the intersection point of the diagonals $AC$ and $BD$ . The circle with center $A$ and radius $AE$ intersects $\Gamma$ in two points $F$ and $G$ . Prove that the line $FG$ is tangent to the circles with diameters $BE$ and $DE$ .

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#2 h Apr 6, 2025, 1:46 PM

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Let $P$ be the image of reflection of $E$ across $AB$ .

Then $APBE$ is a cyclic kite. By radical axis, $AB,FG$ and $PE$ are concurrent at one point, say $I$ .

Hence, $EI\perp AB$ . Similarly, $EJ\perp AC$ where $J$ is the intersection point of $AC$ and $FG$ .

Now we have $AIEJ$ is a rectangle. Hence,

$\angle FIB=\angle AIJ=\angle IAE=\angle IEB,$
so we are done.

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#3 h Apr 6, 2025, 11:07 PM

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Consider inversion about circle with center $A$ and radius $AE.$ Cleary, line $FG$ gets mapped to $\Gamma,$ and $E$ is mapped to itself. Also, we can see that $B$ and feet of altitude from $E$ to $AB$ are swapped. We can notice a similar thing for $D.$ This clearly shows that the circles with diameters $BE$ and $DE$ are mapped to itself. It now suffices to show that these two circles are tangent to $\Gamma,$ but this is easy to see since $BD$ is the diameter of $\Gamma.$

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#4 h Apr 7, 2025, 6:09 AM

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Complex bash with $\Gamma$ as the unit circle, so that
$|a|=|b|=1$ $c = \frac{b^2}a$ $d = -b$ $e = \frac{a+c}2 = \frac{a^2+b^2}{2a}$ Now suppose $z$ is the coordinate of either $F$ or $G$ , so that $|z|=1$ and $|a-z| = |a-e|$ . Squaring both sides,
$-\frac{(a-z)^2}{az} = -\frac{(a^2-b^2)^2}{4a^2b^2}$ $4ab^2(a-z)^2 = z(a^2-b^2)^2$ $4ab^2z^2 - z(a^4+6a^2b^2+b^4) + 4a^3b^2 = 0$ Thus by Vieta's formulas,
$f+g = \frac{a^4+6a^2b^2+b^4}{4ab^2}$ $fg = a^2$ Now let $M$ be the midpoint of $BE$ , so that
$m = \frac{b+e}2 = \frac{(a+b)^2}{4a}$ Then the distance from $M$ to line $FG$ is given by
$\frac12|m-f-g+fg\overline{m}| = \frac12\left|\frac{(a+b)^2}{4a} - \frac{a^4+6a^2b^2+b^4}{4ab^2} + \frac{a(a+b)^2}{4b^2}\right| = \frac{|2a^3b-4a^2b^2+2ab^3|}8 = \frac{|a-b|^2}4$ and this is the same as
$|m-b| = \frac{|b-e|}2 = \frac{|a-b|^2}4$ So the distance from $M$ to line $FG$ is the same as the distance from $M$ to $B$ , and thus the circle with diameter $BE$ (and center $M$ ) is tangent to line $FG$ . The proof for the circle with $DE$ is exactly the same, but with $b$ replaced by $-b$ everywhere. $\blacksquare$

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