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#1 h Feb 25, 2011, 5:33 PM • 2 Y

Y by Adventure10, Mango247

Let $P$ be the intersection of lines $l_1$ and $l_2$ . Let $S_1$ and $S_2$ be two
circles externally tangent at $P$ and both tangent to $l_1$ , and let $T_1$
and $T_2$ be two circles externally tangent at $P$ and both tangent to $l_2$ .
Let $A$ be the second intersection of $S_1$ and $T_1, B$ that of $S_1$ and $T_2, C$ that of $S_2$ and $T_1$ , and $D$ that of $S_2$ and $T_2$ . Show that the points $A,B,C,D$ are concyclic if and only if $l_1$ and $l_2$ are perpendicular.

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#2 h Feb 26, 2011, 4:16 PM • 2 Y

Y by Adventure10, Mango247

Inversion with center $P$ and abitrary power takes $\mathcal{S}_1,\mathcal{S}_2$ into the lines $s_1,s_2$ perpendicular to $\ell_1$ and takes $\mathcal{T}_1,\mathcal{T}_2$ into the lines $\tau_1,\tau_2$ perpendicular to $\ell_2.$ Pairwise lines $s_1,s_2,\tau_1,\tau_2$ meet at the inverses $A'B',C',D'$ of $A,B,C,D.$ Assume that $\ell_1 \perp \ell_2,$ then $A'B'C'D'$ is a rectangle $\Longrightarrow$ $A,B,C,D$ are also concyclic. Converserly, assume that $A,B,C,D$ are concyclic. Since their inverses $A',B',C',D'$ are vertices of a parallelogram, then $A'B'C'D'$ is a rectangle $\Longrightarrow$ $\ell_1 \perp \ell_2.$

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#3 h Feb 26, 2011, 5:38 PM • 2 Y

Y by Adventure10, Mango247

luisgeometra wrote:

Inversion with center $P$ and abitrary power takes $\mathcal{S}_1,\mathcal{S}_2$ into the lines $s_1,s_2$ perpendicular to $\ell_1$ and takes $\mathcal{T}_1,\mathcal{T}_2$ into the lines $\tau_1,\tau_2$ perpendicular to $\ell_2.$ Pairwise lines $s_1,s_2,\tau_1,\tau_2$ meet at the inverses $A'B',C',D'$ of $A,B,C,D$ . Assume that $\ell_1 \perp \ell_2,$ then $A'B'C'D'$ is a rectangle $\Longrightarrow$ $A,B,C,D$ are also concyclic. Converserly, assume that $A,B,C,D$ are concyclic. Since their inverses $A',B',C',D'$ are vertices of a parallelogram, then $A'B'C'D'$ is a rectangle $\Longrightarrow$ $\ell_1 \perp \ell_2$ .

thanks for your solution but can you solve it without inversion? if you can i am waiting your post.

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#4 h Feb 27, 2011, 1:30 AM • 1 Y

Y by Adventure10

Let $X,Z$ be any points on $l_2$ and in $T_1$ and $T_2$ respectively, $Y,W$ are any point on $l_1$ and lies in $S_1,S_2$ . Then $\angle ABD=\angle ABP+\angle DBP=\angle APX+\angle WPD$ . Similarly $\angle ACD=\angle APY+\angle DPZ$ . They both add up to 180 iff $l_1\perp l_2$ .

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#5 h Feb 27, 2011, 9:31 AM • 2 Y

Y by Adventure10, Mango247

oneplusone wrote:

Let $X,Z$ be any points on $l_2$ and in $T_1$ and $T_2$ respectively, $Y,W$ are any point on $l_1$ and lies in $S_1,S_2$ . Then $\angle ABD=\angle ABP+\angle DBP=\angle APX+\angle WPD$ . Similarly $\angle ACD=\angle APY+\angle DPZ$ . They both add up to 180 iff $l_1\perp l_2$ .

Thanks for your solution I thought that this question is hard

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