Let $ABC$ be a triangle with $AB=AC$. Let point $Q$ be on plane such that $AQ \parallel BC$ and $AQ = AB$. Now let the $P$ be the foot of perpendicular from $Q$ to $BC$. Show that the circle with diameter $PQ$ is tangent to the circumcircle of triangle $ABC$.
Problem
Source: Simurgh 2019 - Problem 2
Tags: circumcircle, geometry
DeZade2002
24.02.2019 18:34
Trivially, $AQ$ touches $(ABC)$ at A. Now, let us set $BC$ and $PQ$ as the $x-axis$ and $y-axis$ of the Cartesian plane respectively. Let us set $AB = AC = AQ = a$, $BC = b$. Let $D$ be the foot of perpendicular dropped from $A$ on $BC$. Let $AD = PQ = h$. Then, $P \equiv (0,0)$, $Q \equiv (0,h)$, $D \equiv (-a,0)$, $A \equiv (-a,h)$. Let $X \equiv (0,\frac{h}{2})$ be the midpoint of $PQ$ and let $O$ be the center of $(ABC)$.
It can be written that $\frac{a^{2}b}{4AO} = \frac{bh}{2}$ which implies $AO = \frac{a^{2}}{2h}$.
Again, $A$ divides $DO$ externally into the ratio of $h:AO$. So, if $O \equiv (-a, y_1)$, then $\frac{hy_1}{h - AO} = h$
$\implies y_1 = h - AO = \frac{2h^{2} - a^{2}}{2h}$
Trivial calculations show that $OX = \frac{a^{2} + h^{2}}{2h} = AO + QX$ which completes the proof.
$Q.E.D.$
Yaghi
24.02.2019 19:58
Let $AB=x,BC=y,BP=z$.By the converse of Casey's theorem for $(PQ)$ and degenerated circles $A,B,C$,we should prove that $xy+xz=x(y+z)$,so we are done.
Taha1381
27.04.2019 15:40
Non-computational proof:Apply an inversion with center $A$ and radius $AB^2$ and we are done.
MarkBcc168
27.04.2019 16:36
WLOG $P$ and $B$ lie on different sides w.r.t. $AC$. Let $T$ be the foot from $Q$ to $AP$. Then $AB^2 = AQ^2 = AP\cdot AT$. Thus $$\angle ATC = \angle ACP = 180^{\circ} - \angle ABC$$or $T\odot(ABC)$. Finally, note the tangent at $A$ w.r.t. $\odot(ABC)$ and the tangent at $P$ w.r.t. $\odot(PQ)$ are parallel. As $T\in AP$, we get the desired tangency by homothety.
EulersTurban
24.04.2020 22:33
So we define an inversion $\psi(A,AB)$, basically we are doing a $\sqrt{bc}$ inversion. Now as we see the circumcircle around $\triangle ABC \xrightarrow{\psi} \overline{BC}$. The interesting fact here is since $AB=AC=AQ$, we have that $Q$ belongs to the circle of inversion. From here we see that $Q$ is one of the intersections between the circle with diameter $PQ$ and the circle of inversion. But notice how $\angle AQF = 90$ (here $F$ is the center of the circle of diameter $PQ$), so that means that the circle of inversion and the circle with diameter $PQ$ are orthogonal. So now examine what happens to $P$ under this inversion. We see that $P \in \overline{BC}$, and that $\overline{BC}$ is tangent to the circle with diameter $PQ$ so now we get the inverses touch. Since the inverse of the point $P$ must belong to the circle with diameter $PQ$ (because of orthogonality), since $P$ belongs to the inverse of the circle around $\triangle ABC$, so does the inverse of the point $P$ belong to the circle around $\triangle ABC$. So that means that the circles around $\triangle ABC$ and with diameter $PQ$ touch....
dgreenb801
25.04.2020 04:16
Let $M$ be the midpoint of $BC$. Clearly, $AMPQ$ is a rectangle. Let $K$ be the midpoint of $PQ$, and let $J$ be the circumcenter of $\triangle ABC$ Let $r_1$ be the radius of the circumcircle of $\triangle ABC$, and let $r_2=PK=KQ$ be the radius of the circle with diameter $PQ$. Then $JM= AM-AJ=PQ-AJ=2r_2-r_1$. Then $MC^2=JC^2-JM^2=r_1^2-(2r_2-r_1)^2=4r_1r_2-4r_2^2$. Then $AQ^2=AC^2=AM^2+MC^2=PQ^2+MC^2=(2r_2)^2+(4r_1r_2-4r_2^2)=4r_1r_2$. Then by the Pythagorean Theorem, $JK^2=AQ^2+(AJ-QK)^2=4r_1r_2+(r_1-r_2)^2=(r_1+r_2)^2$. Thus, $JK=r_1+r_2$ is the distance between the centers of the circles with radius $r_1$ and $r_2$, so they must be tangent.