Hints and Solutions to Selected Problems in Part I and Part II

Note: If a radius of an inversion is not specified, then assume that it is arbitrary.

Hint 1-2. $I(A,r)$.

Hint 3. If $r_1\leq r_2\leq r_3$, reduce to $\char93 1$ or $\char93 2$ by replacing $k_1$ by its center, $k_2$ by $k_2^{\prime}(O_2, r_2-r_1)$, and $k_3$ by $k_3^{\prime}(O_2,r_3-r_1)$.

Hint 4. If $K,S,Q$ are the centers if $k,s,q$, respectively, then the wanted geometric place is the incircle of $\triangle KSQ$: use $I(A,r)$.

Hint 5. Ist way: $I(A,r)$; IInd way: if $k_2$ and $k_4$ do not intersect, invert them into concentric circles (cf. Fig. 15-17); IIIrd way: forget about inversion, and notice that $O_1O_2O_3O_4$ is circumscribed around a circle.

Hint 6. $I(A_1,r)$.

Solution 7. Using $I(D,r)$, we have $A_1,B_1,C_1$ collinear and $A_1B_1+B_1C_1=A_1C_1$. Then

$$\frac{AB\cdot r^2}{DA\cdot DB}+\frac{BC\cdot r^2}{DB\cdot DC}= \f... ...cdot r^2}{DA\cdot DC}\,\,\Rightarrow\,\, AB\cdot DC +AD\cdot BC=AC\cdot BD.\qed$$

Hint 8. $I(P,r)$.

Hint 9. If $k_1\cap k_2=\emptyset$, let $I(O,r)$ send them into concentric circles. Consider the cases when $I(k_3)$ is a circle, and when it is a line.

Hint 10-11. $I(P,r)$.

Hint 12. Describe circles $k$ and $k_1$ around the squares, and use inscribed angles to show that the point in question is the intersection point of $k$ and $k_1$, other than $D$.

Hint 13. $I(D,r)$, then $A_1B_1=C_1B_1$.

Hint 14. $I(D,r)$, then $\angle A_1B_1C_1=\beta$ in $\triangle A_1B_1C_1$.

Solution 15. Apply $I(A,r)$ (cf. Fig.18-19.) Then $AD_1B_1C_1$ is a parallelogram, and $E_1\in AB_1^{\rightarrow}$. But $AE_1\cdot AE=r^2=AB_1\cdot AB$, and $AE=2AB$. Hence $AE_1=AB_1/2$, and $E_1$ is the midpoint of $AB_1$, i.e. $AB_1\cap D_1C_1=\{E\}$. Since $D_1,E_1,C_1$ are collinear, then $D,E,C,A$ are concyclic. $\qedsymbol$

Solution 16. Apply $I(O,r)$, and let $\triangle ABC\stackrel{I}{\rightarrow}\triangle A_1B_1C_1$ (cf. Fig.20.) If $\angle AOB=\gamma$, $\angle OA_1B_1=\alpha$, and $\angle OB_1A_1=\beta$, then

$$\frac{OA+OB}{OC}=\frac{OC_1}{OA_1}+\frac{OC_1}{OB_1}= \frac{\text... ...text{sin}\beta}{\text{sin}(\alpha+\gamma/2)}= 2\text{cos}\frac{\gamma}{2}. \qed$$

Hint 17. Let $I(O,r)$ send $k$ and $k^*$ into concentric circles.

Solution 22. Apply $I(O,OP)$ (cf. Fig.21.) Then $I(k)=l$, so that $I(T)=T_1$ and $I(S)=S_1$ are the given points on $l$. Let $k_q(Q,QS=QT)$, and let $X$ be the center of $I(k_q)$. We know that $X\in OQ^{\rightarrow}$. But since $k_q\perp k$, then $I(k_q)\perp l$, i.e. $l$ passes through the center $X$. Thus, $X=Q_1$, and $XT_1=XS_1$ implies that $Q_1$ is the midpoint of $T_1S_1$. $\qedsymbol$

Solution 18. We refer to Fig.22-23 for notation. Let $k_3\cap l=\{B\}$, and let the radius $R$ be chosen so that the circle $k_0(B,R)\perp k$. Apply $I(B,R)$. Then $I(k)=k(O,1)$, $I(l)=l$, $I(k_3)=k_3^{\prime}$ is a line parallel to $l$, and $I(k_1)=k_1^{\prime}(O^{\prime},R^{\prime})$ and $I(k_2)=k_2^{\prime}$ are both circles, touching $l$ and $k_3^{\prime}$, and hence, of the same radii $R^{\prime}$. If $A^{\prime}$ is their point of tangency, from $\triangle O^{\prime}AO$: $(R^{\prime})^2+((R^{\prime})^2-1)^2=((R^{\prime})^2+1)^2 \Rightarrow R^{\prime}=4.$ The wanted distance is $2R^{\prime}-1=7$.$\qedsymbol$

Solution . Let $A_0,B_0,C_0$ be the points of tangency of the inscribed circle $k_P(P,r)$ with $BC,CA,AB$, respectively (cf. Fig.24.) Let $A_1,B_1,C_1$ be the midpoints of the corresponding sides of $\triangle A_0B_0C_0$. Note that $A_1=PA\cap C_0B_0$, and similarly for $B_1$ and $C_1$. Apply $I(P,r)$. Since $PA\perp C_0B_0$, it follows $I(A)=A_1$, and similarly, $I(B)=B_1$, $I(C)=C_1$. Thus, the circumscribed circle $k(O,R)$ goes under $I$ to the circumscribed circle around $\triangle A_1B_1C_1$, $k_1(O_1,r_1)$. Note that $k_1$ is half the size of $k_P$, hence $r_1=r/2$.

This is enough to find the distance between $O$ and $P$. More generally, the distance $d$ between the center $O$ of a circle and the center $P$ of an inversion satisfies:

$$r_1\cdot \vert d^2-R^2\vert=r^2\cdot R,$$ (2)

where $r$ is the radius of the inversion, $R$ is the original radius of the circle, and $r_1$ is the radius of the image circle. Indeed, (cf. Fig.25,) for the diametrically opposite points $S,T\in PO\cap k$, and for their images $S_1,T_1\in PO\cap k_1$ we have:

$$S_1T_1=\frac{ST\cdot r^2}{PS\cdot PT}\,\,\Rightarrow\,\, 2r_1=\frac{2R\cdot r^2}{\vert d-R\vert(d+R)},$$

from where () follows. Substituting $r_1=r/2$, and the obvious relation $R>d$, we get $d=\sqrt{R(R-2r)}$. Note that for $d=0$ we have an equilateral $\triangle ABC$, and $R=2r$. $\qedsymbol$

Solution . Let $A_1$ and $B_1$ be the orthogonal projections of $A$ and $B$ onto $BC$ and $AC$, and let $k_3$ be the circle through $A,B,A_1,B_1$ (cf. Fig. 26.) The radical axis of $k_1$ and $k_2$ is $MN$, the radical axis of $k_1$ and $k_3$ is $AA_1$, and the radical axis of $k_2$ and $k_3$ is $BB_1$. Hence they all intersect in one point $H$, the orthocenter of $\triangle ABC$. Thus, $H\in CC_1\cap MN$.

Let $O_1$ and $O_2$ be the centers of $k_1$ and $k_2$, or equivalently, the midpoints of $BD$ and $AE$. Since $DE\vert\vert AB$, then $O_1O_2\vert\vert AB$. But $MN\perp O_1O_2$, so $MN\perp AB$, i.e. $MN\vert\vert CC_1$. Since the last two intersect in point $H$, it follows that they coincide as lines. $\qedsymbol$

Hint . Set $D\equiv E\equiv C$ in #.

Hint . Consider the radical axis of the two circles and its reflection across the perpendicular bisector of $O_1O_2$ (cf. Fig.27.)

Hint . Consider the intersection of a radical axis of two of the circles with a tangent to these two circles.

Hint 29. $I(E,r=EB)$ and show that $U$ lies on line $AE$, and $V$ lies on line $EC$. ALternatively, find a non-inversive proof that $BVEU$ is a rectangle.

Hint 30. $I(S,r)$.

©1998 by Berkeley Math Circle, Berkeley, CA

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