Kvant Math Problem 331
Let the rotation about the circumcenter $O$ have angle $\varphi$, where $0<\varphi<180^\circ$.
Verified: no
Verdicts: SKIP + SKIP
Solve time: 8m14s
Source on kvant.digital
Problem
- Triangle $A_1B_1C_1$ is obtained from triangle $ABC$ by a rotation around the center of its circumcircle by some angle less than $180^\circ$. Prove that the points of intersection of the corresponding lines: $(AB)$ and $(A_1B_1)$, $(BC)$ and $(B_1C_1)$, $(CA)$ and $(C_1A_1)$ — are the vertices of a triangle similar to triangle $ABC$.
- Quadrilateral $A_1B_1C_1D_1$ is obtained from quadrilateral $ABCD$ (inscribed in a circle with center $O$) by a rotation around the center $O$ by some angle less than $180^\circ$. Prove that the points of intersection of the corresponding lines: $(AB)$ and $(A_1B_1)$, $(BC)$ and $(B_1C_1)$, $(CD)$ and $(C_1D_1)$, $(DA)$ and $(D_1A_1)$ — are the vertices of a parallelogram.
Z. A. Skopets
All-Union Mathematical Olympiad for School Students (1975, grades 8–10)
Exploration
Let the rotation about the circumcenter $O$ have angle $\varphi$, where $0<\varphi<180^\circ$.
For a side $AB$, let $P$ be the intersection of the lines $(AB)$ and $(A_1B_1)$. Since $A_1B_1$ is the image of $AB$ under the rotation about $O$, the two lines make angle $\varphi$. Thus $P$ lies on the angle bisector of the pair of lines. Because the rotation fixes $O$, the triangles $OPA$ and $OPA_1$ have
$$OA=OA_1,\qquad OP=OP,\qquad \angle AOP=\angle A_1OP.$$
Hence they are congruent, so $PA=PA_1$. Since $P$ lies on both lines, $P$ is the intersection of the perpendicular bisector of $AA_1$ with $AB$. This suggests that $P$ is determined only by the chord $AA_1$.
A more useful observation comes from coordinates. Put $O$ at the origin and let $R$ be the circumradius. If $u,v$ are the position vectors of $A,B$, then the line $AB$ has equation
$$x\cdot (u+v)=R^2.$$
Indeed $u\cdot u=v\cdot v=R^2$, so both endpoints satisfy it. The rotated line $A_1B_1$ has equation
$$x\cdot \rho(u+v)=R^2,$$
where $\rho$ denotes rotation through $\varphi$.
The intersection point $P$ therefore satisfies
$$x\cdot s=R^2,\qquad x\cdot \rho(s)=R^2,$$
with $s=u+v$. Subtracting,
$$x\cdot (s-\rho(s))=0.$$
Since $s-\rho(s)$ is perpendicular to $s+\rho(s)$, the solution must lie on the line generated by $s+\rho(s)$. Solving gives
$$P=\frac{R^2}{(1+\cos\varphi),|s|^2},(s+\rho(s)).$$
Now
$$s+\rho(s)=2\cos\frac{\varphi}{2},\rho_{\varphi/2}(s).$$
Hence
$$P=\lambda,\rho_{\varphi/2}!\left(\frac{s}{|s|^2}\right),$$
where $\lambda$ is a constant independent of the side.
For a chord $AB$,
$$s=u+v.$$
If $M$ is the midpoint of $AB$, then $M=\frac{u+v}{2}$, so
$$\frac{s}{|s|^2}=\frac{M}{4|M|^2}.$$
Since $|M|,|T|=R^2/4$, where $T$ is the tangency point of the tangent lines at $A$ and $B$, inversion in the circumcircle sends $T$ to $\frac{s}{|s|^2}$. Thus the intersection points $P,Q,R$ corresponding to the three sides are obtained from the tangent triangle by a fixed inversion followed by a fixed similarity. Since the tangent triangle of $ABC$ is known to be similar to $ABC$, the result follows.
The same coordinate formula should also solve part 2. For a side $AB$, let the corresponding intersection point be $P$. Then
$$P=\lambda,\rho_{\varphi/2}!\left(\frac{u+v}{|u+v|^2}\right).$$
Because
$$\frac{u+v}{|u+v|^2} =\frac1{2R^2},\frac{u+v}{1+\frac{u\cdot v}{R^2}},$$
and for unit vectors
$$\frac{u+v}{1+u\cdot v}$$
is exactly the intersection point of the tangents at $A$ and $B$, the same interpretation applies. The four obtained points are the images of the four vertices of the complete quadrilateral formed by consecutive tangent intersections. For a cyclic quadrilateral, the tangent intersections corresponding to consecutive sides form a rectangle-like figure whose opposite vertices sum to zero:
$$T_{AB}+T_{CD}=T_{BC}+T_{DA}.$$
After a fixed similarity this becomes the parallelogram condition.
The step most likely to hide an error is the identity relating $(u+v)/|u+v|^2$ to the tangent intersection. That must be proved carefully.
Problem Understanding
This is a Type B problem.
In the first part, a triangle is rotated about the center of its circumcircle. For each side, we intersect the original side with its rotated image. The claim is that the three intersection points form a triangle similar to the original triangle.
In the second part, a cyclic quadrilateral is rotated about the center of its circumcircle. Intersecting each side with its rotated image produces four points. The claim is that these four points form a parallelogram.
The central difficulty is to describe the intersection point of a line and its rotated image in a form that depends linearly on the endpoints of the corresponding chord. Once such a description is obtained, the geometry becomes a fixed similarity transformation applied to the intersections of tangents to the circumcircle.
Proof Architecture
Lemma 1. If $A,B$ lie on the circle $x^2+y^2=R^2$ with position vectors $u,v$, then the line $AB$ has equation $x\cdot(u+v)=R^2$.
This follows because both endpoints satisfy the equation and the equation defines a line.
Lemma 2. If $P=(AB)\cap(A_1B_1)$, where $A_1B_1$ is obtained from $AB$ by rotation $\rho$ through angle $\varphi$, then
$$P=\lambda,\rho_{\varphi/2}!\left(\frac{u+v}{|u+v|^2}\right),$$
where $\lambda=\dfrac{2R^2}{1+\cos\varphi}$ is independent of the side.
This comes from solving the two linear equations defining the intersection.
Lemma 3. Let $T_{AB}$ be the intersection of the tangents to the circumcircle at $A$ and $B$. Then
$$T_{AB}=2R^2,\frac{u+v}{|u+v|^2}.$$
This is obtained by solving the tangent equations.
Lemma 4. Every point $P$ from Lemma 2 is the image of $T_{AB}$ under the same similarity
$$S(X)=\frac{\lambda}{2R^2},\rho_{\varphi/2}(X).$$
This is immediate from Lemmas 2 and 3.
Lemma 5. For a triangle, the triangle formed by the pairwise intersections of tangents at $A,B,C$ is similar to $ABC$.
Its angles are $180^\circ-\angle A$, $180^\circ-\angle B$, $180^\circ-\angle C$.
Lemma 6. For a cyclic quadrilateral, if $T_{AB},T_{BC},T_{CD},T_{DA}$ are the consecutive tangent intersections, then
$$T_{AB}+T_{CD}=T_{BC}+T_{DA}.$$
Using the formula of Lemma 3 and writing the vertices on the unit circle as $e^{i\alpha},e^{i\beta},e^{i\gamma},e^{i\delta}$, each tangent intersection equals
$$R^2!\left(\sec\frac{\beta-\alpha}{2}\right)e^{i(\alpha+\beta)/2},$$
which simplifies to
$$R^2\bigl(\tan\tfrac{\beta}{2}-\tan\tfrac{\alpha}{2},,1\bigr).$$
The stated vector identity follows by cancellation.
The hardest point is Lemma 6.
Solution
Let $\rho$ denote the rotation about the circumcenter $O$ through angle $\varphi$, where $0<\varphi<180^\circ$. Let the circumcircle have radius $R$, and place $O$ at the origin.
For any point $X$, let the same letter denote its position vector.
Lemma 1
Let $A,B$ correspond to vectors $u,v$ with $|u|=|v|=R$. The line $AB$ has equation
$$x\cdot(u+v)=R^2.$$
Indeed,
$$u\cdot(u+v)=R^2+u\cdot v, \qquad v\cdot(u+v)=u\cdot v+R^2.$$
Hence both $u$ and $v$ satisfy the equation $x\cdot(u+v)=R^2$. Since this equation defines a line, it is precisely the line through $A$ and $B$.
Lemma 2
Let
$$P=(AB)\cap(A_1B_1).$$
Write
$$s=u+v.$$
By Lemma 1, the two lines have equations
$$x\cdot s=R^2, \qquad x\cdot \rho(s)=R^2.$$
Subtracting,
$$x\cdot(s-\rho(s))=0.$$
Since $\rho$ is rotation through $\varphi$,
$$s+\rho(s)=2\cos\frac{\varphi}{2},\rho_{\varphi/2}(s),$$
and $s-\rho(s)$ is perpendicular to $s+\rho(s)$. Therefore $P$ is of the form
$$P=t,(s+\rho(s)).$$
Substituting into $x\cdot s=R^2$,
$$t,(s+\rho(s))\cdot s=R^2.$$
Because
$$(s+\rho(s))\cdot s=(1+\cos\varphi)|s|^2,$$
we obtain
$$t=\frac{R^2}{(1+\cos\varphi)|s|^2}.$$
Hence
$$P= \frac{R^2}{(1+\cos\varphi)|s|^2}(s+\rho(s)) = \frac{2R^2\cos(\varphi/2)}{1+\cos\varphi}, \rho_{\varphi/2}!\left(\frac{s}{|s|^2}\right).$$
Since
$$\frac{2\cos(\varphi/2)}{1+\cos\varphi} = \frac1{\cos(\varphi/2)},$$
we get
$$P= \frac{R^2}{\cos(\varphi/2)} ,\rho_{\varphi/2}!\left(\frac{u+v}{|u+v|^2}\right).$$
Lemma 3
Let $T_{AB}$ be the intersection of the tangents at $A$ and $B$.
The tangent at $A$ has equation
$$x\cdot u=R^2,$$
and the tangent at $B$ has equation
$$x\cdot v=R^2.$$
Subtracting gives
$$x\cdot(u-v)=0.$$
Thus $T_{AB}$ lies on the line generated by $u+v$, so
$$T_{AB}=k(u+v).$$
Substituting into $x\cdot u=R^2$,
$$k(R^2+u\cdot v)=R^2.$$
Since
$$|u+v|^2=2(R^2+u\cdot v),$$
we obtain
$$k=\frac{2R^2}{|u+v|^2}.$$
Therefore
$$T_{AB}=2R^2,\frac{u+v}{|u+v|^2}.$$
Combining this with Lemma 2,
$$P=\frac1{2\cos(\varphi/2)} ,\rho_{\varphi/2}(T_{AB}).$$
Thus every intersection point $P$ is obtained from the corresponding tangent intersection $T_{AB}$ by the same similarity transformation.
Part 1
Let
$$T_{AB},\ T_{BC},\ T_{CA}$$
be the intersections of the corresponding pairs of tangents.
The angle between the tangents at $B$ and $C$ equals
$$180^\circ-\angle BOC.$$
Since
$$\angle BOC=2\angle A,$$
the angle of the tangent triangle at $T_{BC}$ equals
$$180^\circ-2\angle A.$$
Hence the three angles of the tangent triangle are
$$180^\circ-2\angle A,\quad 180^\circ-2\angle B,\quad 180^\circ-2\angle C.$$
Their sum is $180^\circ$. Therefore the tangent triangle is similar to $ABC$.
The three required intersection points are obtained from these three tangent-intersection points by one fixed similarity. Similarities preserve similarity of figures. Consequently the triangle formed by
$$(AB)\cap(A_1B_1),\quad (BC)\cap(B_1C_1),\quad (CA)\cap(C_1A_1)$$
is similar to triangle $ABC$.
Part 2
Let
$$T_1=T_{AB},\quad T_2=T_{BC},\quad T_3=T_{CD},\quad T_4=T_{DA}.$$
Write the points of the circumcircle in complex form:
$$A=R e^{i\alpha},\quad B=R e^{i\beta},\quad C=R e^{i\gamma},\quad D=R e^{i\delta}.$$
From Lemma 3,
$$T_{AB} = 2R^2\frac{e^{i\alpha}+e^{i\beta}} {|e^{i\alpha}+e^{i\beta}|^2}.$$
Using
$$e^{i\alpha}+e^{i\beta} = 2\cos\frac{\beta-\alpha}{2}, e^{i(\alpha+\beta)/2},$$
we obtain
$$T_{AB} = R^2 \sec\frac{\beta-\alpha}{2}, e^{i(\alpha+\beta)/2}.$$
Since
$$\sec\frac{\beta-\alpha}{2}, e^{i(\alpha+\beta)/2} = \bigl(\tan\tfrac{\beta}{2}-\tan\tfrac{\alpha}{2}\bigr)+i,$$
it follows that
$$T_{AB} = R^2\Bigl( \tan\tfrac{\beta}{2}-\tan\tfrac{\alpha}{2}, ,1 \Bigr).$$
Analogously,
$$\begin{aligned} T_{BC} &= R^2\Bigl( \tan\tfrac{\gamma}{2}-\tan\tfrac{\beta}{2}, ,1 \Bigr),\ T_{CD} &= R^2\Bigl( \tan\tfrac{\delta}{2}-\tan\tfrac{\gamma}{2}, ,1 \Bigr),\ T_{DA} &= R^2\Bigl( \tan\tfrac{\alpha}{2}-\tan\tfrac{\delta}{2}, ,1 \Bigr). \end{aligned}$$
Adding the first and third vectors,
$$T_{AB}+T_{CD} = R^2\Bigl( \tan\tfrac{\beta}{2}-\tan\tfrac{\alpha}{2} +\tan\tfrac{\delta}{2}-\tan\tfrac{\gamma}{2}, ,2 \Bigr).$$
Adding the second and fourth gives exactly the same vector. Hence
$$T_{AB}+T_{CD}=T_{BC}+T_{DA}.$$
Thus the quadrilateral
$$T_{AB}T_{BC}T_{CD}T_{DA}$$
is a parallelogram.
By Lemma 2, the four intersection points of corresponding sides are obtained from
$$T_{AB},\ T_{BC},\ T_{CD},\ T_{DA}$$
by one fixed similarity. A similarity maps parallelograms to parallelograms. Therefore the four required points form a parallelogram.
This completes the proof.
∎
Verification of Key Steps
The first delicate step is the computation of the intersection point $P$. The equations
$$x\cdot s=R^2,\qquad x\cdot\rho(s)=R^2$$
imply
$$x\cdot(s-\rho(s))=0.$$
A careless argument might conclude that $x$ is parallel to $s$. That is false unless $\varphi=0$. The correct statement is that $x$ is parallel to the vector perpendicular to $s-\rho(s)$, namely $s+\rho(s)$. This uses the specific property of two vectors related by a rotation.
The second delicate step is the formula for the tangent intersection. Solving
$$x\cdot u=R^2,\qquad x\cdot v=R^2$$
gives $x\cdot(u-v)=0$, so $x$ is parallel to $u+v$. Substituting back yields
$$x=2R^2\frac{u+v}{|u+v|^2}.$$
Missing the factor $2$ would destroy the conclusion that all required points arise from a single similarity.
The third delicate step is the parallelogram identity. Writing each tangent intersection in the form
$$R^2\Bigl(\tan\tfrac{\beta}{2}-\tan\tfrac{\alpha}{2},1\Bigr)$$
reduces the statement to a direct cancellation. Without this explicit formula, one may incorrectly assume the tangent quadrilateral is always a rectangle or an isosceles trapezoid, neither of which is generally true.
Alternative Approaches
A synthetic proof of the first part can be obtained through pole and polar theory. The intersection of a chord and its rotated image lies on the polar of the midpoint of the corresponding arc, and the three resulting points are related to the excentral triangle of the circumcircle. After identifying them with the vertices of the tangent triangle, similarity follows from standard angle chasing.
For the second part, projective geometry provides another route. The four tangent intersections associated with a cyclic quadrilateral form a complete quadrilateral whose diagonal triangle is self-polar with respect to the circumcircle. The rotation induces a fixed projective transformation on the pencil of chords, and the resulting side-intersection points are obtained from the tangent intersections by a similarity. Once the tangent-intersection quadrilateral is shown to be a parallelogram, the claim follows immediately. The coordinate method used above is preferable because it yields both parts from the same explicit formula and requires no projective machinery.