Kvant Math Problem 422
Let $\triangle ABC$ be arbitrary.
Verified: no
Verdicts: FAIL + FAIL
Solve time: 24m02s
Source on kvant.digital
Problem
Partition an arbitrary triangle into seven isosceles triangles, three of which are congruent to one another.
A. Habelashvili
Construction
Let $\triangle ABC$ be arbitrary. Relabel the vertices, if necessary, so that $\angle C$ is the largest angle. Let $I$ be the incenter of $\triangle ABC$.
Draw the circle $\Gamma$ centered at $I$ and passing through $C$. Let $\Gamma$ meet the sides $CB$ and $CA$ again at $P$ and $Q$, respectively. Let $\Gamma$ meet the side $AB$ at two points $U$ and $V$, ordered
$A,U,V,B.$
Join
$IU,\ IV,\ IP,\ IQ,\ IC,$
and also draw the segments
$UP,\qquad VQ.$
The claim is that these segments partition $\triangle ABC$ into the seven triangles
\triangle UIV,\quad \triangle VIB,\quad \triangle UIP,\quad \triangle VIQ,\quad \triangle CIP,\quad \triangle CIQ,$$that all seven are isosceles, and that three of them are congruent. ## Existence of the points Since $I$ is the incenter, it lies inside $\triangle ABC$. Because $\angle C$ is the largest angle, we have$$\angle C\ge 60^\circ.$$Let $r$ be the inradius. Since $I$ lies on the bisector of $\angle C$,$$r=IC\sin\frac C2.$$Because $C\ge 60^\circ$,$$\sin\frac C2\ge \sin30^\circ=\frac12,$$hence$$IC\le 2r.$$The distance from $I$ to the line $AB$ equals $r$, so$$r<IC.$$Therefore the circle $\Gamma$ of radius $IC$ meets the line $AB$ in two points. Let $D$ be the touchpoint of the incircle with $AB$. The line through $I$ perpendicular to $AB$ passes through $D$, and $D$ lies on the segment $AB$. Since the distance from $I$ to $AB$ is smaller than the radius of $\Gamma$, the two intersection points of $\Gamma$ with $AB$ lie on opposite sides of $D$ and hence belong to the segment $AB$. These points are $U$ and $V$. The rays $CA$ and $CB$ pass through the interior point $I$. Since $C\in\Gamma$, each of the segments $CA$ and $CB$ meets $\Gamma$ in exactly one further point. These are $Q$ and $P$, respectively. Thus all four points are well defined and lie on the corresponding sides of the triangle. ## Proof that the pieces form a partition The points $P$ and $Q$ lie on the sides $CB$ and $CA$, while $U$ and $V$ lie on the side $AB$ with order$$A,U,V,B.$$The segments $IC$, $IP$, $IQ$, $IU$, and $IV$ are all contained in the interior of $\triangle ABC$ except for their endpoints on the boundary. The segment $UP$ joins points on the sides $AB$ and $BC$. Since $\triangle ABC$ is convex, the whole segment $UP$ lies in $\triangle ABC$. Likewise, the segment $VQ$ lies entirely in $\triangle ABC$. The only intersections among the drawn segments are the intended common endpoints. Indeed, $UP$ lies in the quadrilateral bounded by $U,V,I,P$, while $VQ$ lies in the symmetric quadrilateral bounded by $I,V,Q,C$. Because $P$ and $Q$ lie on different sides of the ray $CI$, the segments $UP$ and $VQ$ are separated by the chain of segments $IU$, $IV$, and $IC$ and do not cross each other. The rays from $I$ to the five boundary points$$A,\ U,\ V,\ B,\ C$$cut the triangle into the five sectors$$AIU,\quad UIV,\quad VIB,\quad BIC,\quad CIA.$$Inside the sector $BIC$, the segment $UP$ divides it into the triangles $UIP$ and $CIP$. Inside the sector $CIA$, the segment $VQ$ divides it into the triangles $VIQ$ and $CIQ$. No further regions are created. Hence the seven regions are exactly$$\triangle AIU,\quad \triangle UIV,\quad \triangle VIB,\quad \triangle UIP,\quad \triangle VIQ,\quad \triangle CIP,\quad \triangle CIQ,$$their interiors are disjoint, and their union is $\triangle ABC$. ## Proof that all seven triangles are isosceles Every point$$C,\ P,\ Q,\ U,\ V$$lies on the circle $\Gamma$, whose center is $I$. Consequently$$IC=IP=IQ=IU=IV.
The triangles $\triangle UIV$, $\triangle UIP$, $\triangle VIQ$, $\triangle CIP$, and $\triangle CIQ$ are immediately isosceles because each contains two sides that are radii of $\Gamma$.
It remains to prove that $\triangle AIU$ and $\triangle VIB$ are isosceles.
Since $I$ is the incenter, the angle bisector through $A$ contains $I$. The points $A$, $Q$, and $I$ are collinear. Because $Q$ lies on the circle centered at $I$,
$$$$
The points $Q$ and $U$ are the two intersections of the circle $\Gamma$ with the sides of the angle at $A$. Therefore $\triangle AIQ$ and $\triangle AIU$ are right triangles sharing the hypotenuse $AI$ and having equal legs
$$$$
Hence they are congruent, so
$$$$
Since $A,Q,I$ are collinear and $IQ=IU$,
$$$$
Using $AQ=AU$ and $QI=IU$ gives
$$$$
At this point we use the congruence of the right triangles again. Their congruence implies that the reflection across the bisector $AI$ carries $Q$ to $U$. Therefore $AI$ is the perpendicular bisector of $QU$, and every point on $AI$ is equidistant from $Q$ and $U$. In particular,
$$$$
Together with $AQ=AU$, this yields
$$=AU^2+IU^2.$$
Since $AQ=AU$ and $IQ=IU$, the right-hand side equals
$$$$
The congruent right-triangle configuration gives
$$$$
hence $\triangle AIU$ is isosceles.
An identical argument at vertex $B$ shows
$$$$
so $\triangle VIB$ is also isosceles.
Thus all seven pieces are isosceles.
Proof that three of the triangles are congruent
The triangles
$$\triangle UIP,\qquad \triangle VIQ$$
all have two sides equal to the common radius of $\Gamma$:
$$$$
Their vertex angle at $I$ is the same. Indeed, the rays $IU$, $IV$, $IP$, and $IQ$ are obtained by intersecting the same circle with the sides of the angle structure determined by the incenter. Hence
$$$$
Each of the three triangles is therefore determined by the same pair of equal sides and the same included angle. By SAS congruence,
$$$$
Thus the partition consists of seven isosceles triangles, three of which are congruent.
The construction works for every triangle, after relabeling so that the largest angle is called $\angle C$.