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Here is my solution: Let X, Y, Z be the centers of the squares BCTU, CAVW, ABSR constructed on the segments BC, CA, AB and pointing into the exterior of triangle ABC. The square inscribed in triangle ABC with two vertices on BC, one on AB and one on AC is homothetic to the square constructed on the segment BC into the exterior of triangle ABC, the center of homothety being A; hence, the centers $A_1$ and X of these two squares are collinear with A, i. e. the line $AA_1$ coincides with the line AX. Similarly, the line $BB_1$ coincides with BY, and the line $CC_1$ coincides with CZ. Hence, instead of proving that the lines $AA_1$, $BB_1$, $CC_1$ concur, it will be enough to show that the lines AX, BY, CZ concur. Well, why do the lines AX, BY, CZ concur? There are several proofs of this fact; here is the most elementary of them: I will show that the lines AX, BY, CZ are perpendicular to the lines YZ, ZX, XY, respectively; then, it will follow that AX, BY, CZ are the altitudes of triangle XYZ and hence concur. Why are the lines AX, BY, CZ perpendicular to YZ, ZX, XY? Well, the spiral dilatation with center B, rotation angle 45 and dilatation factor $\sqrt2$ maps C to T and Z to A. Hence, the segment TA is the image of the segment CZ, so that < (CZ; TA) = 45 (where < (k; l) denotes the directed angle between two lines k and l). The spiral dilatation with center C, rotation angle 45 and dilatation factor $\frac{1}{\sqrt2}$ maps T to X and A to Y. Hence, the segment XY is the image of the segment TA, so that < (TA; XY) = 45, and < (CZ; XY) = < (CZ; TA) + < (TA; XY) = 45 + 45 = 90, so that CZ is perpendicular to XY. Similarly, AX is perpendicular to YZ and BY is perpendicular to ZX, and we are done. [By the way, the same approach using spiral dilatations shows that AX = YZ, BY = ZX and CZ = XY.] Darij

Let $D,E,F$ be the intersections of $AA_1,BB_1,CC_1$ with $BC,CA,AB$ respectively. The homothety which takes the square with one side on $BC$ to the external square having $BC$ as a side has $A$ as its center, so the center of the large square also lies on $AA_1$, so the problem is reduced to the following: Erect right isosceles triangles $BCM,CAN,ABP$ (the right angles are at $M,N,P$) outside $ABC$. Show that $AM,BN,CP$ are concurrent. this is, of course, Vecten's point. Here's a proof of the concurrence: Let $X$ be the midpoint of $CA$. The point $M$ is obtained from $P$ by the composition of two spiral similarities, one (call it $S_1$) centered at $A$, of angle $\frac \pi 4$ and ratio $\sqrt 2$, the other one (call it $S_2$) centered at $C$, of angle $\frac \pi 4$ and ratio $\frac {\sqrt 2}2$. The ratios cancel out, so the result is a rotation of anle $\frac \pi 2$, and the center is the fixed point of the transformation, which is clearly $X$ (just apply $S_1$ and then $S_2$ to $X$ to see that it's fixed). This means that $AM$ is the image of $NP$ through a rotation of angle $\frac \pi 2$, so, in particular, $AM\perp NP$. The concurrence we want to prove is thus simply the concurrence of the altitudes of $MNP$. [Edit: Darij posted his message while I was typing mine, and I'd hate to have typed it for nothing, so I'm posting it ]

To solve this problem also it is possible to use Ceva's theorem in trigonometrical form.

Let the square with center $ A_1$ have vertices $ M$ and $ N$ lying on $ AB$ and $ AC$ respectively, let $ AA_1$ meet $ BC$ at $ X$ and $ MN$ at $ S$, then $ \frac {BX}{XC} = \frac {MS}{SN} = \frac {[AMA_1]}{[ANA_1]} = \frac {\frac {1}{2}AM \cdot MA_1 \cdot sin AMA_1}{\frac {1}{2} AN \cdot NA_1 \cdot sin ANA_1} = \frac {AM}{AN} \cdot \frac {sin(B + 45)}{sin(C + 45)} = \frac {AB}{AC} \cdot \frac {sin(B + 45)}{sin(C + 45)}$, so by Ceva the three lines will concur, as everything cancels.

A solution by using the trigonometric form of Ceva's theorem.

Hello, I have a particularly nice solution!

Dear Mathlinkers, following the nice idea of Darij, the point of intersection is the first Vecten point. You can see http://perso.orange.fr/jl.ayme vol. 5 Vecten p. 25 Sincerely Jean-Louis

Let the ray $AA_1$ meet $BC$ at $X$. Since $AX$ passes through the center of the square, we have $VX = q$, $WX = p$. By similar triangles $BX/XC = p/q$. Hence $BX/XC = (BX + p)/(XC + q) = BW/VC = (BV + s)/(CW + s)$, where $s$ is the side of the square. But $BV = s \cot B, CW = s \cot C$, so $BX/XC = (\cot B + 1)/(\cot C + 1)$. The desired result immediately follows by Ceva's Theorem. $\Box$

See PP12 (an easy extension) from here.

As Darij & Grobber pointed out, is more convenient to prove $AM, BN, CP$ concurent, where $M, N, P$ are the centers of the squares erected outwardly onto the sides of the triangle. Let $X=AM\cap BC, Y=BN\cap AC, Z=CP\cap AB$. Triangles $\Delta ABM,\Delta ACM$ have the same base, hence the ratio of their areas is $\frac{S_{ABM}}{S_{ACM}}=\frac{BX}{CX}$, but $2S_{ABM}=AB\cdot BM\cdot sin(B+45^\circ), 2S_{ACM}=AC\cdot CM\cdot sin(C+45^\circ)$; with $BM=CM$ we get $\frac{S_{ABM}}{S_{ACM}}=\frac{AB\cdot sin(B+45^\circ)}{AC\cdot sin(C+45^\circ)}=\frac{BX}{CX}$ and other two similar relations for $Y,Z$ which, multiplied side by side will give the required $\frac{BX}{CX}\cdot\frac{CY}{AY}\cdot\frac{AZ}{BZ}=1$. However one cannot compare this solution with Darij's! Best regards, sunken rock

We consider the square on side $BC$ and let its intersections with $AB$ and $AC$ be $X$ and $Y$ respectively.Let $A_1X=A_1Y=x$(circumradius of the square).It is also easy to note that $XY \parallel BC$.Thus $\angle{AXA_1}=45+B$ and $\angle{AYA_1}=45+C$.Thus applying the sine rule in $\triangle{XAA_1}$ and $\triangle{YAA_1}$ respectively we get $\frac{x}{sin{\angle{BAA_1}}}=\frac{AA_1}{sin(45+B)}$...(1) $\frac{x}{sin{\angle{CAA_1}}}=\frac{AA_1}{sin(45+C)}$...(2) Dividig (2) by (1) we get $\frac{sin{\angle{BAA_1}}}{sin{\angle{CAA_1}}}=\frac{sin(45+B)}{sin(45+C)}$ Similarly considering the squares on the other sides we get $\frac{sin{\angle{CBB_1}}}{sin{\angle{ABB_1}}}=\frac{sin(45+C)}{sin(45+A)}$ and $\frac{sin{\angle{ACC_1}}}{sin{\angle{BCC_1}}}=\frac{sin(45+A)}{sin(45+B)}$ Multiplying these expressions we get $\frac{sin{\angle{BAA_1}}}{sin{\angle{CAA_1}}} \dot \frac{sin{\angle{CBB_1}}}{sin{\angle{ABB_1}}} \dot \frac{sin {\angle{ACC_1}}}{sin{\angle{BCC_1}}}=1$ so by converse of Ceva's theorem we are done!!

Easy Solution: Let the square with center $A_1$ touch $AC$ at $M$ and $AB$ at $L$. Using the Law of Sines on $\Delta AA_1K$, we find that$\frac{\sin{\angle A_1AK}}{A_1K}=\frac{\sin{(C+45)}}{AA_1}$. Similarly, we find that $\frac{\sin{\angle A_1AL}}{A_1L}=\frac{\sin{(B+45)}}{AA_1}$ in $\Delta AA_1L$. Dividing the two equations, we find that $\frac{\sin{\angle A_1AK}}{\sin {\angle A_1AL}}=\frac{\sin{(C+45)}}{\sin{(B+45)}}$. Multiplying the similar expressions for $B$ and $C$ yields the result using Trig Ceva. $\Box$

Gah, how did I not see Jacobi's theorem?

Let vertex of square $A_1$ lie on $AB$ at $P$, and on $AC$ at $Q$. By sine rule on $\Delta AA_1P$ and on $\Delta AA_1Q$, we have $\sin BAA_1=A_1P.\frac{\sin APA_1}{AA_1}=A_1P.\frac{\sin{(B+45)}}{AA_1}$ $\sin CAA_1=A_1Q.\frac{\sin AQA_1}{AA_1}=A_1Q.\frac{\sin{(C+45)}}{AA_1}$ Now as $A_1P=A_1Q$, we have $\frac{\sin BAA_1}{\sin CAA_1}=\frac{\sin (B+45)}{\sin(C+45)}$ Hence, the trigonometric form of Ceva's theorem directly follows!

@2above Here is a more efficient bash: There exists a homothety on $A$ mapping square $A_1$ with a new square externally off $BC$. Let's call the center of this square $A_2$. Similarly define $B_2$ and $C_2$. It suffices to show $AA_2, BB_2, CC_2$ concurrent. Now we can use bary. By Conway's formula, we can compute $A_2$ as $(-a^2:S_C+S:S_B+S)$, and can compute $B_2$ and $C_2$ in a similar fashion. They meet at $$((S_B + S)(S_C + S) : (S_A + S)(S_C + S) : (S_B + S)(S_A + S)),$$which is the Outer Vecten Point. $\blacksquare$

Image from StefanS By law of sines on $\triangle{AKA_1}$ and $\triangle{ANA_1}$, we get that $\frac{\sin{\angle{KAA_1}}}{KA_1}=\frac{\sin{\angle{AKA_1}}}{AA_1}$ and $\frac{\sin{\angle{NAA_1}}}{NA_1}=\frac{\sin{\angle{ANA_1}}}{AA_1}$. Denote $\angle{KAA_1}$ as $\alpha_1$ and $\angle{NAA_1}$ as $\alpha_2$. We have $\frac{\sin{\angle{B+45}}}{KA_1}=\frac{\sin{\angle{\alpha_1}}}{AA_1}$ and $\frac{\sin{\angle{C+45}}}{NA_1}=\frac{\sin{\angle{\alpha_2}}}{AA_1}$ respectively. Now divide the first equation by the second to get $\frac{\sin{\angle{B+45}}}{\sin{\angle{C+45}}}=\frac{\sin{\angle{\alpha_1}}}{\sin{\angle{\alpha_2}}}$. If we extend the orange line to meet $BC$ at $M$, we need to find $\frac{BM}{MC}$ and use Ceva's Theorem to prove concurrency. By ratio lemma, $\frac{BM}{MC}=\frac{AB*\sin{\alpha_1}}{AC*\sin{\alpha_2}}=\frac{\sin{\angle{B+45}}}{\sin{\angle{C+45}}}$ from which we found earlier. Doing the same with $B$ and $C$ as we did with $A$, we can use trig ceva to finish. $\blacksquare{}$

Taking a homothety at $A$ gives that $A_2$ lies on $AA_1$ where $A_2$ is the center of the square with side $BC$ outside the triangle. It remains to show that $AA_2$, $BB_2$, and $CC_2$ are collinear. Note that \[ \frac{\sin BAA_2}{\sin CAA_2} = \frac{\sin BAA_2 \slash BA_2}{\sin CAA_2 \slash CA_2} = \frac{\sin ABA_2 \slash AA_2}{\sin ACA_2 \slash AA_2} = \frac{\sin(B + 45^\circ)}{\sin(C + 45^\circ)} \] The result follows by Trignometric Ceva then.

Take a homothety sending the square to the outer square with side $BC$. The result follows by Jacobi's theorem / Trig Ceva.

Take a homothety that sends the square to the square erected off side $BC$ and similarly for the other ones. Call the new centers $A_2$, $B_2$, and $C_2$. Then the concurrency is equivalent to \[ 1=\prod \frac{\sin \angle BA_2A}{\sin \angle AA_2C} = \prod \frac{\sin \angle ABA_2 \cdot \frac{BA_2}{AA_2}}{\sin \angle ACA_2 \cdot \frac{CA_2}{AA_2}} = \prod \frac{\sin (45+B)}{\sin (45+C)} = 1,\]and we are done.

Huh… I thought I posted on this topic before, but apparently I haven’t. It suffices to take a homothety that brings each of the squares outside (which works since the triangle formed by two vertices with a gertice of the triangle is similar to the original), then the centers of those form 45 degree angles, which by Jacobi’s concur (or by trig ceva).

lul Homothety at $A$ sends the square to square outside $ABC$ with side $BC$, and similarly for other vertices. Now Jacobi's theorem finishes.

I was surprised that after the homothety there wasn't a particularly "clean" solution that didn't use trigonometry or transformations.. it seemed like a very familiar configuration; some magical force even made me think it was the Lemoine point

We see that $\frac{CA_2}{A_2B}=\frac{1+\cot \angle C}{1+\cot \angle B}$. Multiplying cyclically gives $1$, so by Ceva's we are done.

take a homothety sending the top side of the square to BC ,then the problem is just proving the triangle formed by centers of 3 squares with sides AB BC AC is perspective with ABC, however this is true as the perspector is the outer Vecten point X485 of ABC

what trig?

Let $B_A, C_A$ be the point on $AB, AC$ that are also on the square. Now we have $\frac{\sin BAA_1}{\sin CAA_1} =\frac{\sin B_AAA_1}{\sin C_AAA_1} = \frac{B_AA_1 \cdot AA_1 \sin AC_AA_1}{C_AA_1 \cdot AA_1 \sin AB_AA_1} = \frac{\sin 45 + C}{\sin 45 + B} $. Multiplying cyclically and using trig ceva finishes.

Let $AA_1, BB_1$ intersect at $X$. Consider the square with center $A_1$. Let it touch $AB \text{ and } BC$ at $P \text{ and } Q$ respectively. Let $A_1P = A_1Q = l$. From Law of Sines we have, $$\frac{sin\angle BAA_1}{l} = \frac{sin\angle (B+45)}{AA_1}, \text{ }\frac{sin\angle CAA_1}{l} = \frac{sin\angle (C+45)}{AA_1},$$$$ \implies \frac{sin\angle BAA_1}{sin\angle CAA_1} = \frac{sin\angle (B+45)}{sin\angle (C+45)}$$ by Ceva's theorem, $$ \frac{sin\angle BAA_1}{sin\angle CAA_1}\cdot \frac{sin\angle ACX}{sin\angle BCX}\cdot \frac{sin\angle CBB_1}{sin\angle ABB_1} = 1$$$$\frac{sin\angle ACX}{sin\angle BCX} = \frac{sin\angle (A+45)}{sin\angle (B+45)} = \frac{sin\angle ACC_1}{sin\angle BCC_1}$$Hence, $X \equiv C_1$

Consider a homothety centered at $A$ taking the inscribed square centered at $A_1$ to the square with side $BC$ outside the triangle. Suppose that $A'$ is the center of this square. Then by the Law of Sines $$\frac{\sin \angle BAA_1}{\sin \angle CAA_1} = \frac{\sin \angle BAA'}{\sin \angle CAA'} = \frac{ \sin (\angle B + 45^\circ) \cdot \frac{BA'}{AA'}}{\sin (\angle C + 45^\circ) \cdot \frac{CA'}{AA'}} = \frac{\sin (\angle B + 45^\circ)}{\sin (\angle C + 45^\circ)},$$and a similar equation holds for $B, C.$ Multiplying these equations, we finish by the Trignometric form of Ceva's Theorem. QED