For a triangle $T$ and a line $d$, if the feet of perpendicular lines from a point in the plane to the edges of $T$ all lie on $d$, say $d$ focuses $T$. If the set of lines focusing $T_1$ and the set of lines focusing $T_2$ are the same, say $T_1$ and $T_2$ are equivalent. Prove that, for any triangle in the plane, there exists exactly one equilateral triangle which is equivalent to it.
Problem
Source: Turkey Team Selection Test 2018 P9
Tags: geometry, Simson Lines
WizardMath
27.03.2018 11:57
I believe this is well known. The envelope of the Simson lines is the Steiner deltoid of the triangle, whose incircle is the nine-point circle of the triangle. The Steiner deltoid of an equilateral triangle uniquely determines it (easy to see that the correspondence is bijective) so we are done.
kymkozan
27.03.2018 18:15
WizardMath wrote: I believe this is well known. The envelope of the Simson lines is the Steiner deltoid of the triangle, whose incircle is the nine-point circle of the triangle. The Steiner deltoid of an equilateral triangle uniquely determines it (easy to see that the correspondence is bijective) so we are done. Could you please explain this?
WizardMath
27.03.2018 18:58
kymkozan wrote: Could you please explain this? The curve that the Simson lines are all tangent to, turns out to be a deltoid. This deltoid is called the Steiner deltoid of the triangle. Since the deltoid is a hypocycloid with 3 cusps, all deltoids are similar to each other. So the Steiner deltoids of all triangles are similar. In particular, it is similar for an arbitrary triangle and an equilateral triangle. Also, the configuration of an equilateral triangle and its Steiner deltoid should be similar to a fixed configuration, so the equilateral triangle can be reconstructed uniquely, which finishes the proof.
KereMath
28.03.2018 10:57
Lemma: Let $ABC$ be a triangle with orthocentre $H$ and circumcircle $(ABC)$. Suppose $P$ on circle $(ABC)$. Let $l$ be Simson's line of $P$ wrt $ABC$. Prove that $l$ bisects $PH$.
Proof: Let $P_A,P_B,P_C$ be the projections of $P$ on the sides $BC,AC,AB$; let $Q$ be the intersection between $AH$ and the Simson's line of $P$. We want to show that $HQ=PP_A$. If we call $H_A$ the symmetric of $H$ with respect to the $BC$ side, we have that $H_A$ lies on the circumcircle of $ABC$, so, in order to prove that $HQ=PP_A$, it is sufficient to show that $PP_AQH_A$ is an isosceles trapezoid. Let $\alpha=P_CP_AB=P_BP_AC$. Since $PP_AP_CB$ and $PP_ACP_B$ are cyclic quadrilaterals, we have $\alpha=P_CPB=P_BPC$, too.
Now:
$P_AQH_A$=$\pi $−$\alpha$,
and:
$PH_AQ$=$PH_AA$=$PBA$=$PBP_C$=$\pi $−$\alpha$,
Since $PP_A$ and $QH_A$ are both ortogonal to $BC$, $PP_AQH_A$ is an isosceles trapezoid and we have $PP_A=QH$
$QED$.
Lemma 2: Let $S$ be the set of midpoints of $PH$ when $P$ lies on circumcircle of $T$. Prove that $S$ is $9$ $point$ $circle$ of $ABC$.
Proof: Let $H_a$ ,$A'$ be the intersection of $AH$ with $BC$ and circumcircle. Then let $HM$ intersects circumcirle at $X$ when $M$ is midpoint of $BC$.
We know that $BC$ bisects $HA'$ and $HM$. Let the midpoint of $OH$ is $F$. Take the homotethy from $H$. We know $F$ goes $O$. $M$ goes $X$. And $H_a$ goes $A'$. It is easy to see that $9$ $point$ $circle$ bisects $PH$.
$QED$
After some projective things we can get the nine point circle of $T$ and nine point circle of equilateral triangle is same.
Then we can get easily that the center of equilateral triangle is midpoint of $OH$. Then radius of 9 point circle is determined. So rest is easy with angle chasing.
nikolapavlovic
28.03.2018 19:45
We obviously wanna use some "invariant" properties of Simson lines by which i mean a structure which would impose some conditions on the beginning points on the circle.So maybe conurrency,parallelism,perpendicularity are ones we are after.Recall there's a cool property by which the angle two simson lines is equal to the size of the smaller half arc,so technically we lose the parallelism by perpendicularity becomes really nice,in particular we have the two point on both circles are antipodes. Label them $P,Q$ it's well-known that $s(P)$ halfes $HP$,so this motivates a $\mathcal {H}_{0.5}(H)$ it also takes $\mathcal {H}(P),\mathcal{H}(Q)$ to antipodes on the nine-point circle but $s(P)\cap s(Q)$ forms a right angle over $\mathcal{H}(P)\mathcal{H}(Q)$ $\implies$ $s(P)\cap s(Q)$ is on the nine point circle of bot triangles.Of course by picking two more pairs of normal simson lines we see that the nine-point circles of two triangles coincide. Note that by previous paragraph $E_9$ of $\triangle ABC$ is the center of the equilateral triangle and hence its circumcircle is $(E_9,\mathcal{R}(\odot ABC)\cdot 0.5)$.Obviously the above two circles cut each other in two points,i claim that at least one of them is the vertex of the equilateral triangle. Let the above circles cut in $V,W$ if $V,W$ are different from the vertices of equilateral triangle that their simson lines half both $VH,VE_9$ and hence are both parallel to $HE_9$ However two simson lines can't be parallel from the above (orthopolar is never in infinity) .So one of the vertices is fixed and hence we're done.