Bariz - Problem

Source: 17th Durer Competition 2023 E+ P5

Tags: geometry

oVlad

25.11.2023 13:50

For a given triangle $A_1A_2A_3$ and a point $X{}$ inside of it we denote by $X_i$ the intersection of line $A_iX$ with the side opposite to $A_i$ for all $1\leqslant i \leqslant 3.$ Let $P{}$ and $Q{}$ be distinct points inside the triangle. We then say that the two points are isotomic (i.e. they form an isotomic pair) if for all $i{}$ the points $P_i$ and $Q_i$ are symmetric with respect to the midpoint of the side opposite to $A_i.$ Augustus wants to construct isotomic pairs with his favourite app, GeoZebra. He already constructed the vertices and sidelines of a non-isosceles acute triangle when suddenly his computer got infected with a virus. Most tools became unavailable, only a few are usable, some of which even require a fee: Point: select an arbitrary point (with respect to the position of the mouse) on the plane or on a figure (circle or line) - free Intersection: intersection points of two figures (where each figure is a circle or a line) - free Line: line through two points - $5 per use Perpendicular: perpendicular from a point to an already constructed line - $50 per use Circumcircle: circle through three points - $10 per use Agatha selected a point $P{}$ inside the triangle, which is not the centroid of the triangle. Show that Augustus can construct a point $Q{}$ at a cost of at most 1000 dollars such that $P{}$ and $Q{}$ are isotomic. Prove that for any $n\geqslant 1$ Augustus can construct $n{}$ different isotomic pairs at a cost of at most $200 + 10n$ dollars. Note: The parts are unrelated, that is Augustus can’t use his constructions from part a) in part b).

starchan

25.11.2023 18:09

this is an.. interesting.. problem for the sake of mindset, we change dollars to rubles and the triangle from $A_1A_2A_3$ to $ABC$.

We first spend $50$ rubles and drop the altitude from $A$ onto $BC$, say at some point $D$. Now we spend $10$ rubles and draw circle $(ADB)$. We also mark the intersection of this circle with $AC$. Note that this point is $E$, the foot of perpendicular from $B$ onto $AC$. Similarily, we spend $10$ more rubles to get $F$, the foot of altitude from $C$ onto $AB$. So far we have spent $50+10+10= 70$ rubles. We now draw $(DEF)$, the nine point circle of $\triangle ABC ~$. This allows us to mark the midpoints $M_a, M_b, M_c$. So far, we have spent $80$ rubles and we have all the midpoints $M_a, M_b, M_c$. We now draw lines $AM_a$ and $CM_c$, bringing up our cost to $90$ rubles. Note that we can now mark the centroid $G$ as well. We draw circle $(AM_aB)$ while spending $10$ rubles. Our cost so far is exactly $100$ rubles. Finally, using the claim from part (a) we draw the line through $B$ parallel to $CM_c$, spending another $100$ rubles. Let the intersection of this line with $(AM_aB)$ be $X$. Our total cost so far has been $200$ rubles. Claim: If $P$ is a point on $AM_a$ other than $G$, then circle $(BXP)$ meets $AM_a$ at another point $Q$, and this is the isotomic conjugate of $P$ wrt $\triangle ABC ~$. Proof: Before we prove this, we will need to complete the underlying configuration. Let $A'$ denote the reflection of $A$ in $M_a$ and $G'$ the reflection of $G$ in $M_a$. Note that since $AG/AM = 2/3$, $G'$ is actually the midpoint of segment $GA'$. Let $\omega$ denote the circle with diameter $GA'$. The same ratios imply that $A$ and $M_a$ are inverses wrt $\omega$. Moreover, since $BX \parallel CM_c$, it is easy to see that $B, X, G'$ are actually collinear. As a result, $G'B \cdot G'X = G'M_a \cdot G'A$. It follows that $B, X$ are inverses wrt $\omega$. Hence, if $(BXP)$ meets $AM_a$ again at a point $Q$, then points $P, Q$ are inverses wrt $\omega$. As a result \[(PQ;GM_a) = -1\]Projecting through $B$ onto $AC$, we obtain that the midpoint of points $BP \cap AC$ and $BQ \cap AC$ is $BG \cap AC = M_b$. We have a similar conclusion after projecting through $C$, and therefore $P, Q$ are isotomic conjugates wrt $\triangle ABC ~$. With our key claim proven, the end is near. We mark $n$ arbitrary distinct points $P_1, P_2, \dots P_n$ lying strictly inside segment $AG$. For each $1 \leq i \leq n$ we spend $10$ rubles and draw the circle $(BXP_i)$. We also mark the second intersection $Q_i$ of this newly drawn circle with $AM_a$. As a result, we have spent $200+10n$ rubles so far, and have succeeded in creating $n$ pairs of isotomic conjugates, namely: $(P_i, Q_i)$ for all $1 \leq i \leq n$. This finishes the proof for this part $\square$.

oVlad

21.12.2023 14:18

Here's a solution for b) with a much better bound found by my teammate Triangle_Center. We show that we can create $2k^2$ pairs of isotomic points with $100+40k$ dollars, that is, $O(\sqrt{n})$ dollars, in the original wording of the statement. We use the 100 dollars to make the setup. Denote the acute scalene triangle as $ABC$. For 50 dollars we get the perpendicular from $B$ to $AC$, and intersect it with $AC$ at $D$. For 10 dollars we get the circle $(BDC)$ and intersect it with $AB$ again at $E$, the foot of the perpendicular from $C$ to $AB$. For 5 dollars we get the line $CE$, which is the perpendicular from $C$ to $AB$, and we intersect it with $BD$ at $H$, the orthocenter of $ABC$. Next, for 10 dollars we get the circumcircle $\omega$ of $ABC$, and we intersect it with $BD$ and $CE$ at $X$ and $Y$ respectively, which are the reflections of $H$ in $AC$ and $AB$. Now, for 10 dollars we get the circle $(HXY)$, and for another 5 dollars, we get the line $AH$ and their intersection $W$, which is the reflection of $H$ in $A$. With the last 10 dollars we get lines $WX$ and $WY$, and intersect them with $\omega$ to get the points $Z$ and $T$. Observe that the midpoints of $HW$ and $HX$ are on $AC$, thus $WX\parallel AC$, so $Z$ is the reflection of $X$ with respect to the perpendicular bisector of $AC$, and $T$ is the reflection of $Y$ with respect to the perpendicular bisector of $AB$. The setup is done. Take a point $P$ on $AC$, close to $A$, so that it's not the midpoint of $AC{}$. For 10 dollars intersect circle $(PXZ)$ again with $AC$ at $Q$, which is the reflection of $P$ in the midpoint of $AC{}$. For 10 dollars draw the lines $BP$ and $BQ$. Thus, for 20 dollars we get a pair of isotomic cevians from $B$. We do this $k{}$ times for $B$ and $k{}$ times for $C$ thus constructing $2k{}$ cevians from $B$ and $2k{}$ cevians from $C$ at a cost of $40k$. Now, these cevians intersect in $4k^2$ points inside triangle $ABC$, and the isotomic conjugate of each of these points also appears as one of them. Therefore, we have the $2k^2$ pairs we needed. Remark. Let $f(n)$ be the lowest number of dollars required to construct $n{}$ pairs of isotomic points. Then, $f(n)\asymp\sqrt{n}$.