This is also USAMO #5.

guys coord bash worked and took about 20 minutes set P as origin and gamma as y-axis YAY USAMO #5!!!

I did coord bash with x axis. Did anyone use Menelaus/Ceva+harmonic divisions?

BILL9 wrote: guys coord bash worked and took about 20 minutes set P as origin and gamma as y-axis YAY USAMO #5!!! I set P as a point on the y axis, gamma as y axis, and one of the points on the x axis it didn't really work; way too much algebra, but i put it down anyways maybe 1-2 points

I had 3 hours left after 4 and 5 and I gave up after reading the question. Fail and stupid.

Generalizing my above statement, did anyone use projective geometry at all? It seemed like a very good candidate...

Observe that angle A'PC=APC', and same for the other symmetrical angles. Then by law of sines, A'C/sin(A'PC)=A'P/sin(A'CP). Similarly, AC'/sin(APC')=C'P/sin(C'AP). Divide the two equations to get A'C/AC'=A'Psin(C'AP)/C'Psin(A'CP)=A'Psin(BAP)/C'Psin(BCP). Get symmetrical expressions for B'A/BA' and C'B/CB', and multiply them together. You're left with 6 sines, 3 on top and 3 on bottom. Multiply top and bottom by PA*PB*PC. Use Law of Sines again, using triangles PAB, PAC, and PBC, and you can finish with Menelaus's Theorem. (Sorry, this is kinda brief, hope this makes sense.)

Hm. Coordinate bashed it, with $\gamma$ as the x-axis and $P$ as the origin. I got some really nasty coordinates, and said the area was 0 by the Shoelace Theorem, and so the points were collinear. How many points do you think that will get me?

Coordinate bashing with $\gamma$ as the x-axis should work nicely.

Use Menelaus. Then converted the ratio of lengths to ratio of areas. Then convert the areas to 1/2 * a * b * sinx, and it magically cancels out.

My strategy: I used the x-axis. I got an expression for the slope of $A'B'$. I showed that switching the coordinates of $A$ and $B$ in the expression left it unchanged, and since I was running out of time, said that you could show the same for switching $B$ and $C$. This shows that the expression is symmetric, so the slope of $B'C'$ is the same, and they are collinear.

Used Simsun lines. Construct perpendiculars to PA', PB', PC', at points A', B', C', respectively. Call the pairwise intersections D, E, F Then you get a whole bunch of cyclic quads, and you can use them to prove P, D, E, F are cyclic, so A', B', C' form the Simsun line of triangle DEF with respect to point P. Just got this method w/ like 15 min left, spent 5 min trying to get a good diagram (one that fit in the margins), and finished writing w/ 30 sec left

ken961996 wrote: Used Simsun lines. Construct perpendiculars to PA', PB', PC', at points A', B', C', respectively. Call the pairwise intersections D, E, F Then you get a whole bunch of cyclic quads, and you can use them to prove P, D, E, F are cyclic, so A', B', C' form the Simsun line of triangle DEF with respect to point P. Just got this method w/ like 15 min left, spent 5 min trying to get a good diagram (one that fit in the margins), and finished writing w/ 30 sec left From my Geogebra diagram it doesn't look like $P,D,E,F$ are cyclic. Darn. Edit: Okay they are. I drew the diagram wrong.

arrrrrrrrrrrrrrrgh i didn't realize that you could coordinate bash this problem in 10 minutes until after the test. I looked for cyclic quads and similar triangles and couldn't find any

mentioned menelaus... tried areas, didn't seem to get anywhere but aghhh maybe if I had done something a bit different I woulda got the full points... maybe 1 point for mentioning menelaus?

henrypickle wrote: . maybe 1 point for mentioning menelaus? Yeah, I did that too! (even though I don't even know menelaus on the top of my head) I didn't have enough time to do this one though, so I just wrote a few things I found and said menelaus finishes it off.

This was the very first Olympiad geometry problem that I've ever used coordinates on.

i have a question - I bashed this one out with P at origin and $\gamma$ on y-axis, but didn't show much work. How many points will I lose for not showing intermediate algebra? I listed all the equations I started with and ended up with, but I figured the graders couldn't possibly want to see the 40 term equations I worked through in the middle.

AceOfDiamonds wrote: i have a question - I bashed this one out with P at origin and $\gamma$ on y-axis, but didn't show much work. How many points will I lose for not showing intermediate algebra? I listed all the equations I started with and ended up with, but I figured the graders couldn't possibly want to see the 40 term equations I worked through in the middle. you'll get a 0 for sure :\ . The graders don't like seeing stuff like that.

Guidance from MarkBcc168's Handout. Guess I'm learning. Let $X = \overline{AB} \cap \overline{A'B'}$. Then by Dual of Desgargues Involution upon complete quadrilateral $ABA'B'XC$ there exists a unique involution $f$ swapping the reciprocal pairs $(PA, PA')$, $(PB, PB')$ and $(PX, PC)$. Now note that the involution $f'$ which reflects each line about $\gamma$ is an involution and sends $f'(PA) \mapsto PA'$ and $f'(PB) \mapsto PB'$. Hence the reflection about $\gamma$ of $PX$ is $PC$, and thus $X \equiv C'$.

We invoke barycentric coordinates with $\triangle PAA'$ as our reference triangle with $P = (1, 0, 0)$, $A = (0, 1, 0)$, and $A' = (0, 0, 1)$, and let $AA' = a$, $A'P = b$, and $PA = c$. We let $B = (p:q:r)$ and $C = (p_1:q_1:r_1)$ (sorry I must use this aweful notation because it's what I wrote on my paper). Remember that $B$, $A'$ and $C$ are collinear. We will use this later. We can see that $PB'$ is parametrized by $\left(\frac{a^2}{r}: \frac{b^2}{q}: x \right)$. Also notice that $AC$ is given by \[\begin{vmatrix}0&1&0 \\ p_1&q_1&r_1 \\ x&y&z \end{vmatrix} = 0 \Rightarrow p_1z = r_1x\]Thus we can see that \[B' = \left( \frac{a^2}{r_1p} : \frac{b^2}{r_1q} : \frac{a^2}{p_1p}\right)\]Similarly we get that \[C' = \left(\frac{a^2}{p_1r}:\frac{b^2}{q_1r}:\frac{a^2}{p_1p}\right)\]Thus it remains to prove that \[\begin{vmatrix} 0&0&1 \\ \frac{a^2}{r_1p} & \frac{b^2}{r_1q} & \frac{a^2}{p_1p} \\ \frac{a^2}{p_1r}&\frac{b^2}{q_1r}&\frac{a^2}{p_1p} \end{vmatrix}\]Notice that this simplifies to proving that $p_1q = pq_1$. Consider line $BA'C$. This line gives that \[\begin{vmatrix} 0&0&1 \\ p&q&r \\ p_1 & q_1&r_1 \end{vmatrix} = 0 \Rightarrow p_1q = pq_1\]$\blacksquare$

WHAT DDIT on $BCC’B’$ gives an involution swapping $(PC, PC’), (PB,PB’), (PA, PA^*)$ where $A^*=B_1C_1\cap BC$. But the first two involute pairs show that the involution is a reflection about $\gamma$, finishing.

We have : $PB , PB'$ are isogonal with respect to$ \angle APA'$. $PC , PC'$ are isogonal with respect to$ \angle APA'$. Let $A'B' \cap AB = X$. By The second isogonality lemma we have $ PX , PC$ are isogonal with respect to$ \angle APA'$. $ \implies X= C'$.

Shreyasharma wrote: What a cute algebra problem. We complex bash. Set $\gamma$ as the real axis. Also without loss of generality shift till $p = 0$. Then take $a$, $b$ and $c$ as free variables. Let $D = \overline{AP} \cap \overline{BC}$. Similarly define $E$ and $F$. Computing $d$ we find, \begin{align*} d = \frac{ - a(\overline{b}c - b\overline{c})}{\overline{a}(b-c) - a(\overline{b}-\overline{c})} \end{align*}Okay that wasn't too bad. Now similarly we can compute, \begin{align*} e &= \frac{- b(\overline{c}a - c\overline{a})}{\overline{b}(c-a) - b(\overline{c} - \overline{a})}\\ f &= \frac{- c(\overline{a}b - a\overline{b})}{\overline{c}(a-b) - c(\overline{a}-\overline{b})} \end{align*}Note that $A'$, $B'$ and $C'$ are just reflections over the real axis of $D$, $E$ and $F$ so it suffices to show that $D$, $E$ and $F$ are collinear. To show this it suffices to show, \begin{align*} 0 &= \begin{vmatrix} d & \overline{d} & 1\\ e & \overline{e} &1\\ f & \overline{f} & 1 \end{vmatrix}\\ &= \begin{vmatrix} \frac{ - a(\overline{b}c - b\overline{c})}{(\overline{a}b - a\overline{b}) - (\overline{a}c - a\overline{c})} & \frac{ -\overline{a}(\overline{b}c - b\overline{c} )}{(\overline{a}b - a\overline{b}) - (\overline{a}c - a\overline{c})} & 1\\ \frac{ -b(\overline{c}a - c\overline{a})}{ (\overline{b}c - b\overline{c}) - (\overline{b}a - b\overline{a}) } & \frac{-\overline{b}(\overline{c}a - c\overline{a})}{ (\overline{b}c - b\overline{c}) - (\overline{b}a - b\overline{a})} &1\\ \frac{-c(\overline{a}b - a\overline{b})}{(\overline{c}a -c\overline{a}) - (\overline{c}b - c\overline{b})} & \frac{-\overline{c}(\overline{a}b - a\overline{b}}{(\overline{c}a -c\overline{a}) - (\overline{c}b - c\overline{b})} & 1 \end{vmatrix} \end{align*}Then it suffices to show that, \begin{align*} 0 &= \begin{vmatrix} a & \overline{a} & \frac{(\overline{a}b - a\overline{b}) - (\overline{a}c - a\overline{c})}{(\overline{b}c - b\overline{c})}\\ b & \overline{b} & \frac{(\overline{b}c - b\overline{c}) - (\overline{b}a - b\overline{a})}{(\overline{c}a - c\overline{a})}\\ c & \overline{c} & \frac{(\overline{c}a -c\overline{a}) - (\overline{c}b - c\overline{b})}{ (\overline{a}b - a\overline{b})} \end{vmatrix} \end{align*}Now we expand by minors to find, \begin{align*} \sum_{cyc} \frac{(\overline{c}a -c\overline{a}) - (\overline{c}b - c\overline{b})}{ (\overline{a}b - a\overline{b})} \cdot (a\overline{b} - \overline{a}b) &= \sum_{cyc} -\overline{c}a + c\overline{a} + \overline{c}b - c\overline{b}\\ &= 0 \end{align*}and we're done. Can someone explain what expanding by minors is?

signifance wrote: Shreyasharma wrote: What a cute algebra problem. We complex bash. Set $\gamma$ as the real axis. Also without loss of generality shift till $p = 0$. Then take $a$, $b$ and $c$ as free variables. Let $D = \overline{AP} \cap \overline{BC}$. Similarly define $E$ and $F$. Computing $d$ we find, \begin{align*} d = \frac{ - a(\overline{b}c - b\overline{c})}{\overline{a}(b-c) - a(\overline{b}-\overline{c})} \end{align*}Okay that wasn't too bad. Now similarly we can compute, \begin{align*} e &= \frac{- b(\overline{c}a - c\overline{a})}{\overline{b}(c-a) - b(\overline{c} - \overline{a})}\\ f &= \frac{- c(\overline{a}b - a\overline{b})}{\overline{c}(a-b) - c(\overline{a}-\overline{b})} \end{align*}Note that $A'$, $B'$ and $C'$ are just reflections over the real axis of $D$, $E$ and $F$ so it suffices to show that $D$, $E$ and $F$ are collinear. To show this it suffices to show, \begin{align*} 0 &= \begin{vmatrix} d & \overline{d} & 1\\ e & \overline{e} &1\\ f & \overline{f} & 1 \end{vmatrix}\\ &= \begin{vmatrix} \frac{ - a(\overline{b}c - b\overline{c})}{(\overline{a}b - a\overline{b}) - (\overline{a}c - a\overline{c})} & \frac{ -\overline{a}(\overline{b}c - b\overline{c} )}{(\overline{a}b - a\overline{b}) - (\overline{a}c - a\overline{c})} & 1\\ \frac{ -b(\overline{c}a - c\overline{a})}{ (\overline{b}c - b\overline{c}) - (\overline{b}a - b\overline{a}) } & \frac{-\overline{b}(\overline{c}a - c\overline{a})}{ (\overline{b}c - b\overline{c}) - (\overline{b}a - b\overline{a})} &1\\ \frac{-c(\overline{a}b - a\overline{b})}{(\overline{c}a -c\overline{a}) - (\overline{c}b - c\overline{b})} & \frac{-\overline{c}(\overline{a}b - a\overline{b}}{(\overline{c}a -c\overline{a}) - (\overline{c}b - c\overline{b})} & 1 \end{vmatrix} \end{align*}Then it suffices to show that, \begin{align*} 0 &= \begin{vmatrix} a & \overline{a} & \frac{(\overline{a}b - a\overline{b}) - (\overline{a}c - a\overline{c})}{(\overline{b}c - b\overline{c})}\\ b & \overline{b} & \frac{(\overline{b}c - b\overline{c}) - (\overline{b}a - b\overline{a})}{(\overline{c}a - c\overline{a})}\\ c & \overline{c} & \frac{(\overline{c}a -c\overline{a}) - (\overline{c}b - c\overline{b})}{ (\overline{a}b - a\overline{b})} \end{vmatrix} \end{align*}Now we expand by minors to find, \begin{align*} \sum_{cyc} \frac{(\overline{c}a -c\overline{a}) - (\overline{c}b - c\overline{b})}{ (\overline{a}b - a\overline{b})} \cdot (a\overline{b} - \overline{a}b) &= \sum_{cyc} -\overline{c}a + c\overline{a} + \overline{c}b - c\overline{b}\\ &= 0 \end{align*}and we're done. Can someone explain what expanding by minors is? \[ \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} = a_{11} \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33}\end{vmatrix}- a_{12} \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33}\end{vmatrix} + a_{13} \begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32}\end{vmatrix} \]

This generalizes for all $n \times n$ matrices, by considering the alternate sum and difference of the smaller $(n - 1) \times (n - 1)$ matrices. If you have EGMO, it's mentioned in the addendum on linear algebra.

This is also known as the cofactor expansion of the determinant of a matrix. This may be helpful - https://people.math.carleton.ca/~kcheung/math/notes/MATH1107/wk07/07_cofactor_expansion.html

Define $X$ by reverse reconstruction as the intersection of $AB$ and $A'B'$. Observe in quadrilateral $AA'BB'$ there is an involution swapping $PA,PA'$, $PB,PB'$ and $PC,PC'$. From the first two pairs, we see the involution is reflection about the line $\gamma$. This means that $X$ is on the reflection of $PC$, making it the same point as $C'$. Therefore, $A'$, $B'$, $C'$ are collinear.

We employ complex numbers. What a weird setup! Set $P$ be the origin and $\gamma$ be the real axis. Then, $A_1$ is the reflection of $A$ across the real axis. So, \[a_1=\overline{a} , b_1=\overline{b} \text{ and } c_1 = \overline{c}\]Now, note that since $A' = \overline{BC} \cap \overline{PA_1}$, \begin{align*} a' &= \frac{(\overline{b}c-b\overline{c})(p-a_1)-(b-c)(\overline{p}a_1-p\overline{a_1})}{(\overline{b}-\overline{c})(p-a_1)-(b-c)(\overline{p}-\overline{a_1})}\\ &= \frac{\overline{a}(\overline{b}c-b\overline{c})}{\overline{a}(\overline{c}-\overline{b})+a(b-c)} \end{align*}Similarly, we obtain that, \[b' = \frac{\overline{b}(\overline{c}a-c\overline{a})}{\overline{b}(\overline{a}-\overline{c})+b(c-a)}\]and \[c' = \frac{\overline{c}(\overline{b}a-b\overline{a})}{\overline{c}(\overline{b}-\overline{a})+c(a-b)}\]Now, let $E= \frac{a'-b'}{a'-c'}$. Note, that, \begin{align*} E &= \frac{a'-b'}{a'-c'}\\ &= \frac{\frac{\overline{a}(\overline{b}c-b\overline{c})}{\overline{a}(\overline{c}-\overline{b})+a(b-c)} - \frac{\overline{b}(\overline{c}a-c\overline{a})}{\overline{b}(\overline{a}-\overline{c})+b(c-a)}}{\frac{\overline{a}(\overline{b}c-b\overline{c})}{\overline{a}(\overline{c}-\overline{b})+a(b-c)}-\frac{\overline{c}(\overline{b}a-b\overline{a})}{\overline{c}(\overline{b}-\overline{a})+c(a-b)}}\\ \end{align*}We then let, $A=\overline{b}c-b\overline{c}$ , $B=\overline{c}a-c\overline{a}$ and $C=\overline{b}a-b\overline{a}$. Further, let $P=\overline{a}(\overline{c}-\overline{b})+a(b-c)$, $Q=\overline{b}(\overline{a}-\overline{c})+b(c-a)$ and $R=\overline{c}(\overline{b}-\overline{a})+c(a-b)$. Thus, we obtain \begin{align*} E&= \frac{\frac{\overline{a}A}{P}-\frac{\overline{b}B}{Q}}{\frac{\overline{a}A}{P}-\frac{\overline{c}C}{R}}\\ &= \frac{R(\overline{a}AQ-\overline{b}BP)}{Q(\overline{a}AR-\overline{c}CP)} \end{align*}one can do a quick check and verify that $\overline{A}=A$, $\overline{B}=B$, $\overline{C}=C$, $\overline{P}=P,\overline{Q}=Q$ and $\overline{R}=R$. Thus, \[\overline{E}=\frac{R(aAQ-bBP)}{Q(aAR-cCP)}\]we wish to have $E=\overline{E}$. Note the following chain of equivalences, \begin{align*} E &= \overline{E}\\ \frac{R(\overline{a}AQ-\overline{b}BP)}{Q(\overline{a}AR-\overline{c}CP)} &= \frac{R(aAQ-bBP)}{Q(aAR-cCP)}\\ a\overline{a}A^2RQ - \overline{a}cACPQ - a\overline{b}ABPR + \overline{b}cBCP^2 &= a\overline{a}A^2RQ - \overline{a}bABPR -a\overline{c}ACPQ +b\overline{c}BCP^2\\ (\overline{a}b-a\overline{b})ABPR + (\overline{a}c-a\overline{c})ACPQ + (\overline{b}c-b\overline{c})BCP^2 &=0\\ -ABCPR - ABCPQ -ABCP^2 &=0\\ -ABCP(P+Q+R) &=0 \end{align*}So, it suffices to show that $P+Q+R=0$ which is quite clear since, \begin{align*} P+Q+R &= \overline{a}(\overline{c}-\overline{b})+a(b-c) + \overline{b}(\overline{a}-\overline{c})+b(c-a) + \overline{c}(\overline{b}-\overline{a})+c(a-b)\\ &= \overline{a}\overline{c} - \overline{a}\overline{c} + \overline{a}\overline{b} - \overline{a}\overline{b} + \overline{b}\overline{c} - \overline{b}\overline{c} + ab-ab+ac-ac+bc-bc\\ &=0 \end{align*}as desired. So, it is clear that $\frac{a'-b'}{a'-c'} \in \mathbb{R}$. Thus, points $A'$, $B'$ and $C'$ are collinear which was the required result.

$\measuredangle APB=\measuredangle(AP,\gamma)+\measuredangle(\gamma,PB)=-(\measuredangle(A'P,\gamma)+\measuredangle(\gamma,PB'))=-\measuredangle A'PB',$ thus $P$ has an isogonal conjugate $Q$ in quadrilateral $ABA'B',$ thus $A'B'$ is tangent to the inellipse with foci $P,Q.$ Similarly $B'C',C'A'$ tangent as well and now it is easy to check these three must be the same line.

Define $C_1 = A'B' \cap AB$. Then, applying DDIT on quadrilateral $BC'B'CAA'$, we see that the involution is the reflection wrt $\gamma$, so $PC_1, PC$ are symmetric wrt $\gamma \implies C_1 = C'$. $\blacksquare$

construct perpendiculars of $PA'$, $PB'$, and $PC'$ let the perpendiculars at $B'$ and $C'$ intersect at points $A''$, and similarly define $B''$ and $C''$ doing angle chasing yields that $C'A''P$ and $A'C''P$ are similar then, $CA''P=A'C''P$ and thus $A''B''C''P$ are cyclic then, $A'B'C'$ is the simson line of $P$ on circle $A''B''C''$

Define $C'$ as the intersection of $AB$ with the isoptic cubic of $(AA')(BB')$ We want to prove that $C'$ lies on $B'A'$.. First since $APB'=BPA'$ by reflection, we have that $P$ lies on this isoptic cubic. However, since $A'PC = APC'$ by reflection, we have that $P$ lies on the isoptic cubic of $(AA')(CC')$. Additionally, we have that $B$ and the isogonal conjugate of $P$ lie on this isoptic cubic as well, since $B=AC' \cap A'C$. Since they also have a common Newton line (by reflection), this implies that these two isoptic cubics are the same (they share $A,B',B,A',C,M,P, P^*, I,J,\text{(circle points)},\infty_{\text{Newton}}$, ten points), and thus $C,C'$ are isogonal conjugates in $(AA')(BB')$ (since for a points $P, P'$ and quadrilateral $(AC)(BD)$, we have isoptic cubics of $(AC)(BD)$ and $(AC)(XX')$ are equal iff. $X,X'$ are isogonal conjugates. This can also be approached with the theory of perfect isogonal six-point sets; $(AA')(BB')(CC')$ is a perfect isogonal six-point set). Since $C=AB' \cap BA'$, we get $C' = AB \cap B'A'$.

Attachments:

Solution with Desargue's Involution Theorem Firstly observe that because of the construction by reflection, there exists a unique projective involution swapping: $(PA,PA_1) (PB,PB_1) (PC,PC_1)$ Now introduce the point $C' := AB \cap A_1B_1$ We can see that $AB, AB_1, BA_1, A_1B_1$ make up a complete quadrilateral with intersections in $C$ and $C'$ By the dual of Desargue's Involution theorem there exists a unique projective involution swapping: $(PA, PA_1)$ $(PB, PB_1)$ $(PC, PC_1)$ Therefore this must be the same involution mentioned before, so the points $C'$ and $C_1$ must be the same!