id int64 41 3.22M | problem stringlengths 38 4.44k | solution stringlengths 8 38.6k | problem_vector listlengths 1.02k 1.02k | solution_vector listlengths 1.02k 1.02k | last_modified stringdate 2025-08-11 00:00:00 2025-08-11 00:00:00 |
|---|---|---|---|---|---|
599,349 | Let $P$ be a point in the interior of an acute triangle $ABC$, and let $Q$ be its isogonal conjugate. Denote by $\omega_P$ and $\omega_Q$ the circumcircles of triangles $BPC$ and $BQC$, respectively. Suppose the circle with diameter $\overline{AP}$ intersects $\omega_P$ again at $M$, and line $AM$ intersects $\omega_P$ again at $X$. Similarly, suppose the circle with diameter $\overline{AQ}$ intersects $\omega_Q$ again at $N$, and line $AN$ intersects $\omega_Q$ again at $Y$.
Prove that lines $MN$ and $XY$ are parallel.
(Here, the points $P$ and $Q$ are [i]isogonal conjugates[/i] with respect to $\triangle ABC$ if the internal angle bisectors of $\angle BAC$, $\angle CBA$, and $\angle ACB$ also bisect the angles $\angle PAQ$, $\angle PBQ$, and $\angle PCQ$, respectively. For example, the orthocenter is the isogonal conjugate of the circumcenter.)
[i]Proposed by Sammy Luo[/i] | [b][color=red]Claim:[/color][/b] $M$ and $N$ are isogonal conjugates.
[i]Proof.[/i] Redefine $N$ as the isogonal conjugate of $M.$ It suffices to prove $N=(BQC)\cap(AQ).$ First, we claim $CBQN$ is cyclic. Indeed, $$\measuredangle BPC+\measuredangle BQC=\measuredangle BAC+180=\measuredangle BMC+\measuredangle BNC$$ and $\measuredangle BPC=\measuredangle BMC.$ Then, \begin{align*}&\measuredangle MCP=\measuredangle QCN=\measuredangle QBN\\&\implies \measuredangle CBA-\measuredangle NBA-\measuredangle CBQ+\measuredangle MAC=\measuredangle ACB-\measuredangle PCB-\measuredangle ACM+\measuredangle BAN\\&\implies \measuredangle CBA+\measuredangle MAC+\measuredangle ACM-\measuredangle CBQ=\measuredangle ACB+\measuredangle BAN+\measuredangle NBA-\measuredangle PCB\\&\implies \measuredangle CBA+180-\measuredangle CMA-\measuredangle CNQ=\measuredangle ACB+180-\measuredangle ANB-\measuredangle PMB\\&\implies \measuredangle QNA=\measuredangle AMP=90.\end{align*} $\blacksquare$
Take the $\sqrt{bc}$ inversion, denoting $Z^*$ as the inverse of $Z.$
[b][color=red]Claim:[/color][/b] $\omega_P^*=\omega_Q.$
[i]Proof.[/i] Let $P'=\overline{AQ}\cap\omega_Q$ and note $$\measuredangle QP'B=\measuredangle QCB=\measuredangle ACP$$ so $\triangle AP'B\sim\triangle ACP$ or $P^*=P'.$ $\blacksquare$
Hence, $P$ and $Q$ lie on $\omega_Q^*$ and $\omega_P^*,$ respectively. Also, $\measuredangle N^*Q^*A=\measuredangle QNA=90$ so $\overline{PN^*}$ is the diameter of $\omega_Q^*.$ Hence, $Y^*$ is the foot from $P$ to $\overline{AN^*}=\overline{AM}$ so $Y^*=M.$ Similarly, $X^*=N.$ Thus, $$AX^*\cdot AN^*=AB\cdot AC=AY^*\cdot AM^*$$ and $\triangle AX^*Y^*\sim\triangle AM^*N^*.$ $\square$ | [
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599,350 | Let $ABCD$ be a cyclic quadrilateral with center $O$.
Suppose the circumcircles of triangles $AOB$ and $COD$ meet again at $G$, while the circumcircles of triangles $AOD$ and $BOC$ meet again at $H$.
Let $\omega_1$ denote the circle passing through $G$ as well as the feet of the perpendiculars from $G$ to $AB$ and $CD$.
Define $\omega_2$ analogously as the circle passing through $H$ and the feet of the perpendiculars from $H$ to $BC$ and $DA$.
Show that the midpoint of $GH$ lies on the radical axis of $\omega_1$ and $\omega_2$.
[i]Proposed by Yang Liu[/i] | First, let $ E = AB \cap CD $ and $ F = BC \cap DA $ and denote the circumcircle of $ ABCD $ by $ \omega $. Let $ O_1, O_2 $ be the centers of $ \omega_1, \omega_2 $ respectively. Let $ M $ be the midpoint of $ GH $.
Consider the inversion about $ \omega $. It is clear that line $ AB $ goes to $ \omega_1 $ and that line $ CD $ goes to $ \omega_2 $ so $ E $ goes to $ G $. Similarly $ F $ goes to $ H $. Moreover, $ \omega_1 $ and $ \omega_2 $ are the circles with diameters $ EG $ and $ FH $ respectively. Now since $ M $ is the midpoint of $ GH $ and since $ O_1 $ is the midpoint of $ GE $ we have that $ O_{1}M \parallel HE $ and similarly $ O_{2}M \parallel GF $.
But since $ GF $ is the polar of $ E $ with respect to $ \omega $ we have that $ GF \perp OE $ so $ O_{2}M \perp OE $ which implies $ O_{2}M \perp OO_{1} $. Similarly $ O_{1}M \perp OO_{2} $ and so $ M $ is the orthocenter of triangle $ OO_{1}O_{2} $. This means that $ OM \perp O_{1}O_{2} $ so to show that $ M $ is on the radical axis of $ \omega_1 $ and $ \omega_2 $ it suffices to show that $ O $ is on this radical axis.
However, this is clear since both $ \omega_1 $ and $ \omega_2 $ are orthogonal to $ \omega $, so we are done.
Interestingly after doing the inversion step, a complex number solution is somewhat doable as well. | [
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599,351 | In triangle $ABC$ with incenter $I$ and circumcenter $O$, let $A',B',C'$ be the points of tangency of its circumcircle with its $A,B,C$-mixtilinear circles, respectively. Let $\omega_A$ be the circle through $A'$ that is tangent to $AI$ at $I$, and define $\omega_B, \omega_C$ similarly. Prove that $\omega_A,\omega_B,\omega_C$ have a common point $X$ other than $I$, and that $\angle AXO = \angle OXA'$.
[i]Proposed by Sammy Luo[/i] | I will fill in some intermediate steps in XmL's post.
Lemma 1: If $ X $, $ Y $ are the tangency points of the $ A $-mixtilinear incircle with $ AB, AC $ respectively then $ I $ is the midpoint of $ XY $.
[i]Proof:[/i] This is a well-known result, and moreover can be proved immediately with a limiting case of Sawayama's Theorem, but I shall provide a more elementary proof. Let $ B_2, C_2 $ be the midpoints of arcs $ CA, AB $ respectively. Then by Archimedes' Lemma we have that $ A', X, C_2 $ and $ A', Y, B_2 $ are collinear. Then by Pascal's Theorem on hexagon $ A'C_{2}CABB_{2} $ we have that $ X, I, Y $ are collinear. But since $ AX = AY $ and $ AI \perp XY $ this immediately implies the desired result.
Lemma 2: If $ A_1 $ denotes the midpoint of arc $ BAC $ then $ A', I, A_1 $ are collinear
[i]Proof:[/i] Consider the homothety centered at $ A' $ that takes the $ A $-mixtilinear incircle to the circumcircle of triangle $ ABC $. This takes $ X $ to $ C_2 $ and $ Y $ to $ B _2 $ so by Lemma 1 it takes $ I $ to $ M $, the midpoint of $ B_{2}C_{2} $. So it suffices to show that $ A_1, I, M $ are collinear. Some quick angle-chasing yields the fact that $ A_{1}C_{2}IB_{2} $ is a parallelogram which implies the desired result.
Now, continuing as in XmL's post, taking the inversion centered at $ I $ with radius $ \sqrt {A'I * A_{1}I} = \sqrt {B'I * B_{1}I} = \sqrt {C'I * C_{1}I} $, we have that $ \omega_A $ is mapped to the line parallel to $ AI $ passing through the reflection of $ A_1 $ over $ I $. We obtain two similar lines, and then reflecting them about $ I $ we obtain the configuration in XmL's post.
So, letting $ l_A $ be the line through $ A_1 $ parallel to $ AI $ and defining $ l_B, l_C $ similarly it suffices to show that these three lines concur at the circumcenter of the excentral triangle of triangle $ ABC $. Let $ I_A, I_B, I_C $ be the $ A, B, C $-excenters of triangle $ ABC $ respectively. It is well-known that $ A_1 $ is the midpoint of $ I_{B}I_{C} $ and that $ AI \perp I_{B}I_{C} $ so $ l_A $ is the perpendicular bisector of $ I_{B}I_{C} $. This immediately implies the desired result, and so we are done. | [
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599,353 | We are given triangles $ABC$ and $DEF$ such that $D\in BC, E\in CA, F\in AB$, $AD\perp EF, BE\perp FD, CF\perp DE$. Let the circumcenter of $DEF$ be $O$, and let the circumcircle of $DEF$ intersect $BC,CA,AB$ again at $R,S,T$ respectively. Prove that the perpendiculars to $BC,CA,AB$ through $D,E,F$ respectively intersect at a point $X$, and the lines $AR,BS,CT$ intersect at a point $Y$, such that $O,X,Y$ are collinear.
[i]Proposed by Sammy Luo[/i] | My solution:
Since the lines through $ A, B, C $ and perpendicular to $ EF, FD, DE $ are concurrent at the orthocenter of $\triangle DEF $ ,
so $ \triangle DEF $ and $ \triangle ABC $ are orthologic $ \Longrightarrow $ the perpendiculars through $ D, E, F $ to $ BC, CA, AB $ are concurrent at $ X $ .
Since $ AD, BE, CF $ are concurrent ,
so by [b]Terquem theorem[/b] we get $ AR, BS, CT $ are concurrent at $ Y $ .
Since $ \triangle RST $ is the pedal triangle of $ X' $ ($ X' $ is the isogonal conjugate of $ X $ ) ,
so the line passing $ A, B, C $ and perpendicular to $ ST, TR, RS $ are concurrent at $ X $ ,
hence by [b]Sondat theorem[/b] (for $\triangle ABC $ and $\triangle RST $ ) we get $ Y, X', X $ are collinear.
ie. $ X, O, Y $ are collinear
Q.E.D
[b]Remark:[/b]
(1) Easy to see $ X' = AD \cap BE \cap BF $ which is the orthocenter of $\triangle DEF $
(2) Another interesting property in this configuration :
Denote $ O' $ as the circumcenter of $ \triangle ABC $ , then the isogonal conjugate of $ Y $ lie on $ O'H $ .
(see http://www.artofproblemsolving.com/community/c6h366335 ) | [
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599,355 | "Define the Fibanocci sequence recursively by $F_1=1$, $F_2=1$ and $F_{i+2} = F_i + F_{i+1}$ for all(...TRUNCATED) | "Pretty easy. It is [i]well known[/i] that $F_i$ is strictly periodic mod $n$, to say that there exi(...TRUNCATED) | [-0.004518331494182348,0.0016061535570770502,-0.003071886021643877,0.005440608598291874,0.0238755419(...TRUNCATED) | [-0.008906040340662003,-0.033767297863960266,-0.004302870016545057,-0.013291704468429089,0.010613116(...TRUNCATED) | 2025-08-11 |
599,358 | "Let $d$ be a positive integer and let $\\varepsilon$ be any positive real. Prove that for all suffi(...TRUNCATED) | "I'm not sure whether my solution is correct or not, but I'll write it down anyway. \n\nLet $a$ be t(...TRUNCATED) | [-0.0021246045362204313,-0.020619377493858337,-0.0015944798942655325,0.021342255175113678,0.02896595(...TRUNCATED) | [0.038648203015327454,-0.02337401546537876,0.002447495935484767,0.012675275094807148,0.0364719294011(...TRUNCATED) | 2025-08-11 |
599,362 | "Given positive reals $a,b,c,p,q$ satisfying $abc=1$ and $p \\geq q$, prove that \\[ p \\left(a^2+b^(...TRUNCATED) | "Here is a slightly different solution: By AM-GM, we have $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c(...TRUNCATED) | [0.021560685709118843,-0.02538418397307396,0.0031518267933279276,0.0334685742855072,-0.0266567487269(...TRUNCATED) | [0.03342369571328163,-0.01495265681296587,0.0014782147482037544,0.04623185470700264,-0.0286789163947(...TRUNCATED) | 2025-08-11 |
599,363 | "Let $a$, $b$, $c$ be positive reals. Prove that \\[ \\sqrt{\\frac{a^2(bc+a^2)}{b^2+c^2}}+\\sqrt{\\(...TRUNCATED) | "Another proof with expansion...\n\nBy Holder, \\[(\\text{LHS})^2\\left(\\sum a(b^2+c^2)\\right)\\le(...TRUNCATED) | [0.012603342533111572,-0.009711714461445808,0.0035407627001404762,0.06941498070955276,-0.02707886509(...TRUNCATED) | [0.00971190445125103,-0.01850649155676365,-0.0006880834116600454,0.05241920426487923,-0.039252668619(...TRUNCATED) | 2025-08-11 |
599,366 | "Let $ABC$ be a triangle with circumcenter $O$. Let $P$ be a point inside $ABC$, so let the points $(...TRUNCATED) | "[b]Generalization:[/b]\n\nLet $ D, E, F $ be the point on $ BC, CA, AB $ , respectively .\nLet $ K (...TRUNCATED) | [0.049597665667533875,-0.001226920518092811,-0.0035386502277106047,0.029201284050941467,0.0108298333(...TRUNCATED) | [0.05124698951840401,-0.0319947674870491,-0.002840761560946703,0.04044590890407562,0.004170460626482(...TRUNCATED) | 2025-08-11 |
599,368 | "Let $t$ and $n$ be fixed integers each at least $2$. Find the largest positive integer $m$ for whic(...TRUNCATED) | "We claim that $m=n$ is the desired maximum value. Note that the required condition is just a cute w(...TRUNCATED) | [-0.00140288844704628,-0.04768011346459389,-0.001136447419412434,0.022923873737454414,0.000554478378(...TRUNCATED) | [0.028700871393084526,-0.010161207057535648,-0.0048321629874408245,0.01022128015756607,0.01215499360(...TRUNCATED) | 2025-08-11 |
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