ka May Highlights and 2025 AoPS Online Class Information
jlacosta0
May 1, 2025
May is an exciting month! National MATHCOUNTS is the second week of May in Washington D.C. and our Founder, Richard Rusczyk will be presenting a seminar, Preparing Strong Math Students for College and Careers, on May 11th.
Are you interested in working towards MATHCOUNTS and don’t know where to start? We have you covered! If you have taken Prealgebra, then you are ready for MATHCOUNTS/AMC 8 Basics. Already aiming for State or National MATHCOUNTS and harder AMC 8 problems? Then our MATHCOUNTS/AMC 8 Advanced course is for you.
Summer camps are starting next month at the Virtual Campus in math and language arts that are 2 - to 4 - weeks in duration. Spaces are still available - don’t miss your chance to have an enriching summer experience. There are middle and high school competition math camps as well as Math Beasts camps that review key topics coupled with fun explorations covering areas such as graph theory (Math Beasts Camp 6), cryptography (Math Beasts Camp 7-8), and topology (Math Beasts Camp 8-9)!
Be sure to mark your calendars for the following upcoming events:
[list][*]May 9th, 4:30pm PT/7:30pm ET, Casework 2: Overwhelming Evidence — A Text Adventure, a game where participants will work together to navigate the map, solve puzzles, and win! All are welcome.
[*]May 19th, 4:30pm PT/7:30pm ET, What's Next After Beast Academy?, designed for students finishing Beast Academy and ready for Prealgebra 1.
[*]May 20th, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 1 Math Jam, Problems 1 to 4, join the Canada/USA Mathcamp staff for this exciting Math Jam, where they discuss solutions to Problems 1 to 4 of the 2025 Mathcamp Qualifying Quiz!
[*]May 21st, 4:00pm PT/7:00pm ET, Mathcamp 2025 Qualifying Quiz Part 2 Math Jam, Problems 5 and 6, Canada/USA Mathcamp staff will discuss solutions to Problems 5 and 6 of the 2025 Mathcamp Qualifying Quiz![/list]
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Some users don't want to learn, some other simply ignore advises.
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[code]+"first keyword" +"second keyword"[/code]
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[quote][/quote]
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If you do this, too, you are on the best way to get your thread deleted. Write everything clearly, define where your variables come from (and define the "natural" numbers if used). Additionally read your post at least twice before submitting. After you sent it, read it again and use the Edit-Button if necessary to correct errors.
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Let be a triangle with let be the circumcircle of and let be its radius. Point is chosen on such taht and point is the foot of the perpendicular from to . Ray mets again at . Point is chosen on line such that and lie on a line in that order. Finally, let be a point satisfying and . Prove that lies on .
Let be a positive integer. A board with a format is divided in equal squares.Determine all integers ≥3 such that the board can be covered in (or ) pieces so that there is exactly one empty square in each row and each column.
Let and be natural numbers. Every one of the squares of the board is colored either black or white, so that no 2 neighbouring squares are the same color(the board is colored like in chess").In one step we can pick 2 neighbouring squares and change their colors like this: - a white square becomes black; -a black square becomes blue; -a blue square becomes white.
For which and can we ,in a finite sequence of these steps, switch the starting colors from white to black and vice versa.
Source: Serbian selection contest for the IMO 2025
For an table filled with natural numbers, we say it is a divisor table if:
- the numbers in the -th row are exactly all the divisors of some natural number ,
- the numbers in the -th column are exactly all the divisors of some natural number ,
- for every .
A prime number is given. Determine the smallest natural number , divisible by , such that there exists an divisor table, or prove that such does not exist.
Let be a right-angled triangle where ∠=90°.Let be an altitude of that triangle and points and be the midpoints of and , respectively.If is the circumcenter of the triangle , prove that and are paralel.
Given a scalene triangle , let be a point (). Let be the anticomplement of , and let be the -anticomplementary conjugate of . Prove that the line is parallel to the real asymptote of the circular pivotal isocubic with pivot .
is isosceles with . Let be a point in its interior such that . Suppose that the perpendicular bisector of meets the external angle bisector of at , and let be the intersection of the perpendicular bisector of and the external angle bisector of . Prove that are concyclic.
Y bytenplusten, Davi-8191, nguyendangkhoa17112003, Adventure10, Mango247, Rounak_iitr
is isosceles with . Let be a point in its interior such that . Suppose that the perpendicular bisector of meets the external angle bisector of at , and let be the intersection of the perpendicular bisector of and the external angle bisector of . Prove that are concyclic.
Y byVirgil Nicula, Davi-8191, Adventure10, Mango247
Trivial problem for China
Solution: apply an inversion with center and arbitrary power . Let inverse of note that points and are midpoints of arcs and of triangles and so ' and are feet of external angle bisectors of and . also from the condition we have using this fact and external bisector theorem in triangles and with easy calculation we get so and quadrilateral is trapezoid so to prove the problem we have to show that B'A'=Q'A' from the condition that and external bisector theorem we get =×= using this fact and after an easy calculation we get the result that .
This post has been edited 1 time. Last edited by andria, Mar 27, 2015, 6:14 PM
It is clear that is the midpoint of arc of the circumcircle of and similarly is the midpoint of of the circumcircle of Consider the inversion with center and arbitrary radius. I will denote inverted points by their original letter (hopefully the use will be clear). Because of the first observation, we have that and are collinear (in that order). Now we proceed with four metric observations:
- this follows from the given condition
- this follows from the fact that in the original diagram, was isosceles.
- this follows from the fact that in the original diagram, was isosceles.
- this follows from the fact that in the original diagram, was isosceles.
Now we want to show that quadrilateral is cyclic, which is equivalent to showing that But by (3) and (4) we have that so by (2) it suffices to show that But by (3) and (4) again and the facts that and we find that and and now by using (1) and (2) we can simplify to obtain the desired result.
We begin with a well-known lemma that a few above have cited: Lemma
Let be a triangle and let be the midpoint of arc Then lies on the perpendicular bisector of and the external angle bisector of Proof of Lemma. It is clear that since is the midpoint of arc , we have , so lies on the perpendicular bisector of For the second part, denote the midpoint of arc by Then bisects , and since is a diameter of the circle, Therefore, is perpendicular to the internal angle bisector, so lies on the external angle bisector of , as desired.
Now, applying the lemma to the problem at hand, we find that are the midpoints of arcs in , respectively. We will now prove that the lines are concurrent, whence Power of a Point will finish the proof. Let be the point on for which , and let cut at , respectively. We will show that , whence it will follow that ; but first, a few synthetic observations: Since , it follows that is isoceles. Furthermore, we have , so by side-angle-side similarity, By spiral similarity, we also know that In addition, since , it follows by side-angle-side similarity that By spiral similarity, we also have Now, we are ready:
By the Ratio Lemma (Law of Sines) applied to , it follows that Keeping in mind that , to prove that these two ratios are equal, we need only show that Recalling our similar triangles in mind and applying the Ratio Lemma to , we have Because , one final calculation yields Hence, , so , as desired. Now, let us denote the intersection of by Then since lies on the radical axis of and , it follows by Power of a Point that By Power of a Point once again, this implies that are concyclic, as desired.
Y byTripteshBiswas, tenplusten, Adventure10, Mango247, Infinityfun
We easily see that and are cyclic.
Now, pick on with , and note the angle is the complement of half the angle by the isosceles triangle. As angle is the complement of half of and also of , we have that in that and are isogonal. So if meets at , we have .
So , using . This works out to . So then this expression is symmetric, and looking at where meets gives the same result. Thus ,, meet at . So then we have using POP and this finishes the problem.
This post has been edited 2 times. Last edited by yugrey, Apr 4, 2015, 6:31 PM Reason: clean a bit
With an inversion at D radius 1, the problem becomes: "Let 1/DC+1/DB=1/DA and AC/DC = AB/DB and P,Q be the intersections of AC, AB with the external bisector of ADC and ADB. Prove BPQC is cyclic".
Notice CP/PA=CD/DA=CD*(1/DC+1/DB)=1+(CD/DB) Thus CA/PA=CD/DB and this implies PA*CA=(CA^2)/(CD/DB) and similarly QA*BA=(BA^2)(CD/DB) and thus all that remains is to prove CA^2/(CD/DB)=BA^2(CD/DB), which is equivalent to AC/DC=AB/DB, which is given in the problem statement.
This post has been edited 1 time. Last edited by JuanOrtiz, Apr 18, 2015, 6:36 PM
Lemma(Extention of Archemedes Midpoint Theorem): If is the midpoint of arc , then Main Proof: We quickly see that are cyclic quadrilaterals. By radical axis theorem, are concyclic are concurrent. By letting ,, it suffices to show .
Applying the lemma on and midpoint , we have Hence we have
By symmetry, we can similarly derive , and the ratio equality is proven.
Invert around with radius . Denote the inverse by a . = etc imply .
We recognise as the midpoint of arc , so it is on the intersection of it and the perpendicular bisector of . Thus its inverse is the intersection of and circle with as center and as radius.
Now we compute . By a similar computation for the other product get are concyclic. Inverting back, we have the conclusion.
I wonder if a solution along the lines of "...cevians concur hence by radical axis..." exists.
dibyo_99 wrote:
is isosceles with . Let be a point in its interior such that . Suppose that the perpendicular bisector of meets the external angle bisector of at , and let be the intersection of the perpendicular bisector of and the external angle bisector of . Prove that are concyclic.
Invert at . Then lies inside with and ; external bisectors of angles meet lines and again at . It suffices to show are concyclic. Equivalently, we want .
By external bisector theorem, and similarly . The claim now follows.
This post has been edited 1 time. Last edited by anantmudgal09, Mar 25, 2018, 10:29 PM
This problem is trivial with inversion but very nice to solve without it . Here is my synthetic solution.
Clearly quadrilaterals and are cyclic. Extend beyond and beyond to respectively so that . Then . Similarly, we get .
Now construct point such that . Hence . But so or . Similarly if we construct point such that , we also get . Moreover, Similarly . Hence it suffices to show that . But which is symmetric w.r.t. so we are done.
First intersect perpendicular from A , B to AD , BD in K , perpendicular from A , C to AD , CD in J , perpendicular from B , C to BD , CD in I .
Bisector of K , J meet IJ , IK in M ,L .
It's easy to find that D is on the line LM (because AD=DB+DC)
We know that P , K , M and Q , J , L are collinear ( because ADBK , ADCJ are cyclic )
It's easy to see that DPMC , DDQLB are cyclic .
Now we have BPC=BAD+DMC , BQC=DAC+DLB .
we should just prove that BAD+DMC=DAC+DLB and it's equal to prove that JMD=KDL .
If LM intersect KJ at X , we know that XJ.KI=XK.IJ and we want to say that DJ.XK=DK.XJ so it's enough to prove IJ.DK=DJ.IK .
Because AC=AB we can easily prove IJ.DK=DJ.IK so the problem is solved
Long solution but with some nice other properties of the config
Inversion at works. Alternatively, let be the point on for which , and define similarly. Also, let ,. We show that ,, concur. Note that by angle chasing and similarly . Then using these ratios and the Menelaus ratios from we can use Menelaus again to imply the conclusion. Now, by Desargue's theorem on and ,, concur, implying cyclic. Alternatively, by more lengths we can show that , (where is the reflection of over the perpendicular bisector of , which happens to be the intersection of the external angle bisector of with ) and use similarity to get so . Similarly, , so , and are collinear. Let be the arc midpoint of arc of . As ,, meet at a point, it suffices to show that is cyclic. We will in fact show this is true even without the and the conditions.
Let be the center of , let be the center of and let be the center of . Then by a simple angle chase the perpendicular bisector of is the angle bisector of , so the perpendicular bisectors of concur at the incenter of , so we are done.