ka April Highlights and 2025 AoPS Online Class Information
jlacosta0
Apr 2, 2025
Spring is in full swing and summer is right around the corner, what are your plans? At AoPS Online our schedule has new classes starting now through July, so be sure to keep your skills sharp and be prepared for the Fall school year! Check out the schedule of upcoming classes below.
WOOT early bird pricing is in effect, don’t miss out! If you took MathWOOT Level 2 last year, no worries, it is all new problems this year! Our Worldwide Online Olympiad Training program is for high school level competitors. AoPS designed these courses to help our top students get the deep focus they need to succeed in their specific competition goals. Check out the details at this link for all our WOOT programs in math, computer science, chemistry, and physics.
Looking for summer camps in math and language arts? Be sure to check out the video-based summer camps offered at the Virtual Campus that are 2- to 4-weeks in duration. 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)!
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Right-angled triangle if circumcentre is on circle
liberator77
N24 minutes ago
by Ihatecombin
Source: IMO 2013 Problem 3
Let the excircle of triangle opposite the vertex be tangent to the side at the point . Define the points on and on analogously, using the excircles opposite and , respectively. Suppose that the circumcentre of triangle lies on the circumcircle of triangle . Prove that triangle is right-angled.
Let be the circumcircle of triangle , is the midpoint of and be the second intersection of and . Tangent line of at intersects at , let be a transversal of and be intersections of with . Let be a tangent line of . Prove that is cyclic
Let ABC be an isosceles triangle with AB = AC and M the midpoint of BC. Consider a point E outside the triangle such that BE = BM and CE perpendicular to AB. The point of intersection of the perpendicular bisector of segment EB with the circumcircle of triangle AMB, which is on the same side as A with respect to BE, is point F. Show that angle FME = 90°
Let A, B, C, and D be four distinct points on a straight line, in that order. The circles with diameters AC and BD intersect at X and Y. The straight line XY intersects BC at Z. Let P be a point on XY distinct from Z. The straight line CP intersects the circle with diameter AC at C and M, and the straight line BP intersects the circle with diameter BD at B and N. Show that AM, DN, and XY are aligned.
Let C be a point on a semicircle of diameter AB and let D be the mid-length of arc AC. Let E be the projection of point D on BC and F the intersection of AE with the semicircle. Prove that BF bisects segment DE.
Given a non-isosceles triangle ABC inscribed in circle . are the midpoints of respectively. is the midpoint of arc , is symmetric to through . Let be the incenter and tangent of angles of triangle respectively. Prove that have a common point
Welcome to the 1st idk12345678 Math Contest.
You have 4 hours. You do not have to prove your answers.
Post \signup username to sign up. Post your answers in a hide tag and I will tell you your score.*
The contest is attached to the post
Clarifications
specifies the greatest common divisor of and .
In number 3, the probabilities are for the sum of the dice.
*I mightve done them wrong feel free to ask about an answer
is the least angle in . Point is on the arc from the circumcircle of . The perpendicular bisectors of the segments intersect the line at , respectively. Point is the meet point of . Suppose that is the radius of the circumcircle of . Prove that:
is the least angle in . Point is on the arc from the circumcircle of . The perpendicular bisectors of the segments intersect the line at , respectively. Point is the meet point of . Suppose that is the radius of the circumcircle of . Prove that:
Let CT meet circumcircle at P.
∠BPT = ∠A and ∠PBT = ∠PBA + ∠ABM = ∠PCA + ∠BAM = ∠A ---> BT = PT
so CT + BT = CP and CP is chord but 2R is diameter so 2R ≥ CT + BT.
By angle chase, lies on . Moreover, lies on arc as and lies on arc , not containing . However, it is well-known that maximum of is achieved iff is the midpoint of arc (this can proved by calculus), hence equality occurs iff , which is equivalent to passing through .