Geometry

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.

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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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0 replies

jlacosta

May 1, 2025

0 replies

Six variables

Nguyenhuyen_AG 4

N 6 minutes ago by Sunjee

Let $a,\,b,\,c,\,x,\,y,\,z$ be six positive real numbers. Prove that
$\frac{a}{b+c} \cdot \frac{y+z}{x} + \frac{b}{c+a} \cdot \frac{z+x}{y} + \frac{c}{a+b} \cdot \frac{x+y}{z} \geqslant 2+\sqrt{\frac{8abc}{(a+b)(b+c)(c+a)}}.$

4 replies

Nguyenhuyen_AG

Yesterday at 5:09 AM

Sunjee

6 minutes ago

ISI UGB 2025 P7

SomeonecoolLovesMaths 9

N 11 minutes ago by Project_Donkey_into_M4

Source: ISI UGB 2025 P7

Consider a ball that moves inside an acute-angled triangle along a straight line, unit it hits the boundary, which is when it changes direction according to the mirror law, just like a ray of light (angle of incidence = angle of reflection). Prove that there exists a triangular periodic path for the ball, as pictured below.

IMAGE

9 replies

SomeonecoolLovesMaths

Yesterday at 11:28 AM

Project_Donkey_into_M4

11 minutes ago

Trigo relation in a right angled. ISIBS2011P10

Sayan 11

N 12 minutes ago by Project_Donkey_into_M4

Show that the triangle whose angles satisfy the equality
$\frac{\sin^2A+\sin^2B+\sin^2C}{\cos^2A+\cos^2B+\cos^2C} = 2$
is right angled.

11 replies

Sayan

Mar 31, 2013

Project_Donkey_into_M4

12 minutes ago

Expressing polynomial as product of two polynomials

Sadigly 3

N 16 minutes ago by Sadigly

Source: Azerbaijan Senior NMO 2021

Define $P(x)=((x-a_1)(x-a_2)...(x-a_n))^2 +1$ , where $a_1,a_2...,a_n\in\mathbb{Z}$ and $n\in\mathbb{N^+}$ . Prove that $P(x)$ couldn't be expressed as product of two non-constant polynomials with integer coefficients.

3 replies

Sadigly

Yesterday at 9:10 PM

Sadigly

16 minutes ago

No more topics!

Geometry

VicKmath7 2

N Mar 27, 2022 by parmenides51

Source: Bulgaria JTST 2014 P1 day 2

Given is a triangle ABC. Let D be from AB so that <ADC=90° . DM and DN are perpendicular to BC and AC. <CAB=60° and <CBA=45° and H is orthocenter of triangle MNC. Let O be the midpoint of CD, find <COH ?

2 replies

VicKmath7

Aug 27, 2019

parmenides51

Mar 27, 2022

Geometry

G H J

Reveal topic tags

geometry proposed

geometry

G H BBookmark kLocked kLocked NReply

Source: Bulgaria JTST 2014 P1 day 2

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VicKmath7

1389 posts

VicKmath7

#1 h Aug 27, 2019, 10:36 AM • 2 Y

Y by Adventure10, Mango247

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NikoIsLife

9657 posts

NikoIsLife

#2 h Aug 27, 2019, 11:19 AM • 3 Y

Y by VicKmath7, centslordm, Adventure10

Solution

$[asy] size(250); pair A=(-1,0),C=(0,sqrt(3)),B=(sqrt(3),0),D=(0,0),O=(C+D)/2,M=foot(D,B,C),N=foot(D,A,C),X=foot(M,A,C),Y=foot(N,B,C),H=extension(M,X,N,Y); draw(A--B--C--A^^X--M--D--N--Y);dot(O);draw(C--D^^O--H--C,dotted);draw(circle(O,sqrt(3)/2),dotted); label("$A$",A,SW);label("$C$",C,dir(90));label("$B$",B,SE); label("$D$",D,S);label("$M$",M,NE);label("$N$",N,NW);label("$X$",X,NW);label("$Y$",Y,NE); label("$H$",H,dir(90));label("$O$",O,S);[/asy]$
Since $\angle DMC=\angle DNC=90^\circ$ , this means, $D,M,N,C$ lie on the same circle, whose center is $O$ .

Since $H$ is the orthocenter of $\triangle MNC$ , this means,
$\angle MCH=\angle MNY=90^\circ-\angle CMN=90^\circ-\angle CDN=90^\circ-\angle CAB=30^\circ$ and also,
$\angle NCD=90^\circ-\angle CAB=30^\circ$ This shows that $\angle OCH=\angle ACB-\angle MCH-\angle NCD=75^\circ-30^\circ-30^\circ=15^\circ$ .

We have $CH=2OC\cos\angle MNC=2OC\cos75^\circ=OC\cdot\frac{\sqrt6-\sqrt2}2\implies OC=CH\frac{\sqrt6+\sqrt2}2=2CH\cos15^\circ$ .

This shows us that $\triangle CHO$ is isosceles. Hence, $\angle COH=\angle OCH=\boxed{15^\circ}$ .

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parmenides51

30652 posts

parmenides51

#3 h Mar 27, 2022, 11:26 AM

Y by

solved also here

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