perpendicular inside a right triangle, altitude and midpoints

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0 replies

jlacosta

Apr 2, 2025

0 replies

Very tight inequalities

KhuongTrang 2

N 18 minutes ago by SunnyEvan

Source: own

Problem. Given non-negative real numbers $a,b,c$ satisfying $ab+bc+ca=1.$ Prove that $\color{black}{\frac{1}{35a+12b+2}+\frac{1}{35b+12c+2}+\frac{1}{35c+12a+2}\ge \frac{4}{39}.}$ $\color{black}{\frac{1}{4a+9b+6}+\frac{1}{4b+9c+6}+\frac{1}{4c+9a+6}\le \frac{2}{9}.}$ When does equality hold?

2 replies

KhuongTrang

May 17, 2024

SunnyEvan

18 minutes ago

Sum of First, Second, and Third Powers

Brut3Forc3 47

N 31 minutes ago by cubres

Source: 1973 USAMO Problem 4

Determine all roots, real or complex, of the system of simultaneous equations
$\begin{align*} x+y+z &= 3, \\ x^2+y^2+z^2 &= 3, \\ x^3+y^3+z^3 &= 3.\end{align*}$

47 replies

Brut3Forc3

Mar 7, 2010

cubres

31 minutes ago

Modified Sum of floors

prMoLeGend42 2

N 41 minutes ago by cubres

Find the closed form of : $\sum _{k=0}^{n-1} \left\lfloor \frac{ak+b}{n}\right \rfloor$ where $\gcd(a,n)=1$

2 replies

prMoLeGend42

Yesterday at 9:09 AM

cubres

41 minutes ago

6 tangents to 1 circle

moony_ 3

N 43 minutes ago by whwlqkd

Source: own

Let $P$ be a point inside the triangle $ABC$ . $AP$ intersects $BC$ at $A_0$ . Points $B_0$ and $C_0$ are defined similarly. Line $B_0C_0$ intersects $(ABC)$ at points $A_1$ , $A_2$ . The tangents at these points to $(ABC)$ intersect BC at points $A_3$ , $A_4$ . Points $B_3$ , $B_4$ , $C_3$ , $C_4$ are defined similarly. Prove that points $A_3$ , $A_4$ , $B_3$ , $B_4$ , $C_3$ , $C_4$ lie on one conic

3 replies

moony_

Yesterday at 9:30 AM

whwlqkd

43 minutes ago

No more topics!

perpendicular inside a right triangle, altitude and midpoints

parmenides51 2

N Dec 11, 2020 by AllanTian

Source: 2006 Indonedia MO Province P2 q1 OSP

Suppose triangle $ABC$ is right-angled at $B$ . The altitude from $B$ intersects the side $AC$ at point $D$ . If points $E$ and $F$ are the midpoints of $BD$ and $CD$ , prove that $AE \perp BF$ .

2 replies

parmenides51

Dec 11, 2020

AllanTian

Dec 11, 2020

perpendicular inside a right triangle, altitude and midpoints

G H J

Reveal topic tags

geometry

perpendicular

right triangle

G H BBookmark kLocked kLocked NReply

Source: 2006 Indonedia MO Province P2 q1 OSP

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parmenides51

30630 posts

parmenides51

#1 h Dec 11, 2020, 9:00 AM

Y by

Suppose triangle $ABC$ is right-angled at $B$ . The altitude from $B$ intersects the side $AC$ at point $D$ . If points $E$ and $F$ are the midpoints of $BD$ and $CD$ , prove that $AE \perp BF$ .

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EmilXM

378 posts

EmilXM

#2 h Dec 11, 2020, 9:15 AM

Y by

Solution

Let $M$ be intersection point of $AE$ and $BF$ . We have $\triangle BAD \sim \triangle CBD$ . Since $AE$ is $A-median$ of $\triangle BAD$ and $BF$ is $B-median$ of $\triangle CBD$ , we get $\triangle EAD \sim \triangle EBM \Longrightarrow \angle AMB = 90^0 \Longrightarrow AE \perp BF$

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AllanTian

12 posts

AllanTian

#3 h Dec 11, 2020, 10:17 AM

Y by

1. triangle ADB is similar to triangle BDC
2. triangle ADE is similar to triangle BDF
3. <AED=<BEH=<BFD
4. triangle AHE is similar to triangle ADF

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