Hagge-like circles, Jerabek hyperbola, Lemoine cubic

by kosmonauten3114, Jun 2, 2025, 4:05 PM

Let $\triangle{ABC}$ be a scalene oblique triangle with circumcenter $O$ and orthocenter $H$ , and $P$ ( $\neq \text{X(3), X(4)}$ , $\notin \odot(ABC)$ ) a point in the plane.
Let $\triangle{A_1B_1C_1}$ , $\triangle{A_2B_2C_2}$ be the circumcevian triangles of $O$ , $P$ , respectively.
Let $\triangle{P_AP_BP_C}$ be the pedal triangle of $P$ with respect to $\triangle{ABC}$ .
Let $A_1'$ be the reflection in $P_A$ of $A_1$ . Define $B_1'$ , $C_1'$ cyclically.
Let $A_2'$ be the reflection in $P_A$ of $A_2$ . Define $B_2'$ , $C_2'$ cyclically.
Let $O_1$ , $O_2$ be the circumcenters of $\triangle{A_1'B_1'C_1'}$ , $\triangle{A_2'B_2'C_2'}$ , respectively.

Prove that:
1) $P$ , $O_1$ , $O_2$ are collinear if and only if $P$ lies on the Jerabek hyperbola of $\triangle{ABC}$ .
2) $H$ , $O_1$ , $O_2$ are collinear if and only if $P$ lies on the Lemoine cubic (= $\text{K009}$ ) of $\triangle{ABC}$ .

Attachments:

geometry

reflection

Jerabek Hyperbola

Lemoine cubic

collinear

S(an) greater than S(n)

by ilovemath0402, Jun 2, 2025, 3:23 PM

Find all positive integer $n$ such that $S(an)\ge S(n) \quad \forall a \in \mathbb{Z}^{+}$ ( $S(n)$ is sum of digit of $n$ in base 10)
P/s: Original problem

sum of digits

number theory

arithmetric function

Parallel lines on a rhombus

by buratinogigle, Jun 2, 2025, 3:17 PM

Given the rhombus $ABCD$ with its incircle $\omega$ . Let $E$ and $F$ be the points of tangency of $\omega$ with $AB$ and $AC$ respectively. On the edges $CB$ and $CD$ , take points $G$ and $H$ such that $GH$ is tangent to $\omega$ at $P$ . Suppose $Q$ is the intersection point of the lines $EG$ and $FH$ . Prove that two lines $AP$ and $CQ$ are parallel or coincide.

Attachments:

geometry

rhombus

Line bisects a segment

by buratinogigle, Jun 2, 2025, 3:08 PM

Let $ABC$ be a triangle with $AB = AC$ . A circle $(O)$ is tangent to sides $AC$ and $AB$ , and $O$ is the midpoint of $BC$ . Points $E$ and $F$ lie on sides $AC$ and $AB$ , respectively, such that segment $EF$ is tangent to circle $(O)$ at point $P$ . Let $H$ and $K$ be the orthocenters of triangles $OBF$ and $OCE$ , respectively. Prove that line $OP$ bisects segment $HK$ .

Attachments:

geometry

Euler line of incircle touching points /Reposted/

by Eagle116, Apr 19, 2025, 2:48 PM

Let $ABC$ be a triangle with incentre $I$ and circumcentre $O$ . Let $D,E,F$ be the touchpoints of the incircle with $BC$ , $CA$ , $AB$ respectively. Prove that $OI$ is the Euler line of $\vartriangle DEF$ .

FE inequality from Iran

by mojyla222, Apr 19, 2025, 9:20 AM

Find all functions $f:\mathbb{R}^+ \to \mathbb{R}$ such that for all $x,y,z>0$
$3(x^3+y^3+z^3)\geq f(x+y+z)\cdot f(xy+yz+xz) \geq (x+y+z)(xy+yz+xz).$

inequalities

function

Functional inequality

Iran 2nd Round

Tangency of circles with "135 degree" angles

by Shayan-TayefehIR, May 19, 2024, 3:50 PM

For a triangle $\triangle ABC$ with an obtuse angle $\angle A$ , let $E , F$ be feet of altitudes from $B , C$ on sides $AC , AB$ respectively. The tangents from $B , C$ to circumcircle of triangle $\triangle ABC$ intersect line $EF$ at points $K , L$ respectively and we know that $\angle CLB=135$ . Point $R$ lies on segment $BK$ in such a way that $KR=KL$ and let $S$ be a point on line $BK$ such that $K$ is between $B , S$ and $\angle BLS=135$ . Prove that the circle with diameter $RS$ is tangent to circumcircle of triangle $\triangle ABC$ .

Proposed by Mehran Talaei

This post has been edited 2 times. Last edited by Shayan-TayefehIR, May 27, 2024, 10:11 AM

geometry

Elliptic Geometry

by always_correct, Jul 23, 2017, 10:50 PM

always_correct wrote:

Here's a problem of mine!

Denote by $p_{XYZ}$ the perimeter of $\triangle XYZ$ .
Let $ABC$ be a triangle with point $D$ lying on $BC$ beyond $B$ such that $2 \cdot DC = p_{ABC}$ . Collinear points $E,P,Q$ are situated such that $E$ is the midpoint of $\overline{BC}$ , $ED = EP$ , and $EA = EQ$ . Points $R,S$ are the projections from $P$ to $\overline{BC}$ and $Q$ to $\overline{PR}$ , respectively. If $\angle ABC = \angle ACB < 45^\circ$ , prove that $p_{ABC} = p_{SBC}$ .

$[asy] size(10cm); pair A = (0, 0.5), B = (-1, 0), C = (1, 0), D = (-1.118, 0), E = origin, P = (0.52, 0.99), Q = (0.233, 0.443); pair R = (0.52, 0), S = (0.52, 0.443); draw(A--B--C--cycle); draw(D--E--P--cycle); draw(E--A); draw(P--R); draw(Q--S); draw(B--S--C); draw(rightanglemark(P,R,B,1.5)); draw(rightanglemark(P,S,Q,1.5)); label("$A$", A, N); label("$B$", B - (0, 0.05)); label("$C$", C - (0, 0.05)); label("$D$", D - (0, 0.05)); label("$E$", E - (0, 0.05)); label("$P$", P, N); label("$Q$", Q, - (0.05, -.05)); label("$R$", R - (0, 0.05)); label("$S$", S + (0.05, 0)); [/asy]$

It has a very cool solution! (I hope)

Hmm, it seems after a day nobody has a solution for me...

Time to post my own!

$[asy] size(10cm); pair A = (0, 0.5), B = (-1, 0), C = (1, 0), D = (-1.118, 0), E = origin, P = (0.52, 0.99), Q = (0.233, 0.443), Qp = (0.233, 0); pair R = (0.52, 0), S = (0.52, 0.443); draw(A--B--C--cycle); draw(D--E--P--cycle); draw(E--A); draw(P--R); draw(Q--S); draw(B--S--C); draw(Q--Qp); draw(rightanglemark(P,R,B,1.5)); draw(rightanglemark(P,S,Q,1.5)); draw(rightanglemark(Q,Qp,E,1.5)); draw(ellipse(E, 1.118, 0.5), blue+linewidth(1)); draw(circle(E, 0.5), red+linewidth(1)); draw(circle(E, 1.118), green+linewidth(1)); label(scale(0.7)*"$A$", A, N); label(scale(0.7)*"$B$", B - (0, 0.05)); label(scale(0.7)*"$C$", C - (0, 0.05)); label(scale(0.7)*"$D$", D - (0.05, 0.04)); label(scale(0.7)*"$E$", E - (0, 0.05)); label(scale(0.7)*"$P$", P, N); label(scale(0.7)*"$Q$", Q + (0, 0.09)); label(scale(0.7)*"$R$", R - (0, 0.05)); label(scale(0.7)*"$S$", S + (0.05, 0.02)); label(scale(0.7)*"$Q'$", Qp - (0, 0.05)); [/asy]$

Firstly note because $DB + DC = AB + AC$ , that $D, A$ lie on an ellipse with foci $B, C$ .

Let the blue ellipse with foci $B, C$ containing $A$ be $\mathcal{B}$ , the red circle with center $E$ containing $A$ be $\mathcal{R}$ , and the green circle with center $E$ containing $D$ be $\mathcal{G}$ . Finally, let $Q'$ be the projection from $Q$ to $\overline{BC}$ .

Consider the horizontal stretching $\mathcal{S}$ that takes $\mathcal{R} \to \mathcal{B}$ . It is equivalent to a homothety $\mathcal{H}$ at $E$ taking $\mathcal{R} \to \mathcal{G}$ followed by a vertical stretch $\mathcal{V}$ (as $\angle A$ is obtuse) with factor $k =\frac{R}{b}$ where $R$ is the radius of $\mathcal{G}$ and $b$ is the semi-minor axis of $\mathcal{B}$ . In other words,
$\mathcal{S} = \mathcal{V} \circ \mathcal{H}$
As $\triangle ABC$ is isosceles,
$k = \frac{R}{b} = \frac{ED}{EA}$ Now, as $P \in \mathcal{G}$ and $Q \in \mathcal{R}$ , it follows $k = \frac{EQ}{EP} = \frac{RS}{RP}$ by similarity of $\triangle EQQ'$ and $\triangle EPR$ .

Finally, note that $\mathcal{H}(Q) = P$ and as $k = \frac{RS}{RP}$ , $\mathcal{V}(Q) = S$ . Thus
$\mathcal{S}(Q) = \mathcal{V}(\mathcal{H}(Q)) = S$
As $\mathcal{S}(\mathcal{R}) = \mathcal{B}$ it follows $S \in \mathcal{B}$ , and as $B,C$ are the foci of $\mathcal{B}$ , we end up with
$AB + AC = SB +SC$ and we're done!

Orthocenter lies on circumcircle

by whatshisbucket, Jun 26, 2017, 7:03 AM

Let $ABC$ be a triangle with orthocenter $H,$ and let $M$ be the midpoint of $\overline{BC}.$ Suppose that $P$ and $Q$ are distinct points on the circle with diameter $\overline{AH},$ different from $A,$ such that $M$ lies on line $PQ.$ Prove that the orthocenter of $\triangle APQ$ lies on the circumcircle of $\triangle ABC.$

Proposed by Michael Ren

Polish MO Finals 2014, Problem 4

by j___d, Jul 27, 2016, 10:11 PM

Denote the set of positive rational numbers by $\mathbb{Q}_{+}$ . Find all functions $f: \mathbb{Q}_{+}\rightarrow \mathbb{Q}_{+}$ that satisfy
$\underbrace{f(f(f(\dots f(f}_{n}(q))\dots )))=f(nq)$ for all integers $n\ge 1$ and rational numbers $q>0$ .

contests

function

functional equation

Incenter perpendiculars and angle congruences

by math154, Jul 2, 2012, 3:13 AM

$ABC$ is a triangle with incenter $I$ . The foot of the perpendicular from $I$ to $BC$ is $D$ , and the foot of the perpendicular from $I$ to $AD$ is $P$ . Prove that $\angle BPD = \angle DPC$ .

Alex Zhu.

geometry

Submit

there are over 6000! anyway, I'm resting with the whole Euler line problem right now
by always_correct, Aug 6, 2017, 8:16 PM
Have you seen something about Kimberling centers (triangle centers) and center lines?
by alevini98, Aug 5, 2017, 11:19 PM
no its not dead
by always_correct, Jul 12, 2017, 1:19 AM
Wow this is great, is this dead?
by Ankoganit, Jul 3, 2017, 8:41 AM
oh its a math blog.....
by Swimmer2222, Apr 13, 2017, 12:58 PM
Why is everyone shouting
Oh wait...
by ArsenalFC, Mar 26, 2017, 8:45 PM
Shout! ^^
by DigitalDagger, Mar 4, 2017, 3:49 AM
14th shout!
by arkobanerjee, Feb 8, 2017, 2:48 AM
14th shout! be more active
by Mathisfun04, Jan 25, 2017, 9:13 PM
13th shout
by Ryon123, Jan 3, 2017, 3:47 PM
Dang, so OP. Keep it up!
by monkey8, Dec 28, 2016, 6:23 AM
How much time do you spend writing these posts???
by Designerd, Dec 5, 2016, 5:02 AM
good blog!
by AlgebraFC, Dec 5, 2016, 2:09 AM
Nice finally material for calculus people to read thanks
by Math1331Math, Nov 27, 2016, 2:50 AM
Whoa how much time do you spend typing these posts?

They're nice posts!
by MathLearner01, Nov 24, 2016, 3:30 AM
i am remove it please
by always_correct, Nov 22, 2016, 2:18 AM
5th shout!
by algebra_star1234, Nov 20, 2016, 2:41 PM
fourth shout >
by budu, Nov 20, 2016, 4:59 AM
3rd shout!
by monkey8, Nov 20, 2016, 3:24 AM
2nd shout!
by RYang, Nov 20, 2016, 3:01 AM
first shout >
by doitsudoitsu, Oct 9, 2016, 7:54 PM

21 shouts

Mathematical

by kosmonauten3114, Jun 2, 2025, 4:05 PM

by ilovemath0402, Jun 2, 2025, 3:23 PM

by buratinogigle, Jun 2, 2025, 3:17 PM

by buratinogigle, Jun 2, 2025, 3:08 PM

by Eagle116, Apr 19, 2025, 2:48 PM

by mojyla222, Apr 19, 2025, 9:20 AM

by Shayan-TayefehIR, May 19, 2024, 3:50 PM

by always_correct, Jul 23, 2017, 10:50 PM

by whatshisbucket, Jun 26, 2017, 7:03 AM

by j___d, Jul 27, 2016, 10:11 PM

by math154, Jul 2, 2012, 3:13 AM