Perspective Triangles

In the given figure, ABC is an isosceles triangle with AB = AC and ∠BAC = 78°. Point D is chosen inside the triangle such that AD=DC. Find the measure of angle X (∠BDC).

ps: see the attachment for figure

0 replies

orangesyrup

16 minutes ago

0 replies

Lord Evan the Reflector

whatshisbucket 21

N 36 minutes ago by ezpotd

Source: ELMO 2018 #3, 2018 ELMO SL G3

Let $A$ be a point in the plane, and $\ell$ a line not passing through $A$ . Evan does not have a straightedge, but instead has a special compass which has the ability to draw a circle through three distinct noncollinear points. (The center of the circle is not marked in this process.) Additionally, Evan can mark the intersections between two objects drawn, and can mark an arbitrary point on a given object or on the plane.

(i) Can Evan construct* the reflection of $A$ over $\ell$ ?

(ii) Can Evan construct the foot of the altitude from $A$ to $\ell$ ?

*To construct a point, Evan must have an algorithm which marks the point in finitely many steps.

Proposed by Zack Chroman

21 replies

whatshisbucket

Jun 28, 2018

ezpotd

36 minutes ago

Feet of perpendiculars to diagonal in cyclic quadrilateral

jl_ 2

N 38 minutes ago by lw202277

Source: Malaysia IMONST 2 2023 (Primary) P6

Suppose $ABCD$ is a cyclic quadrilateral with $\angle ABC = \angle ADC = 90^{\circ}$ . Let $E$ and $F$ be the feet of perpendiculars from $A$ and $C$ to $BD$ respectively. Prove that $BE = DF$ .

2 replies

jl_

3 hours ago

lw202277

38 minutes ago

The old one is gone.

EeEeRUT 9

N 41 minutes ago by Jupiterballs

Source: EGMO 2025 P2

An infinite increasing sequence $a_1 < a_2 < a_3 < \cdots$ of positive integers is called central if for every positive integer $n$ , the arithmetic mean of the first $a_n$ terms of the sequence is equal to $a_n$ .

Show that there exists an infinite sequence $b_1, b_2, b_3, \dots$ of positive integers such that for every central sequence $a_1, a_2, a_3, \dots,$ there are infinitely many positive integers $n$ with $a_n = b_n$ .

9 replies

EeEeRUT

Apr 16, 2025

Jupiterballs

41 minutes ago

Inequalities

Humberto_Filho 2

N 41 minutes ago by damyan

Source: From the material : A brief introduction to inequalities.

Let a,b be nonnegative real numbers such that $a + b \leq 2$ . Prove that :

$(1+a^2)(1+b^2) \geq (1 + (\frac{a+b}{2})^2)^2$

2 replies

1 viewing

Humberto_Filho

Apr 12, 2023

damyan

41 minutes ago

3 var inequalities

sqing 1

N an hour ago by sqing

Source: Own

Let $a,b> 0$ and $a+b\leq 2ab .$ Prove that
$\frac{ a + b }{ a^2(1+ b^2)} \leq\frac{1 }{\sqrt 2}-\frac{1 }{2}$ $\frac{ a +ab+ b }{ a^2(1+ b^2)} \leq \sqrt 2-1$ $\frac{ a +a^2b^2+ b }{ a^2(1+ b^2)} \leq\frac{\sqrt5 }{2}$

1 reply

sqing

an hour ago

sqing

an hour ago

Divisibility holds for all naturals

XbenX 13

N an hour ago by zRevenant

Source: 2018 Balkan MO Shortlist N5

Let $x,y$ be positive integers. If for each positive integer $n$ we have that $(ny)^2+1\mid x^{\varphi(n)}-1.$ Prove that $x=1$ .

(Silouanos Brazitikos, Greece)

13 replies

XbenX

May 22, 2019

zRevenant

an hour ago

Sum and product of 5 numbers

jl_ 1

N an hour ago by jl_

Source: Malaysia IMONST 2 2023 (Primary) P2

Ivan bought $50$ cats consisting of five different breeds. He records the number of cats of each breed and after multiplying these five numbers he obtains the number $100000$ . How many cats of each breed does he have?

1 reply

jl_

4 hours ago

jl_

an hour ago

Interesting inequalities

sqing 3

N an hour ago by sqing

Source: Own

Let $a,b> 0$ and $a+b\leq 2ab .$ Prove that
$\frac{ 9a^2- ab +9b^2 }{ a^2(1+b^4)}\leq\frac{17 }{2}$ $\frac{a- ab+b }{ a^2(1+b^4)}\leq\frac{1 }{2}$ $\frac{2a- 3ab+2b }{ a^2(1+b^4)}\leq\frac{1 }{2}$

3 replies

1 viewing

sqing

5 hours ago

sqing

an hour ago

a+b+c=2 ine

KhuongTrang 30

N an hour ago by KhuongTrang

Source: own

Problem. Given non-negative real numbers $a,b,c: ab+bc+ca>0$ satisfying $a+b+c=2.$ Prove that $\color{blue}{\frac{a}{\sqrt{2a+3bc}}+\frac{b}{\sqrt{2b+3ca}}+\frac{c}{\sqrt{2c+3ab}} \le \sqrt{\frac{2}{ab+bc+ca}}. }$

30 replies

KhuongTrang

Jun 25, 2024

KhuongTrang

an hour ago

2021 EGMO P1: {m, 2m+1, 3m} is fantabulous

anser 55

N 2 hours ago by NicoN9

Source: 2021 EGMO P1

The number 2021 is fantabulous. For any positive integer $m$ , if any element of the set $\{m, 2m+1, 3m\}$ is fantabulous, then all the elements are fantabulous. Does it follow that the number $2021^{2021}$ is fantabulous?

55 replies

anser

Apr 13, 2021

NicoN9

2 hours ago

Perspective Triangles

MNJ2357 2

N Jan 26, 2020 by Pathological

Source: Korea Winter Program Practice Test 2 P4

$\triangle ABC$ and $\triangle A_1B_1C_1$ are perspective triangles. $(ABB_1)$ and $(ACC_1)$ meet at $A_2 (\neq A)$ . Define $B_2,C_2$ analogously. Prove that $AA_2, BB_2,CC_2$ are concurrent.

2 replies

MNJ2357

Jan 23, 2020

Pathological

Jan 26, 2020

Perspective Triangles

G H J

Reveal topic tags

geometry

G H BBookmark kLocked kLocked NReply

Source: Korea Winter Program Practice Test 2 P4

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MNJ2357

644 posts

MNJ2357

#1 h Jan 23, 2020, 3:33 PM • 2 Y

Y by stroller, Adventure10

$\triangle ABC$ and $\triangle A_1B_1C_1$ are perspective triangles. $(ABB_1)$ and $(ACC_1)$ meet at $A_2 (\neq A)$ . Define $B_2,C_2$ analogously. Prove that $AA_2, BB_2,CC_2$ are concurrent.

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stroller

894 posts

stroller

#2 h Jan 23, 2020, 6:03 PM • 3 Y

Y by MNJ2357, Adventure10, Mango247

mOvInG pOiNtS

Lemma: The map $AO \mapsto (BC_1C)$ , $A_1 \mapsto B_2$ is projective.

Proof: After inversion around $B$ , $A_1 \in (ABA_1)$ and $B_2$ is the intersection of line $AA_1$ with line $CC_1$ . Thus the map becomes projective after inversion. Since inversion preserves cross ratios, the map itself is projective, as desired.

Now we induct on $n$ , the number of points among $A_2, B_2, C_2$ that are equal to $O$ . The base case $n = 1$ is true and this follows from the fact that the radical axes of $(AA_1B), (AA_1C),(BOC)$ concur.

Suppose the statement holds for $n-1$ and we prove it for $n (\le 3)$ . WLOG $A_1 \ne O$ . The maps $A_1 \mapsto BB_2, CC_2$ are both projective, hence it suffices to verify the statement for three positions of $A_1$ . We may take $A_1 = O$ (and have the result from induction hypothesis), $\infty$ (in which case the lines concur at $A$ ) and the point on $AO$ such that $AA_1B_1B$ is cyclic.

why am I such a troll

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Pathological

578 posts

Pathological

#4 h Jan 26, 2020, 11:29 PM • 3 Y

Y by stroller, Pluto1708, Adventure10

Easy Trig

Let $O_{ab}$ denote the circumcenter of $\triangle ABB_1$ , and define $O_{ba}, O_{bc}, O_{cb}, O_{ca}, O_{ac}$ analogously.

For a point $P$ and line $\ell$ , let $f(P, \ell)$ denote the line through $P$ perpendicular to $\ell.$

Note that $AA_2$ is the line $f(A, O_{ab}O{ac}).$ Also, $AP$ is the line $f(A, O_{ba}O_{ca}).$ Since it's obvious that $f(A, O_{ba}O_{ca}), f(B, O_{ab}O_{cb}), f(C, O_{ac}O_{bc})$ are concurrent (at $P$ ), we get from sine Ceva in $\triangle ABC$ that $O_{ab} \cdot O_{bc} \cdot O_{ca} = O_{ba} \cdot O_{cb} \cdot O_{ac}$ .

As this condition is clearly symmetric in $a, b, c$ , it follows that $f(A, O_{ab}O_{ac}), f(B, O_{ba}O_{bc}), f(C, O_{ac}O_{bc})$ are concurrent as well.

$\square$

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