<DPQ =2 <ADC wanted, tangents related

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jlacosta

Apr 2, 2025

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Inspired by bamboozled

sqing 0

5 minutes ago

Source: Own

Let $a,b,c$ be reals such that $(a^2+1)(b^2+1)(c^2+1) = 27.$ Prove that $1-3\sqrt 3\leq ab + bc + ca\leq 6$

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sqing

5 minutes ago

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Range of ab + bc + ca

bamboozled 1

N 14 minutes ago by sqing

Let $(a^2+1)(b^2+1)(c^2+1) = 9$ , where $a, b, c \in R$ , then the number of integers in the range of $ab + bc + ca$ is __

1 reply

+1 w

bamboozled

29 minutes ago

sqing

14 minutes ago

Functional Equation

AnhQuang_67 4

N 14 minutes ago by AnhQuang_67

Find all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying $2\cdot f\Big(\dfrac{-xy}{2}+f(x+y)\Big)=xf(y)+yf(x), \forall x, y \in \mathbb{R}$

4 replies

AnhQuang_67

Yesterday at 4:50 PM

AnhQuang_67

14 minutes ago

Inradius and ex-radii

bamboozled 0

22 minutes ago

Let $ABC$ be a triangle and $r, r_1, r_2, r_3$ denote its inradius and ex-radii opposite to the vertices $A, B, C$ respectively. If $a> r_1, b > r_2$ and $c > r_3$ , then which of the following is/are true?
(A) $\angle{B}$ is obtuse
(B) $\angle{A}$ is acute
(C) $3r > s$ , where $s$ is semi perimeter
(D) $3r < s$ , where $s$ is semi perimeter

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bamboozled

22 minutes ago

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<DPQ =2 <ADC wanted, tangents related

parmenides51 1

N Oct 13, 2021 by toilaDang

Source: 1996 Argentina IberoAmerican Training List p10 https://artofproblemsolving.com/community/c2434876_argentina_training_lists__geom

Let $AB$ and $AC$ be tangent to the circle $\Gamma$ at $B$ and $C$ respectively. Let $D$ be a point on the extension of $AB$ , beyond $B$ , and let $P$ be the second intersection point of $\Gamma$ with the circle circumscribed to the triangle $ACD$ . Let $Q$ be the foot of the perpendicular from $B$ to $CD$ . Prove that $\angle DPQ =2 \angle ADC$ .

1 reply

parmenides51

Oct 12, 2021

toilaDang

Oct 13, 2021

<DPQ =2 <ADC wanted, tangents related

G H J

Reveal topic tags

equal angles

geometry

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Source: 1996 Argentina IberoAmerican Training List p10 https://artofproblemsolving.com/community/c2434876_argentina_training_lists__geom

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parmenides51

30629 posts

parmenides51

#1 h Oct 12, 2021, 8:38 PM

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toilaDang

1 post

toilaDang

#2 h Oct 13, 2021, 10:35 AM

Y by

my solution:
Let $F$ be the midpoint of $BC$ ; $P'$ be the second intersection point of $FE$ with $\Gamma$ .
By power of a point, $FB^2=FE.FP'=FD^2$ ,so $\angle FP'B= \angle EBF$ and $\angle FP'D=\angle EDF$
$\angle DP'B=\angle EBF+\angle EDF=\angle BEC=\angle BP'C$
then $\angle DP'C=2.\angle BP'C=\angle BOC=180^{\circ}-\angle BAC$
$\Rightarrow DP'CA$ is the inscribed quadrilateral
$\Rightarrow P'\equiv P$
easy to prove that $\triangle PED \sim \triangle PCA$
$\Rightarrow \angle DPF=\angle APC=\angle ADC=\angle DQF$ so $DPQF$ is the inscribed quadrilateral
$\Rightarrow \angle DPQ=\angle BFQ=2\angle ADC$

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