Power of a Point with a rectangle

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0 replies

jwelsh

Jul 1, 2025

0 replies

Find the value of angle C

markosa 11

N an hour ago by Razyed

Given a triangle ABC with base BC

angle B = 3x
angle C = x
AP is the bisector of base BC (i.e.) BP = PC
angle APB = 45 degrees

Find x

I know there are multiple methods to solve this problem using cosine law, coord geo
But is there any pure geometrical solution?

11 replies

markosa

5 hours ago

Razyed

an hour ago

Polynomials

Roots_Of_Moksha 7

N 2 hours ago by vanstraelen

The polynomial $x^3 - 3(1+\sqrt{2})x^2 + (6\sqrt{2}-55)x -(7+5\sqrt{2})$ has three distinct real roots $\alpha$ , $\beta$ and $\gamma$ . The polynomial $p(x)=x^3+ax^2+bx+c$ has roots $\sqrt[3]{\alpha}$ , $\sqrt[3]{\beta}$ , $\sqrt[3]{\gamma}$ . Find the integer closest to $a^2 + b^2 + c^2$ .
Answer

22.
I am looking for a good solution.

7 replies

Roots_Of_Moksha

Jul 20, 2025

vanstraelen

2 hours ago

Inequalities

sqing 10

N 3 hours ago by DAVROS

Let $a,b> 0$ and $2a+2b+ab=5.$ Prove that
$\frac{a^4}{b^4}+\frac{1}{a^4}+42ab-a^4\geq 43$ $\frac{a^5}{b^5}+\frac{1}{a^5}+64ab-a^5\geq 65$ $\frac{a^6}{b^6}+\frac{1}{a^6}+90ab-a^6\geq 91$ $\frac{a^7}{b^7}+\frac{1}{a^7}+121ab-a^7\geq 122$

10 replies

sqing

May 28, 2025

DAVROS

3 hours ago

combinatorics

kjhgyuio 1

N 3 hours ago by Alphabeta123

........

1 reply

kjhgyuio

Jun 25, 2025

Alphabeta123

3 hours ago

No more topics!

Power of a Point with a rectangle

elpianista227 2

N Jun 17, 2025 by mathafou

Let $\Gamma$ be a semicircle; let $ABMN$ be its circumscribed rectangle. Extend $BM$ to a point $C$ such that $BM = CM$ ; let $T$ be a point on the circumference of $\Gamma$ such that $CT = 1, BM = 2$ . Find the value of $AB$ .

2 replies

elpianista227

Jun 16, 2025

mathafou

Jun 17, 2025

Power of a Point with a rectangle

G H J

Reveal topic tags

geometry

rectangle

G H BBookmark kLocked kLocked NReply

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elpianista227

64 posts

elpianista227

#1 h Jun 16, 2025, 3:19 PM

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This post has been edited 2 times. Last edited by elpianista227, Jun 16, 2025, 3:30 PM

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elpianista227

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elpianista227

#2 h Jun 16, 2025, 3:21 PM

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Since $BM = CM$ , we have $BM = CM = 2$ . Therefore, $BC = 4$ . We now find the length of $CT$ . Using the Power of a Point Theorem, we have
$CB^2 = (CT)(AC)$ $16 = AC$ Thus, we have that $AC = 16$ and so $AB = \sqrt{AC^2 - BC^2} = \sqrt{16^2 - 4^2} = \sqrt{240} = 4\sqrt{30}$ .

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mathafou

422 posts

mathafou

#3 h Jun 17, 2025, 9:34 AM

Y by

I don't understand

$ABMN$ circumscribed means the semi circle is inside the rectangle
all points inside the rectangle are at a distance $> 2$ to $C$
and to apply PoP to $A, C, T$ they must be in line and $A$ on the circle
which is not said ( $T$ a point , ..)
so what ??
what is the true statement (preferably with a figure) ?

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