Question

Can someone please help me??

Let \(ABC\) be an acute triangle and let \(M\) be the midpoint of \(AC\). A circle \(\omega\) passing through \(B\) and \(M\) meets the sides \(AB\) and \(BC\) again at points \(P\) and \(Q\), respectively. Let \(T\) be the point such that quadrilateral \(BPTQ\) is a parallelogram. Suppose that \(T\) lies on the circumcircle of triangle \(ABC\). Determine all possible values of \(BT/BM\).

Answer 1

Do you perhaps have a screenshot of the problem?

Answer 2

I don't understand this at all.

Answer 3

Anyone?

Answer 4

Please help me

Answer 5

@vocaloid

Answer 6

@alicole123

Answer 7

@minesweeper @meinswepere

Answer 8

They arent showing i cant solve this

Answer 9

In an acute triangle, the circumcenter of the triangle lies on the line O.

Acute triangle.

In geomentry, acute triangle is the type of triangle in which all the three internal angles of the triangle are acute

Given,

Here we need to prove that in the acute triangle ABC, the circumcenter of the triangle MNH lies on the line OH.

Let us consider T be the intersection of σ1 and σ1 other than W.

Now, the Lemma 1: T is on XY.

Now, the Proof is written as We have ∠XTW} = ∠YTW = 90 degres because they intercept semicircles.

Therefore, ∠XTY = ∠XTW + ∠YTW = 180 degrees,

Then XTY is a straight line.

Here we know that NHLB is cyclic because ∠BNH and ∠BLH, opposite and right angles, sum to 180 degree. Furthermore, we are given that NTWB is cyclic.

Therefore, the triangle MNH lies on the line OH.

Answer 10

maybe learn to read before you copy random solutions from the internet

Answer 11

this is 2015 G4 not APMO 2010/4