Given a non-isosceles triangle \(ABC\) with incenter \(I\) and circumcircle \(\omega\). Denote the midpoints of arcs \(BC, AC, AB\) that does not contain the opposite vertex by \(X,Y,Z\) respectively. Denote \(P\) the midpoint of arc \(BC\) containing \(A\). Denote the intersection of \(BP\) and \(ZX\) as \(M\) and the intersection of \(XY\) and \(CP\) as \(N\). Prove that

Quadrilaterals \(BXIM\) and \(XCNI\) are kites.

\(MIN\) are collinear and \(XI\) is perpendicular to \(MN\).

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestThey are two cyclic kites meaning symmetric pair of angles are right,which directly implies the second property

Log in to reply

Yea even i did it that way ... although in the second part i could first prove the second statement and then the collinearity. This was a nice geometry problem @Alan Yan Can you also post some more geometry problems ?

Log in to reply

I m not able to understand the location of point N as intersection of BC and CP makes point C as point N

Log in to reply