In the given figure, P and Q are point of contact. O is the incentre. Line BO produced, meets PQ at G. Find the value of angle AGB?

The reason i shared the question as a note is that i don't know the answer. Try it out, & if you solve, please post a rigorous solution rather than mere guess.

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:

TopNewestCheck out this graphic

Baffling Geometry

With lots of angle-chasing, it can be worked out that, given angles \(a, b\), the following angles are

\(\angle OAP=90-a-b\)

\(\angle OGP=90+a+b\)

\(\angle AOG=90-a\)

\(\angle APG=90+a\)

Hence, \(APGO\) is a cyclic quadrilateral. Moreover, since \(APO\) is a right triangle, \(AO\) is the diameter of the circle that circumscribes the cyclic quadrilateral \(APGO\). Hence, \(AGO\) is also a right triangle, and we have our answer.

Log in to reply

Here is another way to prove it. It is easy to see \( \angle PQC = {\angle BAC} /2 + {\angle ABC}/2 \). So you can get \( \angle BGQ = {\angle BAC} /2 = \angle BAO\), therefore \( \triangle BGQ \sim \triangle BAO \) and \( BG \times BO = BQ \times BA \). Let the point of contact of incircle with AB is D, we have \( BG \times BO = BD \times BA \). So AGOD is cyclic. \( \angle AGB = \angle BDO = 90 ^ \circ \)

Log in to reply

You haven't defined the point \(D\) in this proof. But it is simply the point touching the circle on the segment \(AB\).

Log in to reply

Construct a model diagram with AC and BC AB as tangents

Log in to reply

With any dimensions of radius and lengths of triangle

Log in to reply

AP PG GB

Log in to reply

90

Log in to reply

I took an equilateral triangle and using coordinate geometry found out the value of the angle. Is the method legitimate?

Log in to reply

You've only found it when the triangle is equilateral. You haven't proved that the angle is the same for all triangles. You can't assume that all the triangles have the same angle just because the question asks for a single unique numeric value.

Log in to reply

There's no line that connects with A and G. Therefore, there is no angle AGB.

Log in to reply

There doesn't need to be a line connecting two vertices for an angle to count @Kent Arvin Aquino

Log in to reply

It's from INMO ,right.

Log in to reply

Log in to reply