Circle Inscribed in Triangle: Area Ratios with Inscribed Circle Tangents

[songoku]

Homework Statement

Consider a triangle ABC, where angle A = 60^o. Let O be the inscribed circle of triangle ABC, as shown in the figure. Let D, E and F be the points at which circle O is tangent to the sides AB, BC and CA. And let G be the point of intersection of the line segment AE and the circle O. Set x = AD. Find the ratio of area of triangle ADF and (AG.AE)

Homework Equations

Area of triangle = 1/2 ab sin α
radius is perpendicular to tangent
sine formula
cosine formula

The Attempt at a Solution

I don't have any ideas to begin with. The answer in the manual is √3 / 4

Thanks

 

[scurty]

What do you mean by (AG.AE)? Do you mean their multiplication? i.e. \overline{AG} \cdot \overline{AE}? In any case, you have enough information to solve for the area of \triangle ADF, that should be a good start.

 

[songoku]

What do you mean by (AG.AE)? Do you mean their multiplication? i.e. \overline{AG} \cdot \overline{AE}? In any case, you have enough information to solve for the area of \triangle ADF, that should be a good start.

Yes it is their multiplication.

Let me try. AD = x and AF also = x, so area of ADF = 1/2 x² sin 60^o = 1/4 √3 x².

Then, how to find AG and AE? Thanks

 

[songoku]

anyone please?

 

[NascentOxygen]

You might get inspiration if you mark in things you know.

If you join each of D,F and E all to O you have equal radii of a circle. Furthermore, you are told that D, E and F are points where the circle is tangent to the sides of the triangle. Tangent? What can you mark in that shouts "I'm a tangent here".

I'm sure that at least some of these will be useful.

 

[songoku]

You might get inspiration if you mark in things you know.

If you join each of D,F and E all to O you have equal radii of a circle. Furthermore, you are told that D, E and F are points where the circle is tangent to the sides of the triangle. Tangent? What can you mark in that shouts "I'm a tangent here".

I'm sure that at least some of these will be useful.

I'm not sure what you mean but maybe you are referring that OD, OF and OE are perpendicular to theirs respective tangents?

ADF is equilateral triangle. But I still don't know how to find AG and AE...

 

[rcgldr]

I think he means that the next step is to find the radius of the circle versus the chord length of DF.

 

[songoku]

I think he means that the next step is to find the radius of the circle versus the chord length of DF.

Oh ok. Let me try again:

DF = x and by using cosine rule:
x² = r² + r² - 2.r.r.cos 120^o
x² = 2r² + r²
r = (1 / √3) x

Then, how to relate the radius to find AE or AG? Thanks

 

[rcgldr]

Then, how to relate the radius to find AE or AG? Thanks

I'm not sure what to do next. Hopefully NascentOxygen can offer another hint.

 

[NascentOxygen]

I haven't solved it to completion, but I believe I have all the information needed.

Focus on ΔEOA. Find OE and OA both in terms of x. You don't know any of the angles in ΔEOA, so nominate one as angle θ.

With some effort, you should be able to express EA and EG in terms of x and θ.

Hopefully, there will be complete cancellation of terms involving θ when you find the expression for AG·AE

Good luck!

 

[NascentOxygen]

I haven't solved it to completion, but I believe I have all the information needed.

I had time to finish this today, so it is solved.

Focus on ΔEOA. Find OE and OA both in terms of x. You don't know any of the angles in ΔEOA, so nominate one as angle θ.

Label ∠OEG as θ. Express EO and OA in terms of x and θ, and hence determine EA in terms of x and θ.

In ΔOEG determine EG in terms of x and θ.

Hopefully, there will be complete cancellation of terms involving θ when you find the expression for AG·AE

In the expression, the terms involving θ are found to be equal but opposite, so are eliminated. https://www.physicsforums.com/images/icons/icon14.gif

Good luck!

 

[songoku]

I haven't solved it to completion, but I believe I have all the information needed.

Focus on ΔEOA. Find OE and OA both in terms of x. You don't know any of the angles in ΔEOA, so nominate one as angle θ.

With some effort, you should be able to express EA and EG in terms of x and θ.

Hopefully, there will be complete cancellation of terms involving θ when you find the expression for AG·AE

Good luck!

I had time to finish this today, so it is solved.

Label ∠OEG as θ. Express EO and OA in terms of x and θ, and hence determine EA in terms of x and θ.

In ΔOEG determine EG in terms of x and θ.

In the expression, the terms involving θ are found to be equal but opposite, so are eliminated. https://www.physicsforums.com/images/icons/icon14.gif

Good luck!

Sorry but how to find OA in term of θ and x? In Δ OEA, OE = r and ∠OEA = θ, then there is no other information ...

I think I can continue the working if I am able to find OA

Thanks

 

[NascentOxygen]

Sorry but how to find OA in term of θ and x?

All angles of ∆DOA are known, so express OA in terms of x.

 

[songoku]

All angles of ∆DOA are known, so express OA in terms of x.

Wow, it actually works. Despite that in the middle of working the equation becomes complicated, in the end the terms cancel out. Thanks for the help

 

[NascentOxygen]

Wow, it actually works. Despite that in the middle of working the equation becomes complicated, in the end the terms cancel out. Thanks for the help

My pleasure.

And here's your reward for persistence: