In the previous image \(AM\) and \(CN\) are bisectors and \(P\) lies on the line segment \(MN\). Prove that the distance from \(P\) to line \(AC\) is equal to the sum of the distances from \(P\) to lines \(BA\) and \(BC\).

What happen if \(P\) lies on the line \(MN\) but outside the triangle?

Now draw the circumcircle of triangle \(ABC\) and suppose the ray \(MN\) intersects the circumcircle at \(Q\) (\(N\) is between \(Q\) and \(M\)). Prove that \(\frac{1}{QA}=\frac{1}{QC}+\frac{1}{QB}\).

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:

TopNewestI proved The first 1... and based on that created a problem. Maybe if u likeJorge,,,, you may share this problem. So here's the problem. In an acute angled triangle \(ABC\) we have \(BC=4\),\(AC=5\),\(AB=6\). \(BM\) and \(CN\) are the internal bisectors of \(B\) and \(C\). \(P\) is any point on the segment \(MN\). If \([\triangle PMC]=1\) and \([\triangle PNB]=4\) , find \([\triangle BPC]\) . ( it wud be nice if u mention a little credit to me.. :P )

Log in to reply

Reply if you find this problem interesting! :)

Log in to reply

I think there are too many conditions in your problem. Triangle ABC is fixed and also the points M and N, thus if you choose P in line segment MN such that [BPN]=2 then [CPM] is determined and is not equal to 4. Another idea is to calculate [MNBC] and then find [BPC] (I think the answer will be different of yours).

Try this problem (using your notation):

If [BPN]=2, what is the correct value of [CMP]?

Log in to reply

Then cant we fix \(P\) and say that suppose there exists a point \(P\) on \(MN\) such that ........ ?

Log in to reply

However, solution to #1. Let \(PX \perp BC\), \(PY \perp AC\) and \(PZ \perp AB\). We drop \(MM_1 \perp BC\), \(MZ_1 \perp AB\) and \(NN_1 \perp BC\) and \(NY_1 \perp AC\). Now we have \(\triangle BM_1M \cong \triangle BMZ_1\) (right angle, angle and side equal) and hence \(MZ_1=MM_1\). Similarly we have, \(NN_1 = NY_1\).

We use similarity in triangles \(MPY\) and \(MNY_1\).

\(\dfrac{PY}{NY_1} = \dfrac{MP}{NM}\) and we get \(PY = \dfrac{PM}{NM} \times NY_1\)...(i) Similarly we have \(PZ = \dfrac{NP}{NM} \times MZ_1 \)...(ii)

Now, \(NN_1 , PX,MM_1\) are 3 parallel lines . So we have \(\dfrac{PM}{NM} = \dfrac{M_1X}{M_1N_1}\).... (iii) We join \(N, M_1\) and let that intersect \(PX\) at \(L\).

Using \(\triangle NLP \sim NM_1M\) , we have \(\dfrac{PL}{MM_1} = \dfrac{NP}{NM} \) and thus \(PL = \dfrac{NP}{NM} \times MM_1 = \dfrac{NP}{NM} \times MZ_1\). (as \(MM_1 = MZ_1\) ... (iv)

And, similarly using \(\triangle M_1XL \sim \triangle M_1N_1N\) , we have

\(LX = \dfrac{XM_1}{N_1M_1} \times NN_1 = \dfrac{PM}{NM} \times NY_1\)...(v)

We observe that \((iv) + (v)\) is indeed \((i) + (ii)\) and Hence \(PY + PZ = PL + LX = PX\)

Log in to reply

Is it correct Jorge??

Log in to reply

It seems good, but you misplaced the points, for example you said \(MM_1\) perpendicular to \(BC\).

Log in to reply