The line through the feet of the bisectors

  1. In the previous image AMAM and CNCN are bisectors and PP lies on the line segment MNMN. Prove that the distance from PP to line ACAC is equal to the sum of the distances from PP to lines BABA and BCBC.

  2. What happen if PP lies on the line MNMN but outside the triangle?

  3. Now draw the circumcircle of triangle ABCABC and suppose the ray MNMN intersects the circumcircle at QQ (NN is between QQ and MM). Prove that 1QA=1QC+1QB\frac{1}{QA}=\frac{1}{QC}+\frac{1}{QB}.

Note by Jorge Tipe
5 years, 6 months ago

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I 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 ABCABC we have BC=4BC=4,AC=5AC=5,AB=6AB=6. BMBM and CNCN are the internal bisectors of BB and CC. PP is any point on the segment MNMN. If [PMC]=1[\triangle PMC]=1 and [PNB]=4[\triangle PNB]=4 , find [BPC][\triangle BPC] . ( it wud be nice if u mention a little credit to me.. :P )

Sagnik Saha - 5 years, 5 months ago

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Reply if you find this problem interesting! :)

Sagnik Saha - 5 years, 5 months ago

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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]?

Jorge Tipe - 5 years, 5 months ago

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Then cant we fix PP and say that suppose there exists a point PP on MNMN such that ........ ?

Sagnik Saha - 5 years, 5 months ago

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However, solution to #1. Let PXBCPX \perp BC, PYACPY \perp AC and PZABPZ \perp AB. We drop MM1BCMM_1 \perp BC, MZ1ABMZ_1 \perp AB and NN1BCNN_1 \perp BC and NY1ACNY_1 \perp AC. Now we have BM1MBMZ1\triangle BM_1M \cong \triangle BMZ_1 (right angle, angle and side equal) and hence MZ1=MM1MZ_1=MM_1. Similarly we have, NN1=NY1NN_1 = NY_1.

We use similarity in triangles MPYMPY and MNY1MNY_1.

PYNY1=MPNM\dfrac{PY}{NY_1} = \dfrac{MP}{NM} and we get PY=PMNM×NY1PY = \dfrac{PM}{NM} \times NY_1...(i) Similarly we have PZ=NPNM×MZ1PZ = \dfrac{NP}{NM} \times MZ_1 ...(ii)

Now, NN1,PX,MM1NN_1 , PX,MM_1 are 3 parallel lines . So we have PMNM=M1XM1N1\dfrac{PM}{NM} = \dfrac{M_1X}{M_1N_1}.... (iii) We join N,M1N, M_1 and let that intersect PXPX at LL.

Using NLPNM1M\triangle NLP \sim NM_1M , we have PLMM1=NPNM\dfrac{PL}{MM_1} = \dfrac{NP}{NM} and thus PL=NPNM×MM1=NPNM×MZ1PL = \dfrac{NP}{NM} \times MM_1 = \dfrac{NP}{NM} \times MZ_1. (as MM1=MZ1MM_1 = MZ_1 ... (iv)

And, similarly using M1XLM1N1N\triangle M_1XL \sim \triangle M_1N_1N , we have

LX=XM1N1M1×NN1=PMNM×NY1LX = \dfrac{XM_1}{N_1M_1} \times NN_1 = \dfrac{PM}{NM} \times NY_1...(v)

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

Sagnik Saha - 5 years, 5 months ago

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Is it correct Jorge??

Sagnik Saha - 5 years, 5 months ago

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It seems good, but you misplaced the points, for example you said MM1MM_1 perpendicular to BCBC.

Jorge Tipe - 5 years, 5 months ago

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