In triangle ABC, D, E and F are the points of tangency of incircle with the center of I to BC, CA and AB respectively. Let M be the foot of the perpendicular from D to EF. P is on DM such that DP=MP. If H is the orthocenter of BIC, prove that PH bisects EF.
Problem
Source:
Tags: geometry, geometric transformation, homothety, trigonometry, graphing lines, slope, Iran Lemma
16.05.2009 20:42
khashi70 wrote: In triangle ABC , D , E and F are the points of tangency of incircle with the center of I to BC , CA and AB respectively . Let M be the feet of perpendicular from D to EF and P is on DM such that : DP=MP . If H be the orthocenter of ABC, prove that PH bisects EF. Maybe you have made a typo, check it again.
16.05.2009 21:08
Sorry you're right ... it's OK now .
16.05.2009 21:52
Let CI cut BH at G and BI cut EF ate Q. Then ∠GIB=180∘−∠B+∠C2=90∘−A2=∠EFA. Thus, the quadrilateral BFGI is cyclic and ∠BGI=∠BFI=90∘. So, points B,G and H are collinear. Denote K∈HI∩EF. Let J be midpoint of EF. By Menelaos theorem for △DMK and points P,J,H it's enough to prove that PMPN⋅HDHK⋅JKJM=1⟺HDHK=JMJK (∗). From AF=AE we get IJ⊥EF, so JI∥DM. Therefore JMJK=IDIK. We rewrite the relation (∗) as HDHK=IDIK This is equivalent to prove that (D,I,K,H) is harmonical. QGD is orthic triangle of HBC. So GI is bisector of DGK. From ∠IGH=90∘ we get that GH is exterior bisector of DGK and the result follows. Image not found
16.05.2009 22:25
Sketch of my solution 1. The intersection points of the altitudes from B and C with the corresponding sides in triangle HBC are on the line EF (well known lemma) 2. H, I and D are collinear. (easy with the use of harmonical division in both triangles ABC and HBC) 3. HBC and DEF are similar. (angle chasing) 4. Let the orthocenter of DEF be H' and the midpoint of EF be N. Now from the well known property, H'D is equal to 2 times IN. Using 3, we find that AI/AD=IN/PD. Then the result follows from the fact that IN//PD.
17.05.2009 08:13
Here is another aproach for this problem: Let {K}≡CI∩FE, {G}≡BI∩EF. It is very well- know that BK⊥CK and BG⊥CG (this result can be proved easily by harmonic division). Hence {H}≡BK∩CG. Let J be the midpoint of EF. Denote P the intersection by HJ and DM. We will prove that P is the midpoint of DM. Indeed, let S be the projection of H onto EF, in order to prove P is the midpoint of DM, it is enough to prove H(MJYS)=−1, where {Y}≡HD∩EF, which is equivalent to proving ¯JM¯JY=−¯SM¯SY. Note that J is the midpoint of EF⟹IJ⊥EF⟹IJ//DM, which implies ¯JY¯JM=¯IY¯ID. But (HIYD)=−1⟹¯IY¯ID=−¯HY¯HD=−¯HY¯HY+¯YD=−¯SY¯SY+¯YM=−¯SY¯SM ⟹¯SY¯SM=−¯JY¯JM. Which implies that (MJYS)=−1⟹H(MDPS)=−1, but DM//HS⟹P is the midpoint of DM, which leads to the result of the problem. Our proof is completed then.
18.05.2009 17:47
Sorry for my bad English... Here is the proof without calculation. Lemma:In a triangle ABC,let D be the tangent of incircle (I) to BC,(Ia) be the excircle of A wrt triangle ABC and M be midpoint of the altitude from A wrt triangle ABC,then D,Ia,M are collinear. Proof: Let (Ia) meet BC at N,NIa intersect (Ia) at N and P The homothety center A take (I) to (Ia) also take D to P,then A,D,P are collinear. Then the homothety center D take the altitude from A wrt triangle ABC to NP,it also take M to Ia We conclude that D,Ia,M are collinear. Return to the problem,apply the lemma to the orthic triangle of triangle BHC then P,H,M' are collinear (M' is the midpoint of EF). Q.E.D
20.05.2009 19:55
let K midpoint of EF and r radius of incircle of △ABC. let PH cut IK at K′ so IK′DP=HIHD (i)(because of IK||PD) it is easy to measure this lenght: IK=rsinA2 (1) DM=BDcosB2=2rcosc2cosB2 so DP=rcosB2cosC2 (2) with (1),(2): IKDP=sinA2cosB2cosC2 (3). and ID=BDtanB2,DH=cotC2 so HI=BD(cotC2−tanB2) so HIDH=1−tanB2cotC2=1−tanB2C2=sinA2cosB2cosC2 (4). with (3),(4): IKDP=HIHD (ii). with (i),(ii): PH passer from K. problem solved. Image not found
04.12.2011 10:24
let T be the intersection point of PH and EF, S the intersection point of IH and EF, and R,T′ be the feet of the perpendicular from H and I to the line EF respectively. since HR||DM and P is the midpoint of DM, we have H(RPDM)=−1, so (RTSM)=−1. on the other side, EF is the line joining feet of the altitudes of vertices B and C of triangle HBC respectively, so (HISD)=−1 and therefore (RT′SM)=−1. we cocnlude that T=T′, hence T is the midpoint of EF, and the proof is complete.
06.08.2012 16:30
[asy][asy]import graph; size(12.73cm); real lsf=0.5; pathpen=linewidth(0.7); pointpen=black; pen fp=fontsize(10); pointfontpen=fp; real xmin=-7.08,xmax=5.65,ymin=-3.59,ymax=3.72; pen qqfftt=rgb(0,1,0.2), ttwwqq=rgb(0.2,0.4,0), zzttqq=rgb(0.6,0.2,0), ccqqqq=rgb(0.8,0,0); pair A=(-3,2), B=(-4,-3), C=(3,-3), I=(-1.86,-1.24), D=(-1.86,-3), F=(-3.58,-0.9), H=(-1.86,2.92), B_1=(0.19,0.43), C_1=(-3.19,-0.76), M=(-2.71,-0.59), P=(-2.28,-1.79), X_C=(-4.46,1.37), X_B=(0.75,1.37); D(A--B--C--cycle,linewidth(1.2)+dotted+qqfftt); D(D--(-0.73,0.11)--F--cycle,linewidth(1.2)+dotted+ttwwqq); D(B--C--H--cycle,zzttqq); D(B_1--C_1--D--cycle,blue); D(A--B,linewidth(1.2)+dotted+qqfftt); D(B--C,linewidth(1.2)+dotted+qqfftt); D(C--A,linewidth(1.2)+dotted+qqfftt); D(CR(I,1.76),linetype("2 2")+ttwwqq); D(D--(-0.73,0.11),linewidth(1.2)+dotted+ttwwqq); D((-0.73,0.11)--F,linewidth(1.2)+dotted+ttwwqq); D(F--D,linewidth(1.2)+dotted+ttwwqq); D(B--C,zzttqq); D(C--H,zzttqq); D(H--B,zzttqq); D(B_1--C_1,blue); D(C_1--D,blue); D(D--B_1,blue); D(D--M); D(CR(I,0.9),linewidth(1.6)+linetype("2 2")+blue); D(P--H,linewidth(1.6)+linetype("2 2")+ccqqqq); D(C_1--X_C,linetype("4 4")+blue); D(B_1--X_B,linetype("4 4")+blue); D(CR(H,3.03),linetype("4 4")+blue); D(A); MP("A",(-2.96,2.07),NE*lsf); D(B); MP("B",(-3.96,-2.93),NE*lsf); D(C); MP("C",(3.04,-2.93),NE*lsf); D(I); MP("I",(-1.82,-1.18),NE*lsf); D(D); MP("D",(-1.82,-2.93),NE*lsf); D((-0.73,0.11)); MP("E",(-0.68,0.17),NE*lsf); D(F); MP("F",(-3.54,-0.84),NE*lsf); D(H); MP("H",(-1.82,2.99),NE*lsf); D(B_1); MP("B_1",(0.23,0.49),NE*lsf); D(C_1); MP("C_1",(-3.15,-0.7),NE*lsf); D((-2.15,-0.39)); MP("N",(-2.11,-0.33),NE*lsf); D(M); MP("M",(-2.67,-0.52),NE*lsf); D(P); MP("P",(-2.24,-1.73),NE*lsf); D(X_C); D(X_B); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); [/asy][/asy] Let N be the midpoint of EF, and let B1=EF∩HC, C1=EF∩HB. It is not hard to see, via angle chasing, that BFC1I is cyclic, from which it follows that IC1⊥C1B⟹CC1⊥HB. Similarly BB1⊥HC, so that DB1C1 is the orthic triangle of HBC. Consider △DB1C1. N is the tangency point of its incircle with B1C1 and H is the excenter opposite D. It is well-known that P, N, and H are collinear, implying the conclusion.
13.04.2013 00:45
(Writing the Idea only, calculation is easy).Take F=(1,0),E=(−1,0),I=(0,r),M=(a,0),D=(a,b) now so equation of circle ⊙(DEF) be x2+(y−r)2=r2+1. Now note tangent at (x1,y1) point be xx1+yy1−r(y+y1)=1, using this we get B,C in terms of a,b,r. Now easily we’ll get H in terms of a,b,r. We’ve midpoint of MD is P=(a,b2) , now basically we need to show now slope of the line is tan−1(b2) and that’s quite easy to show. Note here we’re saying , solution is easy by CG because no stuffs involving any power is coming, and that makes our calculation easier. (I got a trig solution too, but there calculation is damn hard to do.)
15.07.2014 22:39
[asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(10cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -1.535709806738513, xmax = 7.699401969463958, ymin = 1.537544229253762, ymax = 7.025735038013770; /* image dimensions */ pen zzttqq = rgb(0.6000000000000006,0.2000000000000002,0.000000000000000); draw((4.120000000000004,6.740000000000007)--(0.6000000000000006,2.120000000000002)--(4.046266301600477,2.770413386903411)--cycle, zzttqq); /* draw figures */ draw((4.120000000000004,6.740000000000007)--(0.6000000000000006,2.120000000000002), zzttqq); draw((0.6000000000000006,2.120000000000002)--(4.046266301600477,2.770413386903411), zzttqq); draw((4.046266301600477,2.770413386903411)--(4.120000000000004,6.740000000000007), zzttqq); draw((2.219650741964543,4.245791598828465)--(4.061766128857575,3.604874537240598)); draw((3.226139448965370,2.615630953617537)--(3.623387848646997,3.757397078019321)); draw((0.6000000000000006,2.120000000000002)--(2.823005108246210,4.751669850437923)); draw((3.424763648806184,3.186514015818429)--(2.823005108246210,4.751669850437923)); draw((3.623387848646997,3.757397078019321)--(4.046266301600477,2.770413386903411)); draw((2.823005108246210,4.751669850437923)--(4.046266301600477,2.770413386903411)); draw((3.623387848646997,3.757397078019321)--(4.046266301600477,2.770413386903411)); draw((4.046266301600477,2.770413386903411)--(2.355699169206497,4.198457018789681)); draw((0.6000000000000006,2.120000000000002)--(3.385861897857764,3.840038173964708)); draw((3.035843169874220,3.623930701373472)--(2.823005108246210,4.751669850437923)); draw((3.226139448965370,2.615630953617537)--(3.035843169874220,3.623930701373472)); /* dots and labels */ dot((4.120000000000004,6.740000000000007),dotstyle); label("A", (4.153046114633986,6.797820137317756), NE * labelscalefactor); dot((0.6000000000000006,2.120000000000002),dotstyle); label("B", (0.6340400478875363,2.175705951202600), NE * labelscalefactor); dot((4.046266301600477,2.770413386903411),dotstyle); label("C", (4.080113346411262,2.822984269179279), NE * labelscalefactor); dot((3.035843169874220,3.623930701373472),dotstyle); label("I", (3.068171187320962,3.679944295796290), NE * labelscalefactor); dot((2.219650741964543,4.245791598828465),dotstyle); label("F", (2.256794140843153,4.299872825689447), NE * labelscalefactor); dot((3.226139448965370,2.615630953617537),dotstyle); label("D", (3.259619703905613,2.668002136705990), NE * labelscalefactor); dot((4.061766128857575,3.604874537240598),dotstyle); label("E", (4.098346538466943,3.661711103740609), NE * labelscalefactor); dot((3.623387848646997,3.757397078019321),dotstyle); label("M", (3.660749929130597,3.807576640186058), NE * labelscalefactor); dot((3.424763648806184,3.186514015818429),dotstyle); label("P", (3.460184816518106,3.242347686459944), NE * labelscalefactor); dot((2.823005108246210,4.751669850437923),dotstyle); label("H", (2.858489478680629,4.810402203248517), NE * labelscalefactor); dot((3.140708435411058,3.925333068034533),dotstyle); label("N", (3.177570339655048,3.980791964715028), NE * labelscalefactor); dot((2.355699169206497,4.198457018789681),dotstyle); label("Jc", (2.393543081260761,4.254289845550244), NE * labelscalefactor); dot((3.385861897857764,3.840038173964708),dotstyle); label("Jb", (3.423718432406743,3.898742600464463), NE * labelscalefactor); dot((2.967592072231618,3.985564490881980),dotstyle); label("X", (3.004355015126078,4.044608136909912), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] Let Jb and Jc be the feet of perpendiculars from B,C on CH,BH respectively. Since CI⊥BH and BI⊥CH, B,I,Jb and C,I,Jc are collinear. Furthermore, since ∠BJcC=∠BJbB=90∘, (BIcIbC) is cyclic and since ID⊥BD,IF⊥BF,IJc⊥BJc, (BDIJcF) is also cyclic. Now, note that ∠FJcB=∠FDB=90∘−∠ABC2 and ∠JbJcC=∠JbBC=∠ABC2, so ∠∠FJcB+∠BJcC+∠JbJcC=180∘, implying F,Jc,Jb are collinear. Similarly, E,Jb,Jc are collinear. Combining them, we see that Jb and Jc both lie on line EF. Since ID⊥BC, H,I,D are collinear. Let this line intersect EF at X. Using cyclic quadrilaterals BDJbH and CDJcH, we deduce that ∠JcJbI=90∘−∠HJbJc=90∘−∠CJbE=∠IJbC, so JbI internally bisects ∠JcJbC. Since IJb⊥JbH, JbH externally bisects ∠JbJbC, so by the angle bisector theorem we have HDID=HXIX. Let N be the midpoint of EF. Note that A and I both lie on the perpendicular bisector of EF, so IN⊥EF, implying (H,I;X,D) is harmonic, or KXIX=DHDI. Plugging these ratios, a simple calculation shows that PMPN⋅DHDX⋅XNMN=1, so the desired result follows by the converse of Menelaus theorem. ◼
03.10.2014 22:15
I will provide a more straightforward and bashy solution.(but a synthetic solution like you have done is obviously more appreciated) Applying the sine rule in △BFD we get FDsinB=s−bcosB2⟹FD=2sinB2(s−b) So MD=FDcosC2=2sinB2cosC2(s−b)⟹DP=sinB2cosC2(s−b) Let K be the midpoint of EF.Then IK=IEsinA2=rsinA2=(s−b)tanB2sinA2 Now easy angle chasing gives ∠ICH=A2 and ∠HIC=90+C2 So applying sine rule in △HIC we get HI=HCsinA2cosC2 Also HD=HCcosB2. Now note that IKPD=(s−b)tanB2sinA2sinB2cosC2(s−b)=sinA2cosB2cosC2 Again HIHD=HCsinA2cosC2HCcosB2=sinA2cosB2cosC2 Thus HIHD=IKPD.Also note that IK∥PD,both being perpendicular to EF. So H,K,P are collinear as desired.
18.06.2015 20:03
My solution: Let BI∩EF=T note that ∠BTF=∠AFE−∠ABI=C2,∠ECI=C2⟶IETC is cyclic so ∠IEC=∠ITC=90 thus C,T,H are collinear similarly B,S,H are collinear now let the perpendicular lines from D to DE,DF intersect EF at K,L note that BT||DL (because both of them are perpendicular to HC) and ∠STB=∠BTD=C2 thus TL=TD so TH is perpendicular bisector of DL similarly HS is perpendicular bisector of DK so H is circumcenter of △DKL now using this problem we get that P,M,H are collinear. DONE
Attachments:

25.08.2015 20:42
Let Q=BI∩EF, N=AI∩EF, and K=BI∩DF. Since ∠CIQ=∠CBI+∠ICB=90∘−∠IAE=∠AEF=∠CEQ and 90∘=∠CDI=∠IEC the pentagon CDIEQ is cyclic with CI as diameter, so ∠IQC=90∘ and hence Q∈CH. Denote by h the homothety centered at H sending I to D, and let R=h(Q). We claim that h(N)=P from which the assertion follows since N is also the midpoint of EF. But obviously h(IN)=DP since IN∥DP, so we just have to show that h(QN)=RP. Note first that h(IQ)=DR so IQ∥DR which means the pentagon DKMQR is cyclic since 90∘=∠QRD=∠DKQ=∠DMQ. Note also that KP is the perpendicular bisector of DM since it is the midline of the right angled triangle DFM, so ∠DKR=90∘−∠KRD=90∘−∠KMD=∠DKP which yields R∈KP, but KP∥FM so we are done.
26.08.2015 07:10
Suppose X,Y are points on BI,CI with ∠BXC=∠BYC=90. It is well known that X and Y lie on EF. Let lines DIH and AI intersect EF at Z, W; then (H,I;D,Z) is harmonic. Projecting from W yields (WH,WI;WM,WD) is harmonic, but WI∥DM, so HW bisects DM as desired.
27.08.2015 19:35
The given problem is equivalent to showing the following problem : Let ABC be a triangle and let AD be A−altitude (D on BC) and let P be the midpoint of AD. Let M be the midpoint of BC. Let O be the circumcentre of △ABC Let the A−tangent to ⊙ABC be lA; lB,lC are defined similarly. A′=lB∩lC B′,C′ are defined similarly. let K be the Lemoine point of △ABC let AO∩PM=X Then X is the orthocentre of triangle B′OC′ Solution : We know that PKM,BKB′,CKC′ are straight lines. if we can show that B′X||AB (and similarly C′X||AC), we will be done. Thus if we show △XAB′ and △CMO are similar, we are done. That is we have to show, B′AOM=AXCM By Menalu's theorem; AXOX⋅OMMA′⋅A′KKA=−1 Since we know all these ratios, the required fractions can be easily shown equal
17.09.2015 03:27
A simple complex bash...Take the in-circle to be the unit circle and observe that If X is the midpoint of EF then X⟶x=e+f2 and the orthocenter of △BIC is H⟶h=2d2(e+f)(d+e)(d+f) and mid point of altitude DM is P⟶p=3d/4+(e+f)/2−ef/d Now applying the criterion of col-linearity in complex numbers we conclude.
09.02.2016 00:51
of △DEF. Let H be the circumconic of △ABC passing through G,I and let this line meet H for a second time at H′. Denote X≡BH′∩CI and Y≡CH′∩BI. It's sufficient to show that H′ is the orthocenter of △BIC. [asy][asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = 939.9590493942901, xmax = 1098.890727397895, ymin = 973.1890994475792, ymax = 1053.622348663321; /* image dimensions */ draw((978.0750118233286,1041.1012964658203)--(972.3534714152007,984.1622953027858)--(1046.4294535107942,984.438698221053)--cycle, red); Label laxis; laxis.p = fontsize(10); xaxis(xmin, xmax, Ticks(laxis, Step = 10., Size = 2, NoZero),EndArrow(6), above = true); yaxis(ymin, ymax, Ticks(laxis, Step = 10., Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ /* draw figures */ draw((978.0750118233286,1041.1012964658203)--(972.3534714152007,984.1622953027858), red); draw((972.3534714152007,984.1622953027858)--(1046.4294535107942,984.438698221053), red); draw((1046.4294535107942,984.438698221053)--(978.0750118233286,1041.1012964658203), red); draw(circle((993.5398909054971,1003.3984720933333), 19.156989626299726), dotted); draw((993.6113585293933,984.2416157770934)--(983.4573969568015,1008.9966124261042), green); pair hyperbolaLeft1 (real t) {return (18.787870262588065*(1+t^2)/(1-t^2),18.787870262588065*2*t/(1-t^2));} pair hyperbolaRight1 (real t) {return (18.787870262588065*(-1-t^2)/(1-t^2),18.787870262588065*(-2)*t/(1-t^2));} draw(shift((992.0859932484519,1022.5002111017292))*rotate(-77.70722080273299)*graph(hyperbolaLeft1,-0.99,0.99), linetype("2 2") + blue); draw(shift((992.0859932484519,1022.5002111017292))*rotate(-77.70722080273299)*graph(hyperbolaRight1,-0.99,0.99), linetype("2 2") + blue); /* hyperbola construction */ draw((988.5343777430974,996.6191141015988)--(993.3926741433031,1042.8525643613414), green); draw((974.4788921062582,1005.3138248852948)--(1013.0946180629153,1021.1531563552602), green); /* dots and labels */ dot((978.0750118233286,1041.1012964658203),linewidth(3.pt) + dotstyle); label("A", (976.7206375238196,1042.4280304735012), N * labelscalefactor); dot((972.3534714152007,984.1622953027858),linewidth(3.pt) + dotstyle); label("B", (972.2981908315454,981.0665826181931), S * labelscalefactor); dot((1046.4294535107942,984.438698221053),linewidth(3.pt) + dotstyle); label("C", (1045.8213670906043,981.6193884547274), S * labelscalefactor); dot((993.6113585293933,984.2416157770934),linewidth(3.pt) + dotstyle); label("D", (993.0284097015808,981.4811869955938), S * labelscalefactor); dot((1005.7657262922834,1018.147004245295),linewidth(3.pt) + dotstyle); label("E", (1006.0193468601364,1019.4865882573275), NE * labelscalefactor); dot((974.4788921062582,1005.3138248852948),linewidth(3.pt) + dotstyle); label("F", (972.2981908315454,1005.5282408848362), NW * labelscalefactor); dot((993.5398909054971,1003.3984720933333),linewidth(3.pt) + dotstyle); label("I", (994.1340213746494,1004.284427752634), N * labelscalefactor); dot((989.0051739983018,1001.0992924544519),linewidth(3.pt) + dotstyle); label("G", (987.0857469588374,1002.4878087838975), N * labelscalefactor); dot((993.3926741433031,1042.8525643613414),linewidth(3.pt) + dotstyle); label("H′", (993.1666111607144,1044.2246494422377), N * labelscalefactor); dot((983.4573969568015,1008.9966124261042),linewidth(3.pt) + dotstyle); label("M", (983.2161061030974,1010.3652919545115), NE * labelscalefactor); dot((988.5343777430974,996.6191141015988),linewidth(3.pt) + dotstyle); label("P", (986.8093440405702,994.7485270724172), S * labelscalefactor); dot((980.8765540564431,1007.9380067633089),linewidth(3.pt) + dotstyle); label("X", (979.2082637882239,1008.9832773631756), NE * labelscalefactor); dot((1013.0946180629153,1021.1531563552602),linewidth(3.pt) + dotstyle); label("Y", (1013.6204271124827,1021.9742145217318), NE * labelscalefactor); dot((990.1223091992708,1011.730414565295),linewidth(3.pt) + dotstyle); label("N", (991.2317907328444,1013.129321137183), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy][/asy] By Pascal's Theorem for CGH′BAI, it follows that F,N,X are collinear. Therefore, X is a common point of EF and CI, so it's well-known (Lemma 8) that ∠BXC=90∘. Similarly, ∠BYC=90∘. Therefore, H′ is the orthocenter of △BIC, as desired.
11.04.2016 09:45
Considering the polar dual of what we need to show that AIB,IAX and the A−midline of ABC are concurrent where X is such that XI⊥IA and X∈BC . Now , we bary it .Take ABC as reference triangle . We only do not know X which turns out as X=(0,k,1−k) where k=ab+bc+ca(a+c)(a+b+c) . The remaining computations are trivial . Hence , done !
12.09.2022 09:47
Let Z be the midpoint of EF and P′=DM∩ZH. From the ceva-menelaus configuration we have −1=(H,I;EF∩DH,D)Z=(P′,∞;M,D)Means P′ is the midpoint of DM. Hence P′≡P.
20.08.2023 18:03
Let X=¯BI∩¯EF and Y=¯CI∩¯EF, and denote the midpoint of ¯EF as N. By Iran Lemma, X and Y are the feet of the C and B altitudes in △BIC, respectively. So, △DXY is the orthic triangle of acute △BHC, hence I is its incenter and H is its D-excenter. Then by the Midpoint of altitudes lemma, we find that P, N, and H are collinear since N is the foot from I onto ¯XY, as desired.
04.09.2023 19:23
Let BI∩CH=K, CI∩BH=L, and let N be the midpoint of EF. Note that I is the antipode of H in (HKL), and K and L are on EF by Iran Lemma. Then since △DEF and △HLK are homothetic, and if M′ is the reflection of M over N, DM′ is the same cevian in △DEF as HN in △HKL (isotomic to altitude), we have PN∥DM′∥HN as desired.
08.09.2023 05:05
Let TB and TC denote the feet from B to CH and the feet from C to BH respectively. Note that BICH is an orthocentric system, so B,I,TB are collinear as well as C,I,TC. By that egmo lemma, BI and EF intersect at some point K for which KB⊥CK. Thus, K=TB, since the intersections of BI and (BC) are B and TB. Now, we have an orthic triangle configuration with D,TB,TC in △HBC. Hence, I is the incenter of △DTBTC. Since TB and TC lie on EF, the foot from I to TBTC is just the midpoint of EF. Hence, in △DTBTC, the midpoint of the D-altitude, P, the intouchpoint to TBTC, which is the midpoint of EF, and the D-excenter, which is H, are collinear, hence PH bisects EF as desired.
26.10.2023 03:09
My sol is crazy long bruh. Just noticed that this actually connects up with 2011 ISL G4. One first considers the following claim. Claim (Partial Iran Lemma) : The feet of the perpendiculars from B to CI and C to BI respectively lie on ¯EF. Let C1=¯EF∩¯CI. Then, note that 2∡EC1I=2∡ECI+2∡C1EC=∡ACB+2∡FEA=∡ACB+∡FAE=∡ACB+∡BAC=∡ABC=2∡FBIThus, ∡EC1I=∡FBI. This means that FBIC1 is cyclic and thus ∡IC1B=90∘ which means that indeed C1 is the foot of the perpendicular from B to CI. Similarly, the foot of the perpendicular from C to BI also lies on EF and we have our claim. Now, we have the following very important claim. Claim (Extended 11SLG4) : Consider a triangle ABC with D,E,F the feet of the perpendiculars from A,B,C to the opposite sides. Let H be its orthocenter and YA be the A−Why Point of △ABC. Let P be the foot of the altitude from D to EF Then, the following points are collinear, 1. A 2. Foot of the perpendicular from H to EF (Z) 3. Midpoint of the altitude from D to EF (M) 4. YA First, consider the inversion centred at A with radius √AD⋅AH. This clearly maps D to H, B to F and C to E (and vice versa). Thus, ¯EF maps to (ABC) and ¯BC maps to (AEF) and thus the A−Ex Point XA maps to the A−Queue Point QA under this inversion. Further, since D and H map to each other, this in fact means that the circle (XAH) remains fixed under this inversion (QA clearly lies on this circle as ∡HQAXA=90∘). Further, Z will map to the intersection of (XAH) and (ABC). The claim is that this intersection is YA. Let A′ be the reflection of A across the perpendicular bisector of BC. Let G be the centroid of △ABC and W the intersection of ¯AM and (ABC). It is well known (from 11SLG4) that YA−D−G−A′ (with A′G=2DG). Simply note that GM⋅GW=AG⋅GW2=A′G⋅GYAGD⋅GYA=GD⋅GYAThus, DMWYA is cyclic. Now, let X′A=¯YAW∩¯BC. Then, X′AD⋅X′AM=X′AYA⋅X′AW=X′AB⋅X′ACThus, X′A≡XA and we have that XA−YA−W. Let HA be the A−Humpty Point. It is well known that XA−H−HA and HHAMD is cyclic. Now, note that from this it follows that, ∡XAYAD=∡WYAD=∡AMD=∡HAMD=∡HAHA=∡XAHDand thus, XAHDYA is cyclic and YA lies on (XAH) as desired. Thus, Z maps to YA in the previously described inversion and we have that A−Z−YA. Now, we deal with the other part of the collinearity. Let N be the midpoint of XAD. We already know that XAYADZ is cyclic. Further, since B,C,HA and YA are cyclic and (XAD;BC)=−1 and we also have, ∡HAYAD=∡HAA′=90∘YA must be the HA−Humpty Point of △XADHA. Further, this implies that HAYA is the HA−median of this triangle which implies HA−YA−N. Thus, ∡NYAD=90∘. Further, MN∥XAP by Midpoint Theorem, which implies that NYADM must indeed be cyclic. Now, ∡DYAM=∡DNM=∡DXAZ=∡DYAZThis clearly implies that YA−M−Z and we put these two together and conclude that indeed A−Z−M−YA are collinear as claimed. By the first claim, EF is the line joining the feet of the altitudes from B and C to the opposite sides. One now notices that by applying the above-proved claim on △BHC, we must have H,M and the foot of the perpendicular from I to EF collinear. But, clearly AI is the perpendicular bisector of EF meaning that, the perpendicular from I to EF passes through the midpoint of EF. Thus, we have that ¯PH bisects ¯EF as was required.
06.02.2024 21:23
Let N be the midpoint of [EF],B′=(BI)∩(EF)and C′=(CI)∩(EF). By the Iran lemma (provable by angle chase), the triangle △DB′C′ with incenter I is the orthic triangle of △HBC. Now since P is the midpoint of the D−altitude in △DB′C′ and H the D−excenter point, the collinearity is well known(4.3 midpoints of Altitudes EGMO). Q.E.D.
13.02.2024 21:53
Let K=BI∩EF and L=CI∩EF. Iran Lemma tells us △DKL is the orthic triangle of △HBC. Hence Midpoint of the Altitude Lemma this triangle with altitude midpoint P, touch point M (midpoint of EF), and excenter K finishes. ◼
05.03.2024 17:10
alright here's the stock solution I guess Let X and Y be the feet from C to ¯BI and from B to ¯CI, which both lie on ¯EF. It follows that DXY is the orthic triangle of HBC, i.e. H is the D-excenter of triangle DXY and I is the incenter. Midpoint of altitudes lemma implies K=¯PH∩¯EF satisfies ∠IKE=90∘, or K is the midpoint of ¯EF.
28.05.2024 05:31
WOAH. Took a "hint" because I heard from somewhere that this is the problem after which the "Iran Lemma" is named. Let X and Y be the feet of the altitudes from B,C to CI,BI, respectively. By the Iran Lemma, we see that X,Y,E,F are collinear. The condition that PH bisects EF is equivalent to PH passing through the foot of the altitude from I to XY. Since DYX is the orthic triangle of △BHC, we see that I is the incenter of △XYD, so the foot from I to XY is the D-intouch point of △DXY. Additionally, it is clear that H is the D-excenter of △DXY. Thus we must show that the midpoint of the D-altitude, the D-excenter, and the D-intouch point are collinear, which is well-known.
12.06.2024 01:10
We use complex bash with the unit circle as the incircle. First, we calculate the circumcenter of (BIC). We know that (AB)(AC)=(AI)(AIA), and by considering angles we can see that (a−b)(a−c)(a−i)=(a−ia). So we have a=2efe+f trivially and cyclic variants. Now we write (a−b)(a−c)(a−i)=4ef(fe+f−dd+e)(ee+f−dd+f)2efe+f=2(ef−de)(ef−df)(e+f)=2ef(d−f)(d−e)(e+f)(d+e)(d+f). Now we find ia=2efe+f(1−(d−e)(d−f)(d+e)(d+f))=2efe+f2de+2df(d+e)(d+f)=4def(d+e)(d+f). The circumcenter of (BIC) is the midpoint of IIa, which is just 2def(d+e)(d+f). Now the orthocenter is just given by b+i+c−2o=2ded+e+2dfd+f−4def(d+e)(d+f)=2d(ed+e+fd+f−2ef(d+e)(d+f))=2d(de+ef+df+ef−2ef(d+e)(d+f))=2d2(e+f)(d+e)(d+f). Now, our strategy will be to show that the midpoint of EF is collinear with P,H. Find the midpoint as e+f2, now find M as 12(d+e+f−efd), and then P is given as 3d+e+f−efd4. Then we desire 2e+2f−4pe+f−2h is real, which is equivalent to showing e+f−3d+efde+f−4d2(e+f)(d+e)(d+f)=(d+e)(d+f)(e+f−3d+efd)((d+e)(d+f)(e+f)−4d2(e+f))=(d+e)(d+f)(e+f−3d+efd)e2f+f2e+e2d+f2d+2def−3d2e−3d2f=(d+e)(d+f)(de+df)which is obviously self conjugating, so we are done.
17.08.2024 10:19
Notice that in ΔBHC, I is the orthocenter due to Iran lemma and it is also the incenter of orthic triangle of ΔBHC. The midpoint of EF is the foot of altitude from I to EF and P is the midpoint of the D altitude of the orthic triangle of BHC and H is the D excenter of this orthic triangle so the desired result follows by midpoint of altitude lemma.
28.10.2024 20:46
Let Q be the midpoint of EF, let S be the foot of the perpendicular from H to EF, and let K be the "Iran lemma point" (the concurrency point of EF, CH, BI). Then −1=(EF∩BC,D;C,B)K=(RD;HI)∞⊥EF=(RM;SQ)H=(D,M;∞DM,HQ∩DM),so HQ bisects MD, as desired.
04.11.2024 21:34
Let N be the midpoint of EF, and note (MD;P∞)N=(MN∩HD,D;HI)=−1.◼
28.12.2024 01:00
Let HB and HC be the feet of the altitudes from B and C to CI and BI, respectively. It's easy to see that FHBIDB and EHCIDC are cyclic. Claim: HB and HC lie on EF. It's clear by ∡HBFI=∡HBBI=90∘−∡BIHB=90∘−∡BIC=−∡BAC2=∡EFIwith HC following analogously. Let N be the intersection of HP with EF. Let HA be the foot of the altitude from H to EF. Let T be the intersection of DH and EF. We will use LinPOP on (BID) and (CID). For any point X let f(X) be Pow(BID)(X)−Pow(CID)(X). LinPOP states that f is linear. It suffices to show that f(N)=f(F)+f(E)2. Since T is on the radical axis of the two circles, f(T)=0. By Menelaus on PNH traversing △MTD, MNTN⋅THDH⋅DPMP=1⟹MNTN=DHTH⟹NTMT=THDH+THTherefore, f(N)=THDH+THf(M)Since ∠DFE=∠CBH and ∠DEF=∠BCH, △DEF∼△HCB∼△HHBHC. Therefore, THDH+TH=d(H,HBHC)2d(H,HBHC)+d(D,EF)=HBHC2HBHC+EFLet MF=u, ME=v, MHB=x, MHC=y. Furthermore, set MD=1. We have 2f(N)=2x+2y2x+2y+u+v⋅(xu−vy)f(F)+f(E)=(u+v)(v+x−u−y)Now, we relate x with v and y with u. Claim: 2x=1v−v; 2y=1u−u. It is simple. We have by orthocenter-incenter duality that △DHBHC∼△ABC implying ∠DHCM=∠C while ∠MDF=90∘−∠EFD=∠C2 so y=cot∠C=1−tan2(∠C2)2tan(∠C2)=1−u22uand the result follows. Therefore, we have 2f(N)=(1v−u+1u−v1u+1v)(xu−vy)=12(u+v−uv(u+v)u+v)(uv−vu)=12(u+v−uv(u+v)u+v)((u−v)(u+v)uv)=12((u+v−uv(u+v))(u−v)uv)=(u+v)((1−uv)(u−v)2uv)=(u+v)(u+uv22uv−v+u2v2uv)=(u+v)(1+v22v−1+u22u)=(u+v)(1−v22v+v−1−u22u−u)=(u+v)(x+v−y−u)and we are done.