Let a triangle $ABC$ be given. On a ruler three segment congruent to the sides of this triangle are marked. Using this ruler construct the orthocenter of the triangle formed by the tangency points of the sides of $ABC$ with its incircle.
Problem
Source: Sharygin 2018
Tags: geometry
Fumiko
04.04.2018 22:46
Using this ruler we can of course construct the incenters and excenters now the homothety center of intouch and excentral triangle lies on the euler line of intouch triangle so with this we can construct the euler line of intouch triangle ,Now homothety at $A$ with ratio $2$ let $B',C'$ be the images of $B,C$ now of course using this ruler we can construct the feet of $A-$nagel ,gergonne cevians of $\Delta AB'C'$ let it be $X,Y$ ,now by homothety $AX\cap BC,AY\cap BC$ are the feet of $A-$ nagel and gergonne cevians of $\Delta ABC$,similarly doing this we can construct the nagel and gergonne points but it is well known that orthocenter of intouch triangle lies in the line connecting nagel and gergonne point so the intersection of the euler line and the above mentioned line gives the orthocenter of intouch triangle $\blacksquare$
TheDarkPrince
04.04.2018 22:54
Detailed solution in #10
anantmudgal09
04.04.2018 23:17
TheDarkPrince wrote: Construct the contact triangle $DEF$ using the rulers. How do we do that? Anyways, my over-complicated solution: TheDarkPrince wrote: Let a triangle $ABC$ be given. On a ruler three segment congruent to the sides of this triangle are marked. Using this ruler construct the orthocenter of the triangle formed by the tangency points of the sides of $ABC$ with its incircle. By lengths, we can construct reflections $B_A, C_A$ of $B$ and $C$ in the angle-bisector of $BAC$. Intersect with $BC$ to get the feet of internal bisectors. Thus, we get $I$ the incenter of $ABC$. Again, by lengths we can dilate $ABC$ at $A$ by factor $2$ and compute the centroid $G_A$ of the new triangle using Cevians from $B$ and $C$. Thus, we get the $A$-median and consequently, the centroid $G$ of triangle $ABC$ and its medial triangle $MNP$. Now intersect $BB_A$ with $AI$. Since we recognise the midpoint of $BB_A$, we can draw a line through $A$ parallel to it. This line is precisely the $A$-external bisector. Thus, we obtain the excentral triangle $I_AI_BI_C$. Now we recognise a line through $B$ parallel to $I_BI_C$ hence we can locate the midpoint $M_a$ of $I_BI_C$ and join it with $M$. Do the same for $N$ and $P$ to obtain the circumcenter $O$ of $\triangle ABC$. Now extend $AI$ to meet this newfound perpendicular bisector of $BC$ at $M'$; define $N', P'$ similarly. Then since we recognise a midpoint $O$ on $M_aM'$, we can draw from $I$ a line perpendicular to $BC$. Thus, we obtain the contact triangle $DEF$. Now we recognise the midpoint $M'$ of $II_A$ hence we can draw from $D$ a line parallel to $AI$; thus perpendicular to $EF$. Consequently, we obtain the orthocenter $H$ of $\triangle DEF$, as required. $\blacksquare$ @below, thanks!
Vrangr
04.04.2018 23:22
anantmudgal09 wrote: TheDarkPrince wrote: Construct the contact triangle $DEF$ using the rulers. How do we do that? That's all I was able to do. Essentially, extend $AB$ to $B'$ such that $AB=BB'$, similarly construct $C'$ on $AC$. $ABC$ is homothetic to $AB'C'$ with ratio $\tfrac12$. Now construct $D'$ on segment $B'C'$, such that length of $B'D'$ is equal to $AB+BC-CA$. Join $AD'$. $D=AD'\cap BC$, similarly construct $E$ and $F$.
suli
05.04.2018 01:00
You can do much more than the problem construction, and you can do it with just one (arbitrary) segment given on the ruler.
[asy][asy]
import graph; size(8.43584207549068cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-3.4684337121794235,xmax=4.9674083633112565,ymin=2.1253189964017882,ymax=9.004893654517447;
pair A=(-2.06,2.6), B=(0.54,2.64), C=(3.14,2.68), D=(1.400805749227354,8.347440739722495), F=(0.9151683046368169,5.127496009475705), G=(0.04099010016542474,6.089163209504114);
draw((xmin,0.015384615384615398*xmin+2.6316923076923078)--(xmax,0.015384615384615398*xmax+2.6316923076923078),linewidth(0.8)); draw((xmin,1.6607232986149676*xmin+6.021089995146833)--(xmax,1.6607232986149676*xmax+6.021089995146833),linewidth(0.8)); draw((xmin,-3.258658851479476*xmin+12.912188793645553)--(xmax,-3.258658851479476*xmax+12.912188793645553),linewidth(0.8)); draw((xmin,6.63034691025868*xmin-0.940387331539688)--(xmax,6.63034691025868*xmax-0.940387331539688),linewidth(0.8)); draw((xmin,0.84953043010595*xmin+4.350032686018257)--(xmax,0.84953043010595*xmax+4.350032686018257),linewidth(0.8)); draw((xmin,-1.1000814194514499*xmin+6.134255657077553)--(xmax,-1.1000814194514499*xmax+6.134255657077553),linewidth(0.8)); draw((xmin,0.015384615384615235*xmin+6.088532592578492)--(xmax,0.015384615384615235*xmax+6.088532592578492),linewidth(0.8));
dot(A,ds); label("$A$",(-1.9994336266198396,2.7507348744123026),NE*lsf); dot(B,ds); label("$B$",(0.6040417725402151,2.779823985017443),NE*lsf); dot(C,linewidth(3.pt)+ds); label("$C$",(3.1929726163976997,2.7652794297148726),NE*lsf); dot(D,ds); label("$D$",(1.4621705353918533,8.495834218927493),NE*lsf); dot((2.0841678707469393,6.120596713666906),ds); label("$E$",(2.1457646346126498,6.2705172576342685),NE*lsf); dot(F,linewidth(3.pt)+ds); label("$F$",(0.9676556551044685,5.20876472054665),NE*lsf); dot(G,linewidth(3.pt)+ds); label("$G$",(0.09498233695026026,6.183249925818847),NE*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy][/asy]
Let $a = BC$ be the length of a marked segment on the ruler. For the following lemmas, we will disregard triangle $ABC$ and re-use letters $A, B, C$ for convenience.
Lemma 1. Given a line $\ell$ and a point $E$ not on $\ell$, we can construct the line through $E$ parallel to $\ell$.
Proof. Choose $A$ on $\ell$. Use the segment of length $a$ to construct points $B, C$ on $\ell$ such that $A, B, C$ are on $\ell$ on that order and $AB = BC = a$. Choose an arbitrary point $D$ on ray $CE$ past $E$. Let $BD$ intersect $AE$ at $F$, and construct $G$ as the intersection of $AD$ and $CF$. We claim that $EG \| AC$. Indeed, Ceva's theorem gives that
$$\frac{DG}{GA} \cdot \frac{AB}{BC} \cdot \frac{CE}{ED} = 1.$$Since $AB = BC$, we can simplify to
$$\frac{DG}{GA} = \frac{DE}{EC}.$$This implies $EG \| AC$, so $EG$ is our desired line.
[asy][asy]
import graph; size(11.6cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-6.4,xmax=5.2,ymin=-2.46,ymax=7.;
pair A=(-2.5,1.28), B=(-0.32,4.32), C=(2.7,1.22), D=(3.420606520539795,4.276839155532233);
draw((xmin,1.3944954128440366*xmin+4.766238532110092)--(xmax,1.3944954128440366*xmax+4.766238532110092),linewidth(0.8)); draw((xmin,-0.011538461538461548*xmin+1.251153846153846)--(xmax,-0.011538461538461548*xmax+1.251153846153846),linewidth(0.8)); draw((xmin,-0.011538461538461548*xmin+4.316307692307692)--(xmax,-0.011538461538461548*xmax+4.316307692307692),linewidth(0.8)); draw((xmin,0.5061709716961572*xmin+2.545427429240393)--(xmax,0.5061709716961572*xmax+2.545427429240393),linewidth(0.8)); label("construct angle bisectors of two lines",(-0.9,0.64),SE*lsf); draw((xmin,-1.9756170462502858*xmin-3.659042615625715)--(xmax,-1.9756170462502858*xmax-3.659042615625715),linewidth(0.8));
dot(A,ds); label("$A$",(-2.42,1.48),NE*lsf); dot(B,ds); label("$B$",(-0.24,4.52),NE*lsf); dot(C,ds); label("$C$",(2.78,1.42),NE*lsf); dot(D,linewidth(3.pt)+ds); label("$D$",(3.5,4.4),NE*lsf); dot((-4.060606520539795,4.3631608444677665),linewidth(3.pt)+ds); label("$E$",(-3.98,4.48),NE*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy][/asy]
Lemma 2. Given two non-parallel lines $\ell_1, \ell_2$, we can construct the two angle bisectors of $\ell_1, \ell_2$.
Proof. Let $A$ be the intersection of $\ell_1, \ell_2$, and choose $B$ on $\ell_1$ such that $AB = a$. We can construct $p$, the parallel to $\ell_2$ through $B$. Let $D, E$ be distinct points on $p$ such that $BD = BE = a$. Choose $C$ on $\ell_2$ such that angle $\angle BAC$ contains $D$; then $BAD$ being isosceles and $BD \| AC$ imply $\angle BAD = \angle BDA = \angle CAD$. Thus, $AD$ bisects $\angle BAC$, so $AD$ is an angle bisector. Similarly, $AE$ is the other angle bisector of $\ell_1, \ell_2$.
[asy][asy]
import graph; size(13cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-7.1399173333333374,xmax=10.652383619047628,ymin=-3.499480190476199,ymax=11.010448;
pair A=(-1.86,1.16), B=(0.82,1.2), C=(0.375796000000001,5.24328838095238), D=(3.5,1.24), F=(3.022291011340267,2.72771538624505), G=(1.0511688083985957,8.866324663229637), H=(1.1517545254355577,2.1270816217532955);
draw((xmin,0.01492537313432837*xmin+1.1877611940298507)--(xmax,0.01492537313432837*xmax+1.1877611940298507),linewidth(0.8)); draw((xmin,-0.9081632653061223*xmin+1.9446938775510203)--(xmax,-0.9081632653061223*xmax+1.9446938775510203),linewidth(0.8)); draw((xmin,0.6936936936936936*xmin+0.631171171171171)--(xmax,0.6936936936936936*xmax+0.631171171171171),linewidth(0.8)); draw((xmin,2.6471582963506695*xmin+6.083714431212245)--(xmax,2.6471582963506695*xmax+6.083714431212245),linewidth(0.8)); draw((xmin,-3.1142712855770314*xmin+12.139949499519608)--(xmax,-3.1142712855770314*xmax+12.139949499519608),linewidth(0.8)); draw((xmin,-0.37776358194316684*xmin+2.562172536801084)--(xmax,-0.37776358194316684*xmax+2.562172536801084),linewidth(0.8)); draw((xmin,0.3211024051216249*xmin+1.7572504735262222)--(xmax,0.3211024051216249*xmax+1.7572504735262222),linewidth(0.8)); draw((xmin,-66.99999999999892*xmin+79.29463482593441)--(xmax,-66.99999999999892*xmax+79.29463482593441),linewidth(0.8)); draw((xmin,-66.99999999999892*xmin+30.42162038095204)--(xmax,-66.99999999999892*xmax+30.42162038095204),linewidth(0.8)); label("construct perpendicular from $C$ to $AB$",(1.9709678095238117,8.280250095238094),SE*lsf);
dot(A,ds); label("$A$",(-1.740874285714286,1.470093523809517),NE*lsf); dot(B,ds); label("$B$",(0.9279708571428585,1.5007699047618979),NE*lsf); dot(C,ds); label("$C$",(0.4985015238095249,5.550052190476186),NE*lsf); dot(D,linewidth(3.pt)+ds); label("$D$",(3.627492380952384,1.408740761904755),NE*lsf); dot((-1.1641761460621378,3.0019558877503085),ds); label("$E$",(-1.0353175238095238,3.3106763809523754),NE*lsf); dot(F,ds); label("$F$",(3.1366702857142887,3.0345889523809464),NE*lsf); dot(G,linewidth(3.pt)+ds); label("$G$",(1.1733819047619063,9.047159619047617),NE*lsf); dot(H,linewidth(3.pt)+ds); label("$H$",(1.2654110476190492,2.298355809523803),NE*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy][/asy]
Lemma 3. Given a point $C$ and line $\ell$, we can construct a line perpendicular to $\ell$ through $C$.
Proof. Choose $A, B, D$ on $\ell$ on that order such that $AB = BD = a$. Construct an arbitrary line $p$ (not equal to $\ell$) through $B$ and let $E$ be on that line such that $BE = a$. Construct another arbitrary line (not $\ell$ or $p$) through $B$ and let $F$ be on that line such that $BF = a$, such that $E$ and $F$ are on the same side of $AD$. Let $AE$ and $DF$ intersect at $G$, and $AF$ and $DE$ intersect at $H$. Since $A, E, F, D$ lie on the circle of radius $a$ centered at $B$, we get $AD$ is a diameter and $\angle AED = \angle AFD = 90^\circ$. Thus, $H$ is the orthocenter of $AGD$, so $GH \perp AB$. Now it suffices to construct a parallel line to $GH$ through $C$ (by Lemma 2), and that parallel line will be perpendicular to $\ell$.
[asy][asy]
import graph; size(11.6cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-4.3,xmax=7.3,ymin=-3.16,ymax=6.3;
pen zzttqq=rgb(0.6,0.2,0.);
pair A=(-0.56,4.56), B=(-2.96,-0.22), C=(5.22,0.), D=(0.12360457548166057,-0.13706686960807268), F=(-1.5758595788389178,2.5367463388124882), I=(0.07397155040430481,1.7083792446317907);
draw(A--B--C--cycle,linewidth(0.8)+zzttqq); draw(D--(1.2174173106087598,3.157746896820771)--F--cycle,linewidth(0.8)+zzttqq);
draw(A--B,linewidth(0.8)+zzttqq); draw(B--C,linewidth(0.8)+zzttqq); draw(C--A,linewidth(0.8)+zzttqq); draw(circle(I,1.8461134303561462),linewidth(0.8)); draw(D--(1.2174173106087598,3.157746896820771),linewidth(0.8)+zzttqq); draw((1.2174173106087598,3.157746896820771)--F,linewidth(0.8)+zzttqq); draw(F--D,linewidth(0.8)+zzttqq); draw((xmin,-4.4980263760252885*xmin+0.4189097711058452)--(xmax,-4.4980263760252885*xmax+0.4189097711058452),linewidth(0.8)); draw((xmin,0.6355956911905837*xmin+2.3839616998170143)--(xmax,0.6355956911905837*xmax+2.3839616998170143),linewidth(0.8)); draw((xmin,0.6355956911905837*xmin+1.6613632459241277)--(xmax,0.6355956911905837*xmax+1.6613632459241277),linewidth(0.8)); draw((xmin,-4.4980263760252885*xmin+2.041105229425838)--(xmax,-4.4980263760252885*xmax+2.041105229425838),linewidth(0.8)); draw((xmin,-0.502092050209205*xmin+1.7455197720314417)--(xmax,-0.502092050209205*xmax+1.7455197720314417),linewidth(0.8)); draw((xmin,-37.181818181818166*xmin+4.45877598239185)--(xmax,-37.181818181818166*xmax+4.45877598239185),linewidth(0.8)); draw((xmin,1.2675438596491229*xmin+1.6146170601280885)--(xmax,1.2675438596491229*xmax+1.6146170601280885),linewidth(0.8));
dot(A,ds); label("$A$",(-0.48,4.76),NE*lsf); dot(B,ds); label("$B$",(-2.88,-0.02),NE*lsf); dot(C,ds); label("$C$",(5.3,0.2),NE*lsf); dot(D,linewidth(3.pt)+ds); label("$D$",(0.2,-0.02),NE*lsf); dot((1.2174173106087598,3.157746896820771),linewidth(3.pt)+ds); label("$E$",(1.3,3.28),NE*lsf); dot(F,linewidth(3.pt)+ds); label("$F$",(-1.5,2.66),NE*lsf); dot(I,linewidth(3.pt)+ds); label("$I$",(0.16,1.82),NE*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy][/asy]
Now, we will use the above Lemmas to construct the desired point. Construct the angle bisectors of $\angle A, \angle B$ which intersect at the incenter $I$. Construct perpendiculars from $I$ to $AB, AC, BC$, intersecting them at $F, E, D$ respectively. Finally, construct the perpendiculars from $D$ to $EF$, and from $E$ to $DF$, intersecting them at the orthocenter of $DEF$, which is what we wanted.
enhanced
05.04.2018 02:00
suli wrote: You can do much more than the problem construction, and you can do it with just one (arbitrary) segment given on the ruler.
[asy][asy]
import graph; size(8.43584207549068cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-3.4684337121794235,xmax=4.9674083633112565,ymin=2.1253189964017882,ymax=9.004893654517447;
pair A=(-2.06,2.6), B=(0.54,2.64), C=(3.14,2.68), D=(1.400805749227354,8.347440739722495), F=(0.9151683046368169,5.127496009475705), G=(0.04099010016542474,6.089163209504114);
draw((xmin,0.015384615384615398*xmin+2.6316923076923078)--(xmax,0.015384615384615398*xmax+2.6316923076923078),linewidth(0.8)); draw((xmin,1.6607232986149676*xmin+6.021089995146833)--(xmax,1.6607232986149676*xmax+6.021089995146833),linewidth(0.8)); draw((xmin,-3.258658851479476*xmin+12.912188793645553)--(xmax,-3.258658851479476*xmax+12.912188793645553),linewidth(0.8)); draw((xmin,6.63034691025868*xmin-0.940387331539688)--(xmax,6.63034691025868*xmax-0.940387331539688),linewidth(0.8)); draw((xmin,0.84953043010595*xmin+4.350032686018257)--(xmax,0.84953043010595*xmax+4.350032686018257),linewidth(0.8)); draw((xmin,-1.1000814194514499*xmin+6.134255657077553)--(xmax,-1.1000814194514499*xmax+6.134255657077553),linewidth(0.8)); draw((xmin,0.015384615384615235*xmin+6.088532592578492)--(xmax,0.015384615384615235*xmax+6.088532592578492),linewidth(0.8));
dot(A,ds); label("$A$",(-1.9994336266198396,2.7507348744123026),NE*lsf); dot(B,ds); label("$B$",(0.6040417725402151,2.779823985017443),NE*lsf); dot(C,linewidth(3.pt)+ds); label("$C$",(3.1929726163976997,2.7652794297148726),NE*lsf); dot(D,ds); label("$D$",(1.4621705353918533,8.495834218927493),NE*lsf); dot((2.0841678707469393,6.120596713666906),ds); label("$E$",(2.1457646346126498,6.2705172576342685),NE*lsf); dot(F,linewidth(3.pt)+ds); label("$F$",(0.9676556551044685,5.20876472054665),NE*lsf); dot(G,linewidth(3.pt)+ds); label("$G$",(0.09498233695026026,6.183249925818847),NE*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy][/asy]
Let $a = BC$ be the length of a marked segment on the ruler. For the following lemmas, we will disregard triangle $ABC$ and re-use letters $A, B, C$ for convenience.
Lemma 1. Given a line $\ell$ and a point $E$ not on $\ell$, we can construct the line through $E$ parallel to $\ell$.
Proof. Choose $A$ on $\ell$. Use the segment of length $a$ to construct points $B, C$ on $\ell$ such that $A, B, C$ are on $\ell$ on that order and $AB = BC = a$. Choose an arbitrary point $D$ on ray $CE$ past $E$. Let $BD$ intersect $AE$ at $F$, and construct $G$ as the intersection of $AD$ and $CF$. We claim that $EG \| AC$. Indeed, Ceva's theorem gives that
$$\frac{DG}{GA} \cdot \frac{AB}{BC} \cdot \frac{CE}{ED} = 1.$$Since $AB = BC$, we can simplify to
$$\frac{DG}{GA} = \frac{DE}{EC}.$$This implies $EG \| AC$, so $EG$ is our desired line.
[asy][asy]
import graph; size(11.6cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-6.4,xmax=5.2,ymin=-2.46,ymax=7.;
pair A=(-2.5,1.28), B=(-0.32,4.32), C=(2.7,1.22), D=(3.420606520539795,4.276839155532233);
draw((xmin,1.3944954128440366*xmin+4.766238532110092)--(xmax,1.3944954128440366*xmax+4.766238532110092),linewidth(0.8)); draw((xmin,-0.011538461538461548*xmin+1.251153846153846)--(xmax,-0.011538461538461548*xmax+1.251153846153846),linewidth(0.8)); draw((xmin,-0.011538461538461548*xmin+4.316307692307692)--(xmax,-0.011538461538461548*xmax+4.316307692307692),linewidth(0.8)); draw((xmin,0.5061709716961572*xmin+2.545427429240393)--(xmax,0.5061709716961572*xmax+2.545427429240393),linewidth(0.8)); label("construct angle bisectors of two lines",(-0.9,0.64),SE*lsf); draw((xmin,-1.9756170462502858*xmin-3.659042615625715)--(xmax,-1.9756170462502858*xmax-3.659042615625715),linewidth(0.8));
dot(A,ds); label("$A$",(-2.42,1.48),NE*lsf); dot(B,ds); label("$B$",(-0.24,4.52),NE*lsf); dot(C,ds); label("$C$",(2.78,1.42),NE*lsf); dot(D,linewidth(3.pt)+ds); label("$D$",(3.5,4.4),NE*lsf); dot((-4.060606520539795,4.3631608444677665),linewidth(3.pt)+ds); label("$E$",(-3.98,4.48),NE*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy][/asy]
Lemma 2. Given two non-parallel lines $\ell_1, \ell_2$, we can construct the two angle bisectors of $\ell_1, \ell_2$.
Proof. Let $A$ be the intersection of $\ell_1, \ell_2$, and choose $B$ on $\ell_1$ such that $AB = a$. We can construct $p$, the parallel to $\ell_2$ through $B$. Let $D, E$ be distinct points on $p$ such that $BD = BE = a$. Choose $C$ on $\ell_2$ such that angle $\angle BAC$ contains $D$; then $BAD$ being isosceles and $BD \| AC$ imply $\angle BAD = \angle BDA = \angle CAD$. Thus, $AD$ bisects $\angle BAC$, so $AD$ is an angle bisector. Similarly, $AE$ is the other angle bisector of $\ell_1, \ell_2$.
[asy][asy]
import graph; size(13cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-7.1399173333333374,xmax=10.652383619047628,ymin=-3.499480190476199,ymax=11.010448;
pair A=(-1.86,1.16), B=(0.82,1.2), C=(0.375796000000001,5.24328838095238), D=(3.5,1.24), F=(3.022291011340267,2.72771538624505), G=(1.0511688083985957,8.866324663229637), H=(1.1517545254355577,2.1270816217532955);
draw((xmin,0.01492537313432837*xmin+1.1877611940298507)--(xmax,0.01492537313432837*xmax+1.1877611940298507),linewidth(0.8)); draw((xmin,-0.9081632653061223*xmin+1.9446938775510203)--(xmax,-0.9081632653061223*xmax+1.9446938775510203),linewidth(0.8)); draw((xmin,0.6936936936936936*xmin+0.631171171171171)--(xmax,0.6936936936936936*xmax+0.631171171171171),linewidth(0.8)); draw((xmin,2.6471582963506695*xmin+6.083714431212245)--(xmax,2.6471582963506695*xmax+6.083714431212245),linewidth(0.8)); draw((xmin,-3.1142712855770314*xmin+12.139949499519608)--(xmax,-3.1142712855770314*xmax+12.139949499519608),linewidth(0.8)); draw((xmin,-0.37776358194316684*xmin+2.562172536801084)--(xmax,-0.37776358194316684*xmax+2.562172536801084),linewidth(0.8)); draw((xmin,0.3211024051216249*xmin+1.7572504735262222)--(xmax,0.3211024051216249*xmax+1.7572504735262222),linewidth(0.8)); draw((xmin,-66.99999999999892*xmin+79.29463482593441)--(xmax,-66.99999999999892*xmax+79.29463482593441),linewidth(0.8)); draw((xmin,-66.99999999999892*xmin+30.42162038095204)--(xmax,-66.99999999999892*xmax+30.42162038095204),linewidth(0.8)); label("construct perpendicular from $C$ to $AB$",(1.9709678095238117,8.280250095238094),SE*lsf);
dot(A,ds); label("$A$",(-1.740874285714286,1.470093523809517),NE*lsf); dot(B,ds); label("$B$",(0.9279708571428585,1.5007699047618979),NE*lsf); dot(C,ds); label("$C$",(0.4985015238095249,5.550052190476186),NE*lsf); dot(D,linewidth(3.pt)+ds); label("$D$",(3.627492380952384,1.408740761904755),NE*lsf); dot((-1.1641761460621378,3.0019558877503085),ds); label("$E$",(-1.0353175238095238,3.3106763809523754),NE*lsf); dot(F,ds); label("$F$",(3.1366702857142887,3.0345889523809464),NE*lsf); dot(G,linewidth(3.pt)+ds); label("$G$",(1.1733819047619063,9.047159619047617),NE*lsf); dot(H,linewidth(3.pt)+ds); label("$H$",(1.2654110476190492,2.298355809523803),NE*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
[/asy][/asy]
Lemma 3. Given a point $C$ and line $\ell$, we can construct a line perpendicular to $\ell$ through $C$.
Proof. Choose $A, B, D$ on $\ell$ on that order such that $AB = BD = a$. Construct an arbitrary line $p$ (not equal to $\ell$) through $B$ and let $E$ be on that line such that $BE = a$. Construct another arbitrary line (not $\ell$ or $p$) through $B$ and let $F$ be on that line such that $BF = a$, such that $E$ and $F$ are on the same side of $AD$. Let $AE$ and $DF$ intersect at $G$, and $AF$ and $DE$ intersect at $H$. Since $A, E, F, D$ lie on the circle of radius $a$ centered at $B$, we get $AD$ is a diameter and $\angle AED = \angle AFD = 90^\circ$. Thus, $H$ is the orthocenter of $AGD$, so $GH \perp AB$. Now it suffices to construct a parallel line to $GH$ through $C$ (by Lemma 2), and that parallel line will be perpendicular to $\ell$.
[asy][asy]
import graph; size(11.6cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(10); defaultpen(dps); pen ds=black; real xmin=-4.3,xmax=7.3,ymin=-3.16,ymax=6.3;
pen zzttqq=rgb(0.6,0.2,0.);
pair A=(-0.56,4.56), B=(-2.96,-0.22), C=(5.22,0.), D=(0.12360457548166057,-0.13706686960807268), F=(-1.5758595788389178,2.5367463388124882), I=(0.07397155040430481,1.7083792446317907);
draw(A--B--C--cycle,linewidth(0.8)+zzttqq); draw(D--(1.2174173106087598,3.157746896820771)--F--cycle,linewidth(0.8)+zzttqq);
draw(A--B,linewidth(0.8)+zzttqq); draw(B--C,linewidth(0.8)+zzttqq); draw(C--A,linewidth(0.8)+zzttqq); draw(circle(I,1.8461134303561462),linewidth(0.8)); draw(D--(1.2174173106087598,3.157746896820771),linewidth(0.8)+zzttqq); draw((1.2174173106087598,3.157746896820771)--F,linewidth(0.8)+zzttqq); draw(F--D,linewidth(0.8)+zzttqq); draw((xmin,-4.4980263760252885*xmin+0.4189097711058452)--(xmax,-4.4980263760252885*xmax+0.4189097711058452),linewidth(0.8)); draw((xmin,0.6355956911905837*xmin+2.3839616998170143)--(xmax,0.6355956911905837*xmax+2.3839616998170143),linewidth(0.8)); draw((xmin,0.6355956911905837*xmin+1.6613632459241277)--(xmax,0.6355956911905837*xmax+1.6613632459241277),linewidth(0.8)); draw((xmin,-4.4980263760252885*xmin+2.041105229425838)--(xmax,-4.4980263760252885*xmax+2.041105229425838),linewidth(0.8)); draw((xmin,-0.502092050209205*xmin+1.7455197720314417)--(xmax,-0.502092050209205*xmax+1.7455197720314417),linewidth(0.8)); draw((xmin,-37.181818181818166*xmin+4.45877598239185)--(xmax,-37.181818181818166*xmax+4.45877598239185),linewidth(0.8)); draw((xmin,1.2675438596491229*xmin+1.6146170601280885)--(xmax,1.2675438596491229*xmax+1.6146170601280885),linewidth(0.8));
dot(A,ds); label("$A$",(-0.48,4.76),NE*lsf); dot(B,ds); label("$B$",(-2.88,-0.02),NE*lsf); dot(C,ds); label("$C$",(5.3,0.2),NE*lsf); dot(D,linewidth(3.pt)+ds); label("$D$",(0.2,-0.02),NE*lsf); dot((1.2174173106087598,3.157746896820771),linewidth(3.pt)+ds); label("$E$",(1.3,3.28),NE*lsf); dot(F,linewidth(3.pt)+ds); label("$F$",(-1.5,2.66),NE*lsf); dot(I,linewidth(3.pt)+ds); label("$I$",(0.16,1.82),NE*lsf);
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle);
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Now, we will use the above Lemmas to construct the desired point. Construct the angle bisectors of $\angle A, \angle B$ which intersect at the incenter $I$. Construct perpendiculars from $I$ to $AB, AC, BC$, intersecting them at $F, E, D$ respectively. Finally, construct the perpendiculars from $D$ to $EF$, and from $E$ to $DF$, intersecting them at the orthocenter of $DEF$, which is what we wanted.
This is actual overkill!!
Drunken_Master
05.04.2018 07:58
Construct contact triangle $DEF$ and incenter $I$. Then let $AI \cap EF=P$ and define $Q,R$ similarly. Let $G$ be intersection of $DP,EQ,FR$. Now, $I$ is circumcenter of $DEF$, $G$ is centroid, and its well known that $G$ divides line joining circumcenter and orthocenter is $2:1$ ratio. So, construct $H$ on $GI$ such that $I,G,H$ lie in that order and $IH=2IG$. $H$ is the desired orthocenter.
Vrangr
05.04.2018 08:03
Drunken_Master wrote: Construct contact triangle $DEF$ and incenter $I$. How do you construct $I$?
TheDarkPrince
05.04.2018 08:21
Construct $B'$ and $C'$ on rays $AB$ and $AC$ respectively such that $AB=BB'$ and $AC=CC'$. Let $M'$ be the midpoint of $B'C'$ which we can construct because $M'B'=BC$. Also let $D'$ be point on $B'C'$ such that $B'D'=AB+BC-AC$. Now let $D=BC\cap AD'$ and $M=BC\cap AM'$. From midpoint theorem, $M$ is the midpoint of $BC$ and $D$ is the touch point of incircle of $ABC$ with $BC$. Construct $E$ and $F$ similarly. Let $AM\cap EF=X$ and $I$ be the incenter of $DEF$. We know that, $X,I,D$ are collinear. Join $XD$. Similarly, join $BY$ and $CZ$ where $Y,Z$ are defined similarly as $X$. The concurrency point of these three line segments give $I$. We know that $IA$ bisects $EF$ perpendicularly. So, we also have the midpoints of $EF,FD$ and $DE$. Let $P,Q,R$ be the midpoints of $EF,FD,DE$ respectively. Let $G$ be the centroid of $DEF$. We also have that $I$ is the orthocenter of $PQR$. So we construct $H_a'$ which is the foot of $P$ on $QR$. Now consider the homothety centered $G$ mapping $DEF$ to $PQR$. Clearly, this homothety will take the foot of $D$ on $EF$ to $H_a'$. Extend $H_a'G$ to meet $EF$ at $H_a$. So, $H_a$ is the foot of perpendicular from $D$ on $EF$. Thus we constructed the orthocenter of $DEF$.
Drunken_Master
05.04.2018 08:27
Vrangr wrote: Drunken_Master wrote: Construct contact triangle $DEF$ and incenter $I$. How do you construct $I$? Angle bisector theorem, if $X$ is $AI \cap BC$ then $BX=BC \left(\frac{AB}{AB+AC} \right)$.
Vrangr
05.04.2018 08:41
Drunken_Master wrote: Angle bisector theorem, if $X$ is $AI \cap BC$ then $BX=BC \left(\frac{AB}{AB+AC} \right)$. How do you contruct this ratio and how do you multiply it, afaik, you need a compass for both. @below, if that's the case, I would've been easily able to complete it.
Drunken_Master
05.04.2018 08:47
Vrangr wrote: Drunken_Master wrote: Angle bisector theorem, if $X$ is $AI \cap BC$ then $BX=BC \left(\frac{AB}{AB+AC} \right)$. How do you contruct this ratio and how do you multiply it, afaik, you need a compass for both. I asked them, they said you can construct any length as long as it is related to sides of the triangle.
fastlikearabbit
05.04.2018 09:19
Drunken_Master wrote: Vrangr wrote: Drunken_Master wrote: Angle bisector theorem, if $X$ is $AI \cap BC$ then $BX=BC \left(\frac{AB}{AB+AC} \right)$. How do you contruct this ratio and how do you multiply it, afaik, you need a compass for both. I asked them, they said you can construct any length as long as it is related to sides of the triangle. You can construct any linear combination with those lengths. (I mean, $xa+yb+zc$, where $x,y,z$ are integers)
Vrangr
05.04.2018 09:45
fastlikearabbit wrote: Drunken_Master wrote: You can construct any linear combination with those lengths. (I mean, $xa+yb+zc$, where $x,y,z$ are integers) We can actually construct it where $x, y, z\in\mathbb{Q}$, by taking various homotheties. However, this doesn't help in costructing $BX$.
Drunken_Master
05.04.2018 10:28
Okay, I lose marks. An alternative is to construct midpoints of $EF,FD,DE$ and let them meet at $I$.
TheDarkPrince
05.04.2018 10:35
How do you construct the midpoints of $EF,FD,DE$?
Drunken_Master
05.04.2018 10:36
TheDarkPrince wrote: How do you construct the midpoints of $EF,FD,DE$? Cant we construct midpoints using ruler? I always thought we can....
fastlikearabbit
05.04.2018 11:23
Drunken_Master wrote: TheDarkPrince wrote: How do you construct the midpoints of $EF,FD,DE$? Cant we construct midpoints using ruler? I always thought we can.... We can construct a parallel to AC through B. So we can construct the midpoint of AE. Similarly, we can construct the midpoint of AF. So we have a line parallel to EF, which means that we can construct the midpoint of EF
Vrangr
05.04.2018 11:28
fastlikearabbit wrote: We can construct a parallel to AC through B. How do we construct a parallel without compass?
fastlikearabbit
05.04.2018 11:35
Vrangr wrote: fastlikearabbit wrote: We can construct a parallel to AC through B. How do we construct a parallel without compass? Take on ray AC a point C’ such that CC’=AC. So C is the midpoint of AC’. Now we can construct a parallel to BC that passes through A
nikolapavlovic
05.04.2018 11:40
hint 1Construct incenter via symmetry,more precisely bunch of A-isosceles triangles
hint 2Remember that lemma where $IO$ is perpendicular to some line connecting points on the sidelengths,use it.
hint 3We want the orthocenter now.Recall that given a triangle and a midpoint of some side you can construct parallels to that side using only straight edge.Now to construct isogonal just use symmetry over $AI,BI,CI$ again
hint 4Obviously midpoints of the sides would be nice.So we want parallel to AH from O.So again we need a midpoint of some segment that contains $AH$
hint 5After the midpoint we basically have the midpoint of any cevian which means we can go cray with the parallels.So construct a tangential triangle using orthocenter and incenter.
hint 6You're basically done.Construct parallel from $D$ where $DI\perp BC$ and $D\in BC$ to $AI$.
Vrangr
05.04.2018 11:42
fastlikearabbit wrote: Take on ray AC a point C’ such that CC’=AC. So C is the midpoint of AC’. Now we can construct a parallel to BC that passes through A That is definitely possible, my question is how to construct midpoint of $EF$ by this method?
fastlikearabbit
05.04.2018 11:43
Vrangr wrote: fastlikearabbit wrote: Take on ray AC a point C’ such that CC’=AC. So C is the midpoint of AC’. Now we can construct a parallel to BC that passes through A That is definitely possible, my question is how to construct midpoint of $EF$ by this method? Look at #6 or read carefully the solution of problem 16 From https://arxiv.org/pdf/1110.1556.pdf
a.h.rajabi
05.04.2018 20:13
I apologize for quality of images in attached file.
Attachments:
problem19-min.pdf (69kb)