Doubly Trisected Triangle

Geometry Level 5

Triangle ABC has A = 4 0 \angle A = 40^{\circ} , B = 6 0 \angle B = 60^{\circ} , C = 8 0 \angle C = 80^{\circ} . Points M , N M,N trisect the side B C BC and points P , Q P,Q trisect the side A C AC . The lines A M , A N , B P , B Q AM, AN, BP, BQ intersect at the points S , T , U , V S,T,U,V as shown in the figure below, dividing the triangle into 9 regions. Determine the smallest possible value of [ A B C ] + [ S T U V ] [ABC] + [STUV] such that both [ A B C ] [ABC] and [ S T U V ] [STUV] are positive integers.

Details and assumptions

[ P Q R S ] [PQRS] refers to the area of figure P Q R S PQRS .


The answer is 79.

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7 solutions

Ang Yan Sheng
May 20, 2014

It is easy to see that if [ S T U V ] [ A B C ] = p q \frac{[STUV]}{[ABC]}=\frac{p}{q} in lowest terms then the minimal value of [ A B C ] + [ S T U V ] [ABC]+[STUV] , when both terms are positive integers, is p + q p+q . Hence we will find [ S T U V ] [ A B C ] \frac{[STUV]}{[ABC]} .

Firstly, let CU intersect AB at D. Then by Ceva's Theorem, C Q Q A A D D B B N N C = 1 \frac{CQ}{QA}\!\!\frac{AD}{DB}\!\!\frac{BN}{NC}=1 , so A D = D B AD=DB and D is the midpoint of AB. Similarly if CS intersects AB at E then E is the midpoint of AB, so D=E, and CUSD is a straight line.

Now by Menelaus's Theorem on triangle CPB and transversal ATN, we have P T T B = P A A C C N N B = 1 3 1 2 = 1 6 \frac{PT}{TB}=\frac{PA}{AC}\!\!\frac{CN}{NB}=\frac13\!\!\frac12=\frac16 , so T B P B = 1 1 + P T T B = 1 1 + 1 6 = 6 7 \frac{TB}{PB}=\frac1{1+\frac{PT}{TB}}=\frac1{1+\frac16}=\frac67 .

Thus [ A T B ] = T B P B [ A P B ] = T B P B A P A C [ A B C ] = 6 7 1 3 [ A B C ] = 2 7 [ A B C ] [ATB]=\frac{TB}{PB}\![APB]=\frac{TB}{PB}\!\!\frac{AP}{AC}\![ABC]=\frac67\!\!\frac13\![ABC]=\frac27\![ABC] .

Similarly, we can get D S S C = 1 4 \frac{DS}{SC}=\frac14 , and [ B S C ] = 2 5 [ A B C ] [BSC]=\frac25\![ABC] ; also, N U U A = 1 3 \frac{NU}{UA}=\frac13 , and [ C U A ] = 1 4 [ A B C ] [CUA]=\frac14\![ABC] .

Therefore,

[ U T S ] = [ A B C ] [ A P B ] [ B S C ] [ C U A ] = ( 1 2 7 2 5 1 4 ) [ A B C ] = 9 140 [ A B C ] \begin{aligned} [UTS]&=[ABC]-[APB]-[BSC]-[CUA]\\&=\left(1-\frac27-\frac25-\frac14\right)[ABC]\\&=\frac9{140}\![ABC]\end{aligned} .

An analogous argument gives [ U V S ] = 9 140 [ A B C ] [UVS]=\frac9{140}\![ABC] , so [ S T U V ] = [ U T S ] + [ U V S ] = 9 70 [ A B C ] [STUV]=[UTS]+[UVS]=\frac9{70}\![ABC] , and the answer to the original question is 9 + 70 = 79 9+70=79 .

The ratio of areas is independent of the angles that are given.

Calvin Lin Staff - 7 years ago
Tingxuan Wang
May 20, 2014

Draw AR parallel to BC, note the intersection with BS,BV as X,Y. Since AR parallel to BC, so AX:BC=AP:PC=1:2, AY:BC=AQ:QC=2:1. By using intercept theorem several times, we would be able to conclude that BV:VU:UQ=12:9:7, BS:ST:TP=21:9:5. Same thing applies for AS:SV:VM=21:9:5, AT:TU:UN=12:9:7. Hence, the area of STUV over the area of triangle ABC would be (the area of triangle SUV+the area of triangle STU):the area of triangle ABC=[2/3×9/(12+9+7)×9/(9+21)+2/3×9/(12+9+7)×21/(9+21)×9/(9+12)]/1=2×9/140=9/70. Therefore, we can conclude that the smallest possible integral value of [ABC] and [STUV] must be 70 and 9 respectively so the sum of them is 79.

Carlos Yee
May 20, 2014

Let h C , h_C, h U h_U and h Q h_Q be the distance from the vertexes C , C, U U and Q , Q, respectively, to the segment A B AB ,

Now, we can see that the triangles A B U ABU and N Q U NQU are similar, since Q N QN is parallel to A B AB , and the ratio of similarity is 3:1.

From this we have that h U = 3 4 h Q = 3 4 2 3 h C = 1 2 h C . h_U=\frac{3}{4}\cdot h_Q=\frac{3}{4}\cdot \frac{2}{3} \cdot h_C=\frac{1}{2}\cdot h_C.

Using this, we get that [ A B U ] = 1 2 [ A B C ] . [ABU]=\frac{1}{2}[ABC].

A similar argumente leads to [ A B S ] = 1 5 [ A B C ] [ABS]=\frac{1}{5}[ABC] , using the similarity of the triangles A B S ABS and M P S MPS and its similarity ratio 3:2.

In particular we have that P S = 2 5 B P PS=\frac{2}{5}\cdot BP and A S = 3 5 A M . AS=\frac{3}{5}\cdot AM.

Now we use the Generalized Angle Bisector Theorem (GABT) in the triangle A C M ACM and we get C N N M = A C A M sin ( C A N ) sin ( N A M ) \frac{CN}{NM}=\frac{AC}{AM}\cdot\frac{\sin(\angle CAN)}{\sin(\angle NAM)} .

Since M N = N C , MN=NC, we get A M A C = sin ( C A N ) sin ( N A M ) . \frac{AM}{AC}=\frac{\sin(\angle CAN)}{\sin(\angle NAM)}.

We use again GABT in the triangle A P S , APS, we get

P T T S = A P A S sin ( P A T ) sin ( T A S ) = A P A S sin ( C A N ) sin ( N A M ) = A P A S A M A C = 5 9 . \frac{PT}{TS}=\frac{AP}{AS}\cdot\frac{\sin(\angle PAT)}{\sin(\angle TAS)}=\frac{AP}{AS}\cdot\frac{\sin(\angle CAN)}{\sin(\angle NAM)}=\frac{AP}{AS}\cdot\frac{AM}{AC}=\frac{5}{9}.

From this we have that P T = 5 14 P S = 1 7 B P , PT=\frac{5}{14}\cdot PS=\frac{1}{7}\cdot BP, wich implies that [ A T P ] = 1 7 [ A B P ] [ATP]=\frac{1}{7}[ABP] and

[ A B T ] = 6 7 [ A B P ] = 2 7 [ A B C ] . [ABT]=\frac{6}{7}\cdot[ABP]=\frac{2}{7}\cdot[ABC].

By symmetry [ A B V ] = 2 7 [ A B C ] . [ABV]=\frac{2}{7}\cdot[ABC].

Using all these results we get that [ S T U V ] = [ A B U ] [ A B T ] [ A B V ] + [ A B S ] [STUV]=[ABU]-[ABT]-[ABV]+[ABS] = ( 1 2 2 7 2 7 + 1 5 ) [ A B C ] = 9 70 [ A B C ] , \;\;\;\;\;\;\;\;\;\;\;\;=(\frac{1}{2}-\frac{2}{7}-\frac{2}{7}+\frac{1}{5})\cdot[ABC]=\frac{9}{70}\cdot[ABC],

thus, [ S T U V ] [STUV] is integer if and only if [ A B C ] [ABC] is a multiple of 70.

So [ A B C ] + [ S T U V ] = 79 70 [ A B C ] = 79 k [ABC]+[STUV]=\frac{79}{70}[ABC]=79 \cdot k and its smallest value is 79.

Mihai Nipomici
May 20, 2014

Ratio theorem if a cevian divide the side in ratio a/b then the ratio of 2 small triangles are a/b.From menelaus in MAC and transversal PSB we have AS/SM=3/2 ( )and from menelaus in MAC and transversal QVB we have AV/VM=6( ). From( )and( ) we haveAS:SV:VM=21:9:5.In the same way for CAN and the same transversals we get AT:TU:UN=12:9:7.Because its simetric from angle A to B we have BS:ST:TP=21:9:5 and BV:VU:UQ=12:9:7.Now [AVU]=[ABQ] (VU/BQ)=[ABC] (AQ/AC) (VU/BQ)=[ABC] (2/3) (9/28)=[ABC] 3/14.In the same way [AST]=[ABC] 3/35.From this we have [STUV]=[AVU]-[AST]=[ABC] ((3/14)-(3/35))=[ABC]*9/70.And because [STUV]and[ABC] integer and their ratio is 9/70 and we are asked for the smalest integer satisfing the ratio.They are 9 and 70and answer is 70+9=79

Superman Son
May 20, 2014

The Menelaus Theorem states that when a transversal cuts a triangle at 3 points on the three sides the ratio becomes (-1) so , here in this given problem we take the triangle BCP taking ASM as the transversal , so we get that (BM:MC) (CA:AP) (PS:SB)=(-1) Now putting values of (BM:MC),(CA:AP) as (1:2) and (3:1) respectively we get (PS:SB)=(2:3) again we take the triangle BPC but this time taking ATN as the transversal,so we get that (BN:NC) (CA:AP) (PT:TB)=(-1) Now putting values of (BN:NC),(CA:AP) as (2:1) and (3:1) respectively we get (PT:TB)=(1:6) so aggregating we got (PT:TS:SB)=(5:9:21) Now again this time we take triangle BQC and AVM as transversal , so we get that (BM:MC) (CA:AQ) (QV:VB)=(-1) Now putting values of (BM:MC),(CA:AQ)as (1:2) and (3:2) respectively we get (QV:VB)=(4:3) ,again this time we take triangle BQC and AUN as the transversal ,so we get that (BN:NC) (CA:AQ) (QU:UB)=(-1) Now putting values of (BN:NC),(CA:AQ)as (2:1) and (3:2) respectively we get (QU:UB)=(1:3) so aggregating we got (QU:UV:VB)=(7:9:12) NOW,([PQRS] refers to the area of figure PQRS.)

[STUV]=[AMN] - [AST] - [VUMN], [STUV]=[AMN] - [AST] -{ [AMN] - [AVU] }, [STUV]=(1/3)of[ABC] - (1/3)of[ABC] {(9)/(5 + 9 + 21)} - { (1/3)of[ABC] - (2/3)of[ABC] {(9)/(7 + 9 + 12)} }, [STUV]=(9/70)of[ABC] so,when both [ABC] and [STUV] are positive integers and it is the minimum we get [ABC]+[STUV] = 9+70 = 79

Calvin Lin Staff
May 13, 2014

Notice that B M U \triangle BMU , M N U \triangle MNU , and N C U \triangle NCU all have the same area, since they have the same base and height, let this area be x x . Similarly, [ A P U ] = [ P Q U ] = [ C Q U ] = y [APU] = [PQU] = [CQU] = y . [ B C Q ] = [ A B C ] 3 = [ A C N ] [BCQ] = \frac{[ABC]}{3} = [ACN] , so 3 x + y = [ A B C ] 3 = x + 3 y 3x + y = \frac{[ABC]}{3} = x + 3y . This gives x = y = [ A B C ] 12 x = y = \frac{[ABC]}{12} , so [ U N C Q ] = [ A B C ] 6 [UNCQ] = \frac{[ABC]}{6} .

If we replace U U with S S in the above argument, again using areas x , y x,y , we will get that 3 x + 2 y = 2 [ A B C ] 3 = 2 x + 3 y 3x + 2y = \frac{2[ABC]}{3} = 2x + 3y . This gives x = y = 2 [ A B C ] 15 x = y = \frac{2[ABC]}{15} . Since [ A B S ] + [ B M S ] = [ A B C ] 3 [ABS] + [BMS] = \frac{[ABC]}{3} , and [ B M S ] = 2 [ A B C ] 15 [BMS] = \frac{2[ABC]}{15} , we have [ A B S ] = [ A B C ] 5 [ABS] = \frac{[ABC]}{5} .

If we replace S S with T T and repeat this again, we will find that [ A P T ] = [ A B C ] 21 [APT] = \frac{[ABC]}{21} . By symmetry, [ B M V ] = [ A B C ] 21 [BMV] = \frac{[ABC]}{21} . Since [ A P T ] + [ P T U Q ] + [ C N U Q ] = [ A B C ] 3 [APT] + [PTUQ]+[CNUQ]=\frac{[ABC]}{3} , we have [ P T U Q ] = 5 [ A B C ] 42 [PTUQ] = \frac{5[ABC]}{42} . We can similarly deduce that [ B S V ] = 3 [ A B C ] 35 [BSV] = \frac{3[ABC]}{35} , and then use these to deduce that [ S T U V ] = 9 [ A B C ] 70 [STUV] = \frac{9[ABC]}{70} . For this to be an integer, it must be the case that [ A B C ] [ABC] is a multiple of 70 70 , which gives [ A B C ] = 70 , [ S T U V ] = 9 [ABC] = 70, [STUV] = 9 and [ A B C ] + [ S T U V ] = 79 [ABC] + [STUV] = 79 as the smallest possible value.

Note: The equation [ S T U V ] = 9 [ A B C ] 70 [STUV] = \frac{9[ABC]}{70} holds for all triangles and is independent of the value of A \angle A , B \angle B and C \angle C .

Debjit Mandal
May 20, 2014

By using menelaus' theorem on triangle ANC with transversal BP and BQ, we can find that, ST=3/7 BS, TP=5/21 BS, and VU=3/4 BV, UQ=7/12 BV. Let, angle PBQ=ß. Then, [BUT]-[BSV]=[STUV]. By sine rule, we can find that, [STUT]=3/4 BS BV sin ß. Then, Similarly, we can find that, [BPQ]=35/6 BS BV sin ß=1/3*[ABC]. By simple trigonometrical calculation with the informations angle B=60 degree, angle A=40 degree and angle C=80 degree

Wrong triangles for menelaus. Uncertain about how to continue.

Calvin Lin Staff - 7 years ago

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