November 2020

 

11/05/2020

image-20201117165628598

Prove 1x=1a+1b1a+b

Proof:

image-20201119073808749

Take the three centers of the circles:

image-20201119081713180

From the Law of Cosines:

cosα=(a+r)2+b2(a+br)22b(a+r)=1ar+brabb(a+r)

cosβ=(b+r)2=a2(a+br)22a(b+r)=2br+araba(b+r)

From the Law of Sines:

b+rsinα=a+rsinβ(a+r)2(1cos2α)=(b+r)2(1cos2β)

(a+r)2(2ar+brab)2b2=(b+r)2(2br+arab)a2

(a+r)2(b+r)2=a2(2ar+brab)2b2(2br+arab)2a2b2

a2b2(a+b+2r)(ab)=(2a2r+abra2b+2b2r+abrab2)(2a2r+abra2b2b2rabr+ab2)

a2b2(a+b+2r)(ab)=[2(a2+b2+ab)rab(a+b)][2r(a+b)(ab)ab(ab)]

a2b2(a+b+2r)=[2(a2+b2+ab)rab(a+b)][2(a+b)rab]

2a2b2r+a2b2(a+b)=4(a+b)(a2+b2+ab)r2[2ab(a2+b2+ab)+2ab(a+b)2]r+a2b2(a+b)

a2b2=2(a+b)(a2+b2+ab)rab(2a2+2b2+3ab)

(a+b)(a2+b2+ab)r=ab(a+b)2r=ab(a+b)a2+b2+ab

r=ab(a+b)a2+b2+aba+br=a2+b2+abab

a+br=(a+b)2ab11r=a+bab1a+b

1r=1a+1b1a+b

Proof 2

Use Stewart’s Theorem

(a+r)2a+(b+r)2b=(a+b)[(a+br)2+ab]

a3+2a2r+ar2+b3+2b2r+br2=(a+b)[a2+b2+3ab2(a+b)r+r2]

2(a2+b2)r+2(a+b)2r=ab(b+3a)+ab(a+3b)=4ab(a+b)

r=ab(a+b)a2+ab+b2a+br=a2+ab+b2aba+br=(a+b)2ab1

1r=a+bab1a+b1r=1a+1b1a+b


11/06/2020

image-20201117033820379

Solution:

image-20201117033950167

Make equilateral triangle ABE with point C and E on the same side of AB. Easy to know :

EA=AB=EB, CAE=20, CBE=40.

CB=AB=EB,CBE=40CEB=70CEA=10

CAE=CBA=20,EA=AB,AC=BDCAEDBC

DCB=CEA=10CDA=DCB+CBD=10+20=30


11/07/2020

image-20201117005107051

Solution:

image-20201117010801435

Let O be the circumcenter of ABP. Then AOP=2ABP=20 and

BOP=2BAP=40AOB=60. This and AO=BO shows that AOB is equilateral.

So OAB=60 and AO=AB. By assumption AB=AC, so AO=AC and

AOC=9012OAC=9012140=20=AOP

Therefore, O,P,C are collinear and we find

APC=OAP+AOP=80+20=100


11/08/2020

image-20201117003236412

Solution:

image-20201117003450428


image-20201117003321013

Solution:

DE2+BE2=14, DE+BC=1

DEAC=BEBCDE(1DE)=BE2

DE=DE2+BE2=14ABC=30


11/09/2020

problem image

The distance from a point D inside equilateral triangle ABC to the vertices are a, b and c respectively. Solve the area for SABC

Solution 1:

solution 1 image

Rotate BDC by 60 to AEC so we have AE=b.

Then cosADE=a2+c2b22ac,

sinADE=1(a2+c2b2)24a2c2

=(2ac+a2+c2b2)(2aca2c2+b2)2ac

=(a+c+b)(a+cb)(b+ac)(ba+c)2ac

Then we know

cosADC=cos(ADE+60)=cosADEcos60sinADEsin60

=a2+c2b24ac3(a+b+c)(a+bc)(b+ca)(c+ab)4ac

=a2+c2AC22ac

AC2=a2+b2+c22+3(a+b+c)(a+bc)(b+ca)(c+ab)2

SABC=34AC2

=38(a2+b2+c2)+38(a+b+c)(a+bc)(b+ca)(c+ab)

Solution 2:

Solution 2 image

Rotate BDC by 60 to AEC , CDA by 60 to BFA, ADB by 60 to CGB.

Easy to find that CG=AF=a,AE=BG=b,BF=CE=c.

So SAFBGCE=2SABC=3SCDG+SCDE+SBDG+SADF

According to Heron’s formula :

SCDG=(a+b+c)(a+bc)(b+ca)(c+ab)4

And we know

SCDE=34c2, SBDG=34b2, SADF=34a2

So we have

SABC=38(a2+b2+c2)+38(a+b+c)(a+bc)(b+ca)(c+ab)


11/10/2020

image-20201116165041554

ABC is equilateral. Point D is outside ABC and DAC=10, DCA=20. Prove BC=BD

Make circumcircle of ACD, suppose the center of the circle is B. Then we know

  1. Point D and B cannot be on the same side of line AC, so B and B are on the same side of line AC.

  2. AB=DB=CB,ABC=ABD+DBC =2ACD+2DAC=40+20=60 So ABC is also equilateral.

From 1 and 2 we know B is the same as B, then BC=BD.


11/12/2020

image-20201116221941530

A semicircle is constructed on line segment AB. Another semicircle is constructed on chord CD, intersecting AB at P and Q. If AP=3,PQ=7, and QB=2, then find the length CD.

Solution 1:

image-20201116221408123

POKDOQ,PO=3,DO=6,OQ=4 KO=2CD=62+8222=96

Solution 2:

OE=12,PE=72,O1E2+OE2=OO12, O1E2+PE2=PO12=CO12=OO1214+494=OO12+12

OO12+CO12=CO2=62CO12=24CD=224=96


11/13/2020

image-20201113073810590

Point D is outside of circle O with diameter MN. From D make two lines DM and DB with points D, A and B on the circle O. Extend OD and NA to intersect at point C. Prove that CDDB.


11/14/2020

Let ABC be an acute triangle with circumcircle ω, and let H be the intersection of the altitudes of ABC. Suppose the tangent to the circumcircle of HBC at H intersects ω at points X and Y with HA=3, HX=2, and HY=6. the area of ABC can be written in the form mn, where m and n are positive integers, and n is not divisible by the square of any prime. Find m+n.

image-20201116164550124


11/15/2020

Find all primes p to make p32p+4 is a perfect square.

Solution:

If q=3k+1 or 3k1, then q2=1(mod3) But if p=3k+1, p32p+4=0(mod3), if p=3k1, p32p+4=2(mod3) only when p=3k, p32p+4=1(mod3) and p must be prime, so k=1,p=3 is the only one.


11/16/2020

A scale model of a building is 8 inches wide and 27 inches tall. It is placed against a wall. What is the length of the shortest pole that will reach the wall above it from the level ground?

image-20201116160912095

Solution 1:

Use trigonometry it is easier to get CG=8cosα+27sinα, it gets minimum value when

8sinαcos2α27cosαsin2α=0tanα=32CGmin=1313

Note: to avoid trigonometrical approach, it would be a hard work to solve this problem, seems.

Solution 2:

Get another solution based on another problem on 11/30/2020:

Suppose AF=a,AD=b,FG=x,CD=y,CG=l

ya=bxxy=ab

l2=(a+x)2+(b+y)2=(a+x)2+(b+abx)2

l2 gets minimum value when

2(a+x)+2(b+abx)(abx2)=0

(x3ab2)(x+a)=0

x=ab23,y=a2b3

lmin=a2+3aab23+3ba2b3+b2


11/17/2020

How many sequences of integers {a1,a2,,a7} are there for which 1ai1 for every i, and a1a2+a2a3+a3a4+a4a5+a5a6+a6a7=4?


11/18/2020

There are 12 students in a classroom: 6 of them are Democrats and 6 of them are Republicans. Every hour the students are randomly separated into four groups of three for political debates. If a group contains students from both parties, the minority in the group will change his/her political alignment to that of the majority at the end of the debate. What is the expected amount of time needed for all 12 students to have the same political alignment, in hours? (HMMT November 2017 Team Round Problem 7)

Solution:


11/22/2020

image-20201122163753898

Point I is the incenter of ABC and point M bisects side BC. Extend line IM and intersects circumcircle of ABC at point D. Point E and F bisects the arc AB and BC. Line DE intersects AB at point U, Line DF intersects BC at point V. Prove: BU=CV

Proof:

Point E bisects the arc AB

EBU=BDEEBUEBD

BUBE=BDDEBU=BDBEDE

Point F bisects the arc BC

CVCF=CDDFCV=CDCFDF

sinBDF=sinBCF,sinCBE=sinCDE

BDBEDFCDDECF=BDDFsinBDFBEBCsinCBECDDEsinCDECFBCsinBCF

=SBDFSCDESBCESBCF

SBCE=2SCEM=2IMIDSCDE

SBCF=2SBFM=2IMIDSBDF

BDBEDFBDDFBEBC=1BU=CV


11/23/2020

In a single-elimination tournament consisting of 29=512 teams, there is a strict ordering on the skill levels of the teams, but Joy does not know that ordering. The teams are randomly put into a bracket and they play out the tournament, where the better team always beats the worse team. Joy is then given the results of all 511 matches and must write a list of teams such that she can guarantee that the third-best team is on the list. What is the minimum possible length of Joy’s list? (Shared by Brady from HMMT Guts Test November 2020)

Solution:

After understanding the model, easy to know that, checking the result of all 511 matches, the only team who won 9 matches is the first team in the skill level list.

The second best team must be one of those teams who lost to the first team - because only the best team can win it. There are 9 teams in this set.

The third best team has two situations:

  • It could be one of the teams who lost to the first team - it could be in the 9-team-set.
  • Or, it could be one of the teams who lost to any team from the 9-team-set.

Check the teams in the 9-team-set:

  • There is a team lost in the 1st round, and 0 team lost to it.
  • There is a team lost in the 2nd round, and 1 team lost to it.
  • There is a team lost in the 3rd round, and 2 teams lost to it.
  • There is a team lost in the 9th round, and 8 teams lost to it.

So totally there are 1+2++8=36 teams lost to any team from the 9-team-set.

To include two situations for the third best team, we need the list length no shorter than 9+36=45 to guarantee that the third best team in it.


11/24/2020

image-20201124223551777

Point D bisects side BC in ABC and DEAB,DFAC,BF=CE , prove: AB=AC

Proof:

image-20201125032132835

Let point M and N bisects BD and DC respectively.

BD=DC=2ME=2NFME=NF,CM=BN

and we already know BF=CE, so

BNFCMEECB=FBC

so ECBFBCEBC=FCB


11/25/2020

image-20201125113526475

CD and BE are angle bisectors of ABC and CD=BE. Prove AB=AC.

Proof 1:

Let AC=b,AB=c,BC=a,CD=e,BE=f

From Length of Angle Bisector:

e2=ab(a+b)2((a+b)2c2)

f2=ac(a+c)2((a+c)2b2)

e=fbbc2(a+b)2=cb2c(a+c)2

(bc)+bcb(a+b)2c(a+c)2(a+b)2(a+c)2=0

(bc)+bca2(bc)+2a(b2c2)+(b3c3)(a+b)2(a+c)2=0

(bc)(1+bca2+2ab+2ac+a2+ab+c2(a+b)2(a+c)2)=0

(bc)(a+b)2(a+c)2+bc[(a+b+c)2ab](a+b)2(a+c)2=0

a+b+c>a,a+b+c>b(a+b+c)2ab>0

bc=0AB=AC

Proof 2:

image-20201128151708916

Draw DCF=8=BEC=7 and make CE=CF. Then connect D and F.

CD=BE,CE=CF,7=8BCEDFC

BC=DF,1=4

x=2+3=1+3=4+3=6+7=5+7=5+8

4+3=5+8y=z

Draw FGAB,BHCF, BC=DF,y=zDFGBCH

DG=CH,FG=BHBFGBFH

BHFG is a parallelogram m=n

DBC=DFC=ECB

Proof 3:

image-20201128160143152

Assume ABC is not isosceles. Let ABC>ACB.

CD=BE,BC=BC,CE>BD (if two triangles have two pairs of corresponding sides congruent and their included angles are not congruent, then the greater third side is opposite the greater included angle).

Through point E, construct EGAB,DGBE, so BDGE is a parallelogram

BE=DG,CD=BECD=DG

(g+g)=(c+c),g=b

b+g=c+c

b>cg<cCE<EGCE<BD

The assumption of the inequality of ABC and ACB leads to two contradictory results: CE>BD and CE<BD.

Therefore, ABC is isosceles.


11/27/2020

A sphere is centered at a point with integer coordinates and passes through the three points (2,0,0),(0,4,0),(0,0,6), but not the origin (0,0,0). If r is the smallest possible radius of the sphere, compute r2. (HMMT General, November 2020)

Solution:

Suppose the center of the sphere is (x0,y0,z0)

(x02)2+y02+z02=r2

x02+(y04)2+z02=r2

x02+y02+(z06)2=r2

x024x0+4+y02+z02=x02+y028y0+16+z02=x02+y02+z0212z0+36

x0=2y03=3z08x0=3z08,y0=3z052, z0 is odd

(3z08)2+(3z052)2+(z04)2=r2

49z02270z0+425=4r2

If z0R, when z0=27098, 4r2 get minimum value.

The closest odd integer to 27098 is

z0=3x0=1,y0=2r2=14

but x02+y02+z02=14=r2 so this is not solution.

The second closest odd integer to 27098 is

z0=1x0=5,y0=1r2=51

and z02+y02+z02=27r2

So the solution is r2=51.


11/30/2020

Solve over the integers: 615+x2=2y

Solution:

Easy to see x1 mod 2. The last digit of x2 could be 1,5,9 thus the last digit of 615+x2 could be 0,4,6, but the last digit of 2y could be 2,4,6,8 so the common subset of the last digit of 2y is 4,6, and this means y0 mod 2.

Suppose y=2k, then 615=22kx2=(2k+x)(2kx). We know that 615=41×5×3, so {2k+x=412kx=152k=28no integer solution,{2k+x=1232kx=52k=64k=6,x=59,y=12{2k+x=2052kx=32k=104no integer solution{2k+x=6152kx=12k=308 no integer solution

So the answer is x=59,y=12


11/30/2020

image-20201130023450021

l and c are known. Find x and y. This problem seems similar to a previous one.

Solution:

image-20201130024443003

yc=cxxy=c2

l=xy+x2+xy+y2

w=x+yl=xw+yw=w(x+y)

l2=w(x+y+2xy)=w(w+2c)

w2+2cwl2=0w=2c+4c2+4l22=c2+l2c

So x,y are roots of t2(c2+l2c)t+c2=0:

x,y=c2+l2c±l22c22cc2+l22

y=c2+l2c+l22c22cc2+l22