,
are all polynomials such that ![\[ P(x^5)+xQ(x^5)+x^2R(x^5)=(x^4+x^3+x^2+x+1)S(x),\]](http://latex.artofproblemsolving.com/1/6/a/16a71abc110ece558d427a07d7be2d1f72148024.png)
prove that 
is a factor of 
.
such that the equality 
holds for all pairs of real numbers 
.
is an obvious solution. Now, let 
.
or 
.
.
Y by AwesomeToad, infiniteturtle, tastymath75025, forthegreatergood, hwl0304, minimario, TheCrafter, AlcumusGuy, yan_justin98, ahaanomegas, TheMaskedMagician, bcp123, abishek99, fz0718, Royalreter1, happiface, gobathegreat, capttawish, MSTang, efang, mathtastic, va2010, chezbgone, professordad, numbersandnumbers, AkshajK, Amir Hossein, aTastyNoodle, butter67, El_Ectric, EpicSkills32, HYP135peppers, mentalgenius, raptorw, SHARKYBOY, tau172, flamefoxx99, pi37, gamjawon, csmath, v_Enhance, DanielL2000, jkoj25, hamup1, DigitalKing257, swamih, sturdyoak2012, fireonice18, lazorpenguin27143, pedronr, sicilianfan, bobthesmartypants, jh235, hungrycap, miru99, zmyshatlp, blasterboy, mikechen, WalkerTesla, magicarrow, checkmatetang, scfliu, cobbler, 15Pandabears, Einstein314, jhoer, spartan168, Binomial-theorem, dantx5, DrMath, tchebytchev, bluecarneal, C-bass, VUThanhTung, mihajlon, tkhalid, WorstIreliaNA, MathLearner01, MathStudent2002, jam10307, Higgsboson79, knittingfrenzy18, mxgo, RadioActive, droid347, blippy1998, MathSlayer4444, joey8189681, FlakeLCR, Mathlete55, WhaleVomit, linga, mathmaster2012, rjiang16, ninjasrule34, bestwillcui1, CaptainFlint, Tommy2000, thkim1011, mathboxboro, countingarithmetic, zachman99323, uwotm8, Eat_Sleep_Math, pandyhu2001, Anish_S, fireclaw105, AdBondEvent, buratinogigle, jonyj1005, chemistrygirl, Kingofmath101, mathman523, rkm0959, mathwizard888, stan23456, champion999, Michael_Huang23, david2006322, NormanWho, Rubaiya, hesa57, azmath333, coolrg, phi_ftw1618, mathmaster2000, ac_math, Eugenis, ImpossibleCube, mathguy623, harvey2014, Mathaddict11, math101010, yojan_sushi, Cube1, whatshisbucket, flamesofpi, muti66, Einsteinhead, blep, spin8, PiDude314, Gibby, 62861, gradysocool, poonicle, wonder23, ThisIsASentence, Temp456, mjkkra, Wave-Particle, skys, Python54, yegkatie, rt03, brianapa, Rotack00, Not_a_Username, claserken, hoangbo5h5, sw993, yrnsmurf, Mathcat1234, mathguy5041, CountofMC, mathisawesome2169, techguy2, bookworm2003, illogical_21, Heisenberg09, frogs, acegikmoqsuwy2000, niyu, liant, talkon, JasperL, nukelauncher, shawnee03, ghghghghghghghgh, GeneralCobra19, jeffisepic, eashang1, thedoge, neep, shootingstar8, blitzkrieg21, Mathisfun04, bel.jad5, WireCat, eisirrational, mukes070304, tenplusten, Mathuzb, arulxz, sunfishho, rafa2be, Vrangr, tdeng, QWERTYphysics, maXplanK, Wizard_32, franchester, sriraamster, mingxu, andyxpandy99, tapir1729, Ultroid999OCPN, mathleticguyyy, opptoinfinity, dchenmathcounts, Stormersyle, StoryScene, MathInfinite, arkobanerjee, expiredkiwi1313, smartninja2000, vedonidesimath, Mathcollege, Williamgolly, nikhil.mudumbi, PROA200, Lol_man000, ghu2024, Toinfinity, hALAL_cartDonut, Lcz, OliverA, Jay17, Awesome_guy, Zorger74, Aryan-23, tree_3, nikenissan, Jerry122805, badguy, vsamc, OlympusHero, Spaced_Out, mzeng1325, eagles2018, DSL13, MindCalculator, mathisverynice, Pi-rate, b0bby, tenebrine, lneis1, Didier, Taco12, player01, JVAJVA, centslordm, math31415926535, RetroTurtle, phoenixfire, DofL, andyloo666, Lamboreghini, samrocksnature, MasterInTheMaking, ike.chen, jhu08, Math_fj, Upwgs_2008, pog, Bimikel, Jndd, PEKKA, mahaler, asimov, megarnie, Cheerfulfrog, rrc08, mathking999, ASweatyAsianBoie, ryanbear, peelybonehead, huashiliao2020, happypi31415, v4913, s12d34, Adventure10, Mango247, iamhungry, ostriches88, anduran, Sedro, and 12 other users
Geometry is a fascinating subject. Although it's the Olympiad concentration with the least relevance to real life, in my opinion it's also the most beautiful - and therefore one of the most difficult ones to learn. Specifically, transitioning from computational geometry to Olympiad geometry is kinda hard because, well, Olympiad geometry problems are usually very hard! There are a boatload of problem sets of sorts floating around the internet, but many of these mass collections of problems are for the experienced Olympiad problem solver; if one wants an easy geometry problem, he or she usually has to scavenge the Contests page.
So, for my 4000th post, I decided to fix this issue. Here is my first problem set released on AoPS, 100 Geometry Problems: Bridging the Gap from AIME to USAMO!
A few things to note:
Enjoy the problems!
So, for my 4000th post, I decided to fix this issue. Here is my first problem set released on AoPS, 100 Geometry Problems: Bridging the Gap from AIME to USAMO!
A few things to note:
- The list starts with mostly computational problems (like around AMC #15-ish) and ends with mostly proof problems. However, this does not mean that the two classes of problems are completely segregated. The transition is (I hope) as smooth as possible, with problems of both types being interlinked with each other. That being said, it's very difficult to compare difficulties between calculations and proofs, so some problems may seem out-of-place. Feel free to skip around as needed.
- Not everything is self-contained; more specifically, while some key lemmas that are used for later problems are themselves problems in the packet, others are not. The hope here is that you'll encounter them at some point during your journey anyway.
- There are no solutions to these problems included with the document. That being said, the majority of problems have sources - use the Contests page to your advantage here!
Enjoy the problems!
![[asy]
size(150);
defaultpen(linewidth(0.8)+fontsize(10pt));
pair A=(36*1/3,36*sqrt(8)/3),B=origin,C=(40,0),D=(18,0),M=(20,0);
path circ=circumcircle(A,B,C);
pair out=2*D-A,E=intersectionpoint(D--out,circ);
draw(A--B--C--A--E--M^^circ^^B--E--C);
draw(rightanglemark(E,M,C,50));
draw(anglemark(B,A,E,100)^^anglemark(E,A,C,100));
label("$A$",A,N);
label("$B$",B,SW);
label("$C$",C,SE);
label("$D$",D,SW);
label("$E$",E,S);
label("$M$",M,N);
[/asy]](http://latex.artofproblemsolving.com/c/0/6/c068f34033f0e503863bf8a074190806420f9a7e.png)
be the midpoint of 
.
is a cyclic quad, 
,
is isosceles. (This is well-known.)
.
and 
.
,
,![\[\dfrac{BD}{DE}=\dfrac{AB}{BE}\implies \dfrac{BE}{DE}=\dfrac{AB}{BD}=\dfrac{36}{18}=2.\]](http://latex.artofproblemsolving.com/8/a/1/8a1aa5f9dcaecb6fbcc1fbf52b3f5ee9dbf2ac7a.png)
Therefore 
.
Solving this gives 
.
,
,
,
,
.
and 
the tangency points of the incircle of the triangle with 
and 
respectively. Furthermore, let the line parallel to 
passing through 
respectively. Then it is easy to see that 
and 
are both equilateral, so it is easy to compute that 
and 
.
by some simple angle chasing. Furthermore, it is obvious that 
and that 
,
,
.
.![\[\dfrac x{2\sqrt 3/3}=\dfrac{8\sqrt 3/3}x\implies x=\dfrac{4\sqrt 3}3,\]](http://latex.artofproblemsolving.com/0/9/3/0937a07badae000bfe64fa59618e34104e77920b.png)
so 
.
,
.![[asy]
size(150);
defaultpen(linewidth(0.8)+fontsize(10pt));
pair A=dir(100),B=dir(220),C=dir(320),D=reflect(origin,(1,0))*A,Ap=foot(A,B,C),Bp=foot(Ap,A,C),Cp=foot(Ap,A,B);
pair l=2*Cp-Bp,r=2*Bp-Cp;
path chord=l--r;
pair[] chords=intersectionpoints(chord,unitcircle);
draw(unitcircle^^A--B--C--cycle^^A--D^^Cp--Ap--Bp^^chords[0]--chords[1]);
label("$A$",A,dir(origin--A));
label("$B$",B,dir(origin--B));
label("$C$",C,dir(origin--C));
label("$D$",D,dir(origin--D));
label("$A'$",Ap,SE);
label("$B'$",Bp,dir(65));
label("$C'$",Cp,dir(110));
label("$X$",chords[0],dir(chords[1]--chords[0]));
label("$Y$",chords[1],dir(chords[0]--chords[1]));
draw(rightanglemark(A,Ap,C,2)^^rightanglemark(Ap,Bp,A,2)^^rightanglemark(Ap,Cp,A,2));
[/asy]](http://latex.artofproblemsolving.com/c/7/d/c7dd1e8dabfdc9067e73b27e6d90c88d58a5c844.png)
.
.
is isosceles and 
bisects 
.
,
.
,
and 
.
,
and by Fact 5 we win.
touch 
at 
,
be the 
,
,
.
and 
,
.
be any point on the extension of 
,![\[\angle AGF=\angle FCX=\angle FCG=\angle FAG,\]](http://latex.artofproblemsolving.com/e/1/1/e110dd1b08e8df19c2f453a42b2e77afdea90083.png)
so 
is isosceles as desired.![[asy]
size(250);
defaultpen(linewidth(0.8)+fontsize(11pt));
path circ1=circle(origin,5),circ2=circle((8,0),3);
pair A=5*dir(130),C=5*dir(179),D=(8,0)+3*dir(5), P=(5,0);
path circ3=circle((A+(8,0))/2,1/2*arclength(A--(8,0)));
pair[] x = intersectionpoints(circ2,circ3);
draw(circ1^^circ2^^x[0]--A--x[1]);
pair[] e=intersectionpoints(A--x[0],circ1), f=intersectionpoints(A--x[1],circ1);
draw(e[0]--P--f[0]);
label("$A$",A,dir(origin--A));
label("$C$",C,dir(origin--C));
label("$D$",D,dir(5));
label("$N$",x[0],dir((8,0)--x[0]));
label("$M$",x[1],dir((8,0)--x[1]));
label("$F$",e[0],dir(origin--e[0]));
label("$E$",f[0],dir(210));
label("$P$",P,dir(0));
[/asy]](http://latex.artofproblemsolving.com/1/7/1/17102d632c2aeb9d3dd9efb3ee45a828b44e4c14.png)
with 
be 
with 
.
be the common tangent to both circles, with 
on the same side of 
as 
are not drawn in the figure.) Note that 
and 
since they equal 
and 
,
and 
by vertical angles, we have 
.
,
and 
by Power of a Point, so![\[\left(\dfrac{ME}{NF}\right)^2=\left(\dfrac{PE}{PF}\right)\left(\dfrac{EY}{EX}\right)=\left(\dfrac{PE}{PF}\right)\left(\dfrac{PE}{PF}\right)=\left(\dfrac{PE}{PF}\right)^2.\]](http://latex.artofproblemsolving.com/7/d/2/7d2759862b465097bd1b276dad237a233f1df028.png)
,![[asy]
size(250);
defaultpen(linewidth(0.8));
pair P=(3,0),C=dir(130),X=(1,0);
path circTan = circle((1.5,0),1.5),line=P--C;
real slope=(P.y-C.y)/(P.x-C.x);
pair x[]=intersectionpoints(unitcircle,circTan),A=intersectionpoint(line,unitcircle),pointbwDE=(x[1].x-2,x[1].y-2*slope);
path lineDE=pointbwDE--x[1];
pair E=intersectionpoint(lineDE,unitcircle),M=extension(A,C,x[0],E);
draw(unitcircle^^x[0]--P--x[1]--E--cycle^^C--P^^M--origin--A^^P--origin--x[0]);
dot(origin);
label("$A$",A,dir(origin--A));
label("$B$",x[0],dir(origin--x[0]));
label("$C$",C,dir(origin--C));
label("$D$",x[1],dir(origin--x[1]));
label("$E$",E,dir(origin--E));
label("$P$",P,dir(origin--P));
label("$M$",M,dir((C+x[0])/2));
label("$O$",origin,S);
label("$X$",X,NE);
[/asy]](http://latex.artofproblemsolving.com/b/7/5/b7556e025b1584d7baffa0e97d77d0a4be47a87d.png)
be the center of the circle, let 
,
intersects circle 
and 
,
is a kite, which means that 
,
and 
have the same measure. Thus the measure of arc 
,
,
.
,
,
.
is cyclic, so 
.
,
bisects ![[asy]
size(230);
defaultpen(linewidth(0.8)+fontsize(11pt));
pair A=dir(100),B=dir(210),C=dir(330),M=waypoint(A--B,.6),N=waypoint(A--C,.27),P=extension(M,N,B,C);
path circum=circumcircle(A,M,N);
pair[] q=intersectionpoints(circum,unitcircle);
draw(P--C--A--B^^unitcircle^^N--P--q[1]^^circum);
draw(B--q[1]--C^^M--q[1]--N,linetype("4 4"));
dot(origin);
label("$A$",A,dir(origin--A));
label("$B$",B,SW);
label("$C$",C,SE);
label("$M$",M,dir(280));
label("$N$",N,NE);
label("$Q$",q[1],dir(origin--q[1]));
label("$P$",P,S);
label("$O$",origin,dir(260));
[/asy]](http://latex.artofproblemsolving.com/d/7/2/d72a3db004921f24b7ac0be73b8ca99a0bbc3a03.png)
.
is cyclic.
is cyclic, 
,
.![\[\angle QBM\equiv\angle QBA=\angle QCA\equiv \angle QCN,\]](http://latex.artofproblemsolving.com/d/3/d/d3dfee025012966982de80ed6df41a1e9f4c60a3.png)
so 
.
,
.
be the center of 
.
,
,
,
,
.
and so 
,
is an isosceles trapezoid. This implies 
.
and 
demonstrate that 
,
is tangent to 
.![\[\angle QNM=\angle QCB=\angle PQB=\angle PMB=\angle AMN.\]](http://latex.artofproblemsolving.com/1/d/0/1d0e61ef0545dad20b5a8f114adb1cab9242087d.png)
Thus 
and 
,
.
is tangent to 
,
we have 
cyclic, which implies 
.
,
we have 
.![\[DC\cdot DA=DE^2=DF^2\implies \dfrac{DF}{DC}=\dfrac{DA}{DF}.\]](http://latex.artofproblemsolving.com/c/7/7/c77bdc4498794a70bd0cf3640dafa639bf5e2f92.png)
Therefore 
.
,
,
,
are collinear.
,
to be the second intersection point of 
with 
,
,
and 
.
and 
,
and 
.
.
By PoP we see that 
,![\[\overline{PA_2}\cdot\overline{A_4X_2}+\overline{A_3X_1}\cdot\overline{PA_1}=0\implies\dfrac1{\overline{A_3X_1}\cdot\overline{PA_1}}+\dfrac1{\overline{PA_2}\cdot\overline{A_4X_2}}=0.\]](http://latex.artofproblemsolving.com/e/1/c/e1c6fbe8ac52e5bc44e90bc991d7d4b257164385.png)
Next, clever manipulation gives 
Through the fact that 
Thus 
Similarly, we get 
Finally, note that 
is equal to the power of the point 
with respect to 
.![\[\frac{1}{O_1A_1^2-r_1^2}+\frac{1}{O_2A_2^2-r_2^2}+\frac{1}{O_3A_3^2-r_3^2}+\frac{1}{O_4A_4^2-r_4^2}=0,\]](http://latex.artofproblemsolving.com/f/7/e/f7e49792c2099b1fbc4193a5aa99fb204b62ed62.png)
.
so 
is a cyclic quadrilateral. In addition, note that the circumcircle of 
passes through 
.
,
must be the radical axis of the circumcircles of 
.
and 
gives that 
and 
In addition, note that 
,
.
,
is cyclic, so 
.
,
are collinear as desired.
be the radii of the circles centered at 
respectively. We then have the following system of equations ![\[\begin{cases} BC = 9 = r_b + r_c \\ AX = r_a = 6 + r_b \\ AY = r_a = 5 + r_c. \end{cases}\]](http://latex.artofproblemsolving.com/0/5/f/05ff82aaa2431589982e38762d9e4bec6832b817.png)
and 
![[asy]
unitsize(1cm);
pair A = (0,4);
pair B = (0,0);
pair C = (3,0);
draw(A--B--C--cycle,linewidth(0.9));
draw(rightanglemark(A,B,C),blue);
pair D = (0,1.5);
pair E = foot(D,A,C);
draw(D--E--C--B--cycle,red+linewidth(0.91));
draw(rightanglemark(D,E,C),blue);
dotfactor = 4.5;
dot("$A$",A,NW);
dot("$B$",B,SW);
dot("$C$",C,SE);
dot("$D$",D,W);
dot("$E$",E,NE);
//diagram by happiface
[/asy]](http://latex.artofproblemsolving.com/9/c/c/9cc0f4fed66d80200dcbda9ab531b96270a79ef8.png)
quadrilateral 
is cyclic, from which the result follows due to same inscribed arcs.![[asy]
unitsize(.32cm);
pair B = (0,0);
pair C = (20.14,0);
pair A = intersectionpoint(Circle(B,14), Circle(C, 18));
draw(A--B--C--cycle);
pair I = incenter(A,B,C);
draw(Circle(I,abs(I-A)),purple+linewidth(.91));
pair X = intersectionpoints(Circle(I,abs(I-A)),B--C)[0];
label("$X$",X,1.5SW+.5E);
pair Y = intersectionpoints(Circle(I,abs(A-I)),C--B)[1];
draw(X--Y,blue+linewidth(.91));
pair D = foot(I,B,C);
pair E = foot(I,A,B);
draw(I--D,dashed+orange);
draw(I--E,dashed+orange);
draw(I--X,dashed+heavygreen);
draw(I--A,dashed+heavygreen);
markscalefactor=0.07;
draw(rightanglemark(I,D,X));
draw(rightanglemark(I,E,A));
dotfactor=4.5;
dot("$A$",A,NW);
dot("$B$",B,SW);
dot("$C$",C,SE);
dot("$I$",I,NE);
dot(X,red);
dot(Y,red);
label("$Y$",Y,1.5SE+.5W);
dot("$D$",D,S);
dot("$E$",E,NW);
//diagram by happiface, I hope you enjoy the unnecessary coloring lol
[/asy]](http://latex.artofproblemsolving.com/f/8/8/f88be6852e03330ba065ac13ebdbebeab6dad4c5.png)
to 
and 
we have that 
by HL congruence (both are clearly right triangles). It is also easy to show that 
So, 
and thus 
.
is a cyclic quadrilateral, thus we get 
.
.
are points of the same circle and 
,
must go through the circumcenter of circle 
.
and hence 
,
be a point on the same side of 
.
so 
are collinear. Hence 
![[asy]
//100 Geo problems #29 diagram by happiface
unitsize(.35cm);
defaultpen(fontsize(10pt));
pair A = (-5,0);
pair B = (0,12);
pair C = (9,0);
pair M = midpoint(A--C);
pair H = foot(B,A,C);
draw(A--B--C--cycle);
real r1 = abs(M-B), r2 = abs(M-A), r3 = r1-r2;
pair P = M*(r3/r1) + B*(r2/r1);
draw(A--P--C--cycle, orange+linewidth(0.91));
draw(B--H--M--cycle,red+linewidth(0.9));
markscalefactor=0.08;
draw(rightanglemark(A,P,C),linewidth(.89));
draw(rightanglemark(B,H,M));
label("5",A--H,S);
label("7",M--C,S);
label("7",P--M,E);
dotfactor=4.4;
dot("$A$",A,SW);
dot("$B$",B,1.5N);
dot("$C$",C,SE);
dot("$M$",M,1S);
dot(P,blue);
label("$P$",P,NE);
dot("$H$",H, 1S);
[/asy]](http://latex.artofproblemsolving.com/1/3/4/134e82786c7698fbb3ce4b4a4d00c1503b26f4e0.png)
.
.
to 
We also have 
and 
.
we get 
.
is a median to the hypotenuse of 
we have 
.![\[ P = \tfrac{7}{\sqrt{148}}B + \tfrac{\sqrt{148}-7}{\sqrt{148}}M = \left(\frac{5(7) + 7(\sqrt{148}-7)}{\sqrt{148}}, \frac{12(7)}{\sqrt{148}}\right) = \left(\frac{7\sqrt{37}-7}{\sqrt{37}}, \frac{42}{\sqrt{37}}\right)\]](http://latex.artofproblemsolving.com/b/6/b/b6ba9a62cf798efbe5d2f4a1c56ec64c4e52f90a.png)
(see, it all worked out nicely in the end), and computing 
gives ![$[APC] = \frac{294}{\sqrt{37}}$](http://latex.artofproblemsolving.com/c/2/e/c2e8715dadb94fb25fea09b00d328d688b747fc2.png)
so 
.
in the denominator, just add that in at the end. Then squaring some binomials and rearranging terms won't be very bad at all.
is 84, and that 
is 7. All that remains is to find 
,
to find 
.
is immediate from there.
and 
.
with legs 
and 
with a hypotenuse 
.
and 
.
.
by Star Trek Lemma. 
and 
,
,
is similar to 
,
.
,
(Someone should clarify this step). Now, the distances are 
,
.
,
Then we have 
and therefore 
Via AA Similarity, we can obtain 
Letting 
we apply this similarity to get 
Subtracting, we get 
The Pythagorean Theorem gives 
or 
Solving for 
gives 
and this is the answer, since 
.
and 
be 
.
and 
Therefore, 
Furthermore, note that 
The answer is 
the height of triangle 
which corresponds to 
.
are isosceles doesn't cause contradictions, we do so. We now have a 
triangle, making the height 
.
just like in the solution, but since we cannot find the individual areas of DCN or DAM we must find their sum. Their sum is 
![[asy]
pair A, B, C, D, N;
C=(-1, 0);
A=(cos(2pi/7), sin(2pi/7));
N=(cos (pi/10), sin (pi/10));
B=(cos (2pi/3), sin (2pi/3));
D=(3.692, -0.743);
draw(circle((0, 0), 1));
draw(D--A--N--C--D--B--C--A--B--D);
label(A, "A", NE);
label(B, "B", NW);
label(C, "C", W);
label(D, "D", NE);
label(N, "N", SSW);
[/asy]](http://latex.artofproblemsolving.com/1/9/2/1928a10c530aac48de3c61f694f8edd2c1dfe937.png)
be the quadrilateral in the problem, and 
)
.
.
.
.
,
.
,
(by the angle bisector theorem) which is the same as 
also lies on the angle bisector of 
and 
.
and 
(By 
)
.
,
must be cyclic, which means 
