Difference between revisions of "2017 USAJMO Problems/Problem 5"

(Solution 1)
(Solution 1: added diagram)
 
Line 4: Line 4:
  
 
==Solution 1==
 
==Solution 1==
 +
<asy>
 +
import olympiad;
 +
 +
unitsize(100);
 +
 +
pair pA = dir(120);
 +
pair pB = dir(225);
 +
pair pC = dir(315);
 +
pair pO = origin;
 +
pair pH = orthocenter(pA, pB, pC);
 +
pair pM = midpoint(pB--pC);
 +
pair dD = bisectorpoint(pB, pA, pC);
 +
pair pD = extension(pA, dD, pB, pC);
 +
pair pN = intersectionpoints(pM--(3*pO-2*pM), circumcircle(pB, pH, pC))[0];
 +
pair dHprime = foot(pH, pB, pC);
 +
pair pHprime = 2*dHprime-pH;
 +
pair dAprime = foot(pA, pB, pC);
 +
pair pAprime = 2*dAprime-pA;
 +
pair dNprime = foot(pN, pB, pC);
 +
pair pNprime = 2*dNprime-pN;
 +
 +
draw(pA--pB--pC--cycle);
 +
draw(unitcircle, blue);
 +
draw(pH--pHprime, magenta);
 +
draw(pA--pD, red);
 +
draw(circumcircle(pB, pH, pC), blue);
 +
draw(pA--pH, blue);
 +
draw(pA--pN, magenta);
 +
draw(pA--pO, blue);
 +
draw(pD--pNprime, magenta);
 +
 +
dot("A", pA, NW);
 +
dot("B", pB, W);
 +
dot("C", pC, E);
 +
dot("O", pO, SW, blue);
 +
dot("H", pH, SW, blue);
 +
dot("N", pN, NE, magenta);
 +
dot("M", pM, S, red);
 +
dot("D", pD, S, red);
 +
dot("H", pHprime, SW, magenta);
 +
dot("A", pAprime, SW);
 +
dot("N", pNprime, S, magenta);
 +
</asy>
 +
(original diagram by [[User:integralarefun|integralarefun]])
 +
 
It's well known that the reflection of <math>H</math> across <math>\overline{BC}</math>, <math>H'</math>, lies on <math>(ABC)</math>. Then <math>(BHC)</math> is just the reflection of <math>(BH'C)</math> across <math>\overline{BC}</math>, which is equivalent to the reflection of <math>(ABC)</math> across <math>\overline{BC}</math>. Reflect points <math>A</math> and <math>N</math> across <math>\overline{BC}</math> to points <math>A'</math> and <math>N'</math>, respectively. Then <math>N'</math> is the midpoint of minor arc <math>\overarc{BC}</math>, so <math>A, D, N'</math> are collinear in that order. It suffices to show that <math>\angle AA'N'=\angle ADO</math>.
 
It's well known that the reflection of <math>H</math> across <math>\overline{BC}</math>, <math>H'</math>, lies on <math>(ABC)</math>. Then <math>(BHC)</math> is just the reflection of <math>(BH'C)</math> across <math>\overline{BC}</math>, which is equivalent to the reflection of <math>(ABC)</math> across <math>\overline{BC}</math>. Reflect points <math>A</math> and <math>N</math> across <math>\overline{BC}</math> to points <math>A'</math> and <math>N'</math>, respectively. Then <math>N'</math> is the midpoint of minor arc <math>\overarc{BC}</math>, so <math>A, D, N'</math> are collinear in that order. It suffices to show that <math>\angle AA'N'=\angle ADO</math>.
  

Latest revision as of 11:05, 7 June 2024

Problem

Let $O$ and $H$ be the circumcenter and the orthocenter of an acute triangle $ABC$. Points $M$ and $D$ lie on side $BC$ such that $BM = CM$ and $\angle BAD = \angle CAD$. Ray $MO$ intersects the circumcircle of triangle $BHC$ in point $N$. Prove that $\angle ADO = \angle HAN$.

Solution 1

[asy] import olympiad;  unitsize(100);  pair pA = dir(120); pair pB = dir(225); pair pC = dir(315); pair pO = origin; pair pH = orthocenter(pA, pB, pC); pair pM = midpoint(pB--pC); pair dD = bisectorpoint(pB, pA, pC); pair pD = extension(pA, dD, pB, pC); pair pN = intersectionpoints(pM--(3*pO-2*pM), circumcircle(pB, pH, pC))[0]; pair dHprime = foot(pH, pB, pC); pair pHprime = 2*dHprime-pH; pair dAprime = foot(pA, pB, pC); pair pAprime = 2*dAprime-pA; pair dNprime = foot(pN, pB, pC); pair pNprime = 2*dNprime-pN;  draw(pA--pB--pC--cycle); draw(unitcircle, blue); draw(pH--pHprime, magenta); draw(pA--pD, red); draw(circumcircle(pB, pH, pC), blue); draw(pA--pH, blue); draw(pA--pN, magenta); draw(pA--pO, blue); draw(pD--pNprime, magenta);  dot("\(A\)", pA, NW); dot("\(B\)", pB, W); dot("\(C\)", pC, E); dot("\(O\)", pO, SW, blue); dot("\(H\)", pH, SW, blue); dot("\(N\)", pN, NE, magenta); dot("\(M\)", pM, S, red); dot("\(D\)", pD, S, red); dot("\(H'\)", pHprime, SW, magenta); dot("\(A'\)", pAprime, SW); dot("\(N'\)", pNprime, S, magenta); [/asy] (original diagram by integralarefun)

It's well known that the reflection of $H$ across $\overline{BC}$, $H'$, lies on $(ABC)$. Then $(BHC)$ is just the reflection of $(BH'C)$ across $\overline{BC}$, which is equivalent to the reflection of $(ABC)$ across $\overline{BC}$. Reflect points $A$ and $N$ across $\overline{BC}$ to points $A'$ and $N'$, respectively. Then $N'$ is the midpoint of minor arc $\overarc{BC}$, so $A, D, N'$ are collinear in that order. It suffices to show that $\angle AA'N'=\angle ADO$.

Claim: $\triangle AA'N' \sim \triangle ADO$. The proof easily follows.

Proof: Note that $\angle BAA'=\angle CAO=90^{\circ}-\angle ABC$. Then we have $\angle A'AN'=\angle BAD-\angle BAA'=\angle CAD-\angle CAO=\angle DAO$. So, it suffices to show that \[\frac{AA'}{AN'}=\frac{AD}{AO}\rightarrow AA'\cdot AO=AN'\cdot AD.\] Notice that $\triangle ABA' \sim \triangle AOC$, so that \[\frac{AB}{AA'}=\frac{AO}{AC}\rightarrow AA'\cdot AO=AB\cdot AC.\] Therefore, it suffices to show that \[AB\cdot AC=AN'\cdot AD\rightarrow \frac{AB}{AN'}=\frac{AD}{AC}.\] But it is easy to show that $\triangle BAN'\sim \triangle DAC$, implying the result. $\blacksquare$

Solution 2

[asy] size(9cm); pair A = dir(130);  pair B = dir(220);  pair C = dir(320);  draw(unitcircle, lightblue);  pair P = dir(-90);  pair Q = dir(90); pair D = extension(A, P, B, C);  pair O = origin;  pair M = extension(B, C, O, P);  pair N = 2*M-P;  draw(A--B--C--cycle, lightblue);  draw(A--P--Q, lightblue);  draw(A--N--D--O--A, lightblue);  draw(A--D--N--O--cycle, red);  dot("$A$", A, dir(A));  dot("$B$", B, dir(B));  dot("$C$", C, dir(C));  dot("$P$", P, dir(P));  dot("$Q$", Q, dir(Q));  dot("$D$", D, dir(225));  dot("$O$", O, dir(315));  dot("$M$", M, dir(315));  dot("$N$", N, dir(315)); [/asy]

Suppose ray $OM$ intersects the circumcircle of $BHC$ at $N'$, and let the foot of the A-altitude of $ABC$ be $E$. Note that $\angle BHE=90-\angle HBE=90-90+\angle C=\angle C$. Likewise, $\angle CHE=\angle B$. So, $\angle BHC=\angle BHE+\angle CHE=\angle B+\angle C$. $BHCN'$ is cyclic, so $\angle BN'C=180-\angle BHC=180-\angle B-\angle C=\angle A$. Also, $\angle BAC=\angle A$. These two angles are on different circles and have the same measure, but they point to the same line $BC$! Hence, the two circles must be congruent. (This is also a well-known result)

We know, since $M$ is the midpoint of $BC$, that $OM$ is perpendicular to $BC$. $AH$ is also perpendicular to $BC$, so the two lines are parallel. $AN$ is a transversal, so $\angle HAN=\angle ANO$. We wish to prove that $\angle ANO=\angle ADO$, which is equivalent to $AOND$ being cyclic.

Now, assume that ray $OM$ intersects the circumcircle of $ABC$ at a point $P$. Point $P$ must be the midpoint of $\stackrel{\frown}{BC}$. Also, since $AD$ is an angle bisector, it must also hit the circle at the point $P$. The two circles are congruent, which implies $MN=MP\implies ND=DP\implies$ NDP is isosceles. Angle ADN is an exterior angle, so $\angle ADN=\angle DNP+\angle DPO=2\angle DPO$. Assume WLOG that $\angle B>\angle C$. So, $\angle DPO=\angle APO=\frac{\angle B+\angle C}{2}-\angle C=\frac{\angle B-\angle C}{2}$. In addition, $\angle AON=\angle AOP=\angle AOB+\angle BOP=2\angle C+\angle A$. Combining these two equations, $\angle AON+\angle ADN=\angle B-\angle C+2\angle C+\angle A=\angle A+\angle B+\angle C=180$.

Opposite angles sum to $180$, so quadrilateral $AOND$ is cyclic, and the condition is proved.

-william122

See also

2017 USAJMO (ProblemsResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6
All USAJMO Problems and Solutions