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

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

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

Page

Toolbox

Search

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

Latest revision as of 11:05, 7 June 2024

Contents

Problem

Solution 1

Solution 2

See also

@@ Line 3: / Line 3: @@
 Let <math>O</math> and <math>H</math> be the circumcenter and the orthocenter of an acute triangle <math>ABC</math>. Points <math>M</math> and <math>D</math> lie on side <math>BC</math> such that <math>BM = CM</math> and <math>\angle BAD = \angle CAD</math>. Ray <math>MO</math> intersects the circumcircle of triangle <math>BHC</math> in point <math>N</math>. Prove that <math>\angle ADO = \angle HAN</math>.
-==Solution==
+==Solution 1==
-THIS SOLUTION HAS NO DIAGRAM, SOMEONE WHO HAS ASYMPTOTE SKILLS PLEASE HELP. <math>-1</math>!!! because of the diagram rule
+<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>.
+<b>Claim:</b> <math>\triangle AA'N' \sim \triangle ADO</math>. The proof easily follows.
+<b>Proof:</b> Note that <math>\angle BAA'=\angle CAO=90^{\circ}-\angle ABC</math>. Then we have <math>\angle A'AN'=\angle BAD-\angle BAA'=\angle CAD-\angle CAO=\angle DAO</math>. So, it suffices to show that <cmath>\frac{AA'}{AN'}=\frac{AD}{AO}\rightarrow AA'\cdot AO=AN'\cdot AD.</cmath> Notice that <math>\triangle ABA' \sim \triangle AOC</math>, so that <cmath>\frac{AB}{AA'}=\frac{AO}{AC}\rightarrow AA'\cdot AO=AB\cdot AC.</cmath> Therefore, it suffices to show that <cmath>AB\cdot AC=AN'\cdot AD\rightarrow \frac{AB}{AN'}=\frac{AD}{AC}.</cmath> But it is easy to show that <math>\triangle BAN'\sim \triangle DAC</math>, implying the result. <math>\blacksquare</math>
+==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 <math>OM</math> intersects the circumcircle of <math>BHC</math> at <math>N'</math>, and let the foot of the A-altitude of <math>ABC</math> be <math>E</math>. Note that <math>\angle BHE=90-\angle HBE=90-90+\angle C=\angle C</math>. Likewise, <math>\angle CHE=\angle B</math>. So, <math>\angle BHC=\angle BHE+\angle CHE=\angle B+\angle C</math>.