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Difference between revisions of "2002 Indonesia MO Problems/Problem 4"

(created page w/ solution & categorization)
 
(added diagram)
 
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== Solution ==
 
== Solution ==
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 +
<asy>
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size(5cm);
 +
 +
draw(unitcircle);
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pair A, C, B;
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A = (-0.6, 0.8);
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C = (0.6, 0.8);
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B = (0.8, 0.6);
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dot(A);
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dot(C);
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dot(B);
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label("$A$", A, NW);
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label("$C$", C, N);
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label("$B$", B, SE);
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draw(C--B--A);
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pair Bp;
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Bp = (0.88, 0.8);
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dot(Bp, blue);
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label("$B'$", Bp, NE, blue);
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draw(A -- Bp);
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pair midAB, dirAB;
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midAB = (A + B)/2;
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dirAB = (B - A);
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path perpAB;
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perpAB = (midAB - dir(90)*dirAB) -- (midAB + dir(90)*dirAB);
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draw(perpAB, blue);
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pair Ep, dirABp;
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path perpABp;
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Ep = (A + Bp)/2;
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dirABp = (Bp - A);
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perpABp = (Ep - dir(90)*dirABp) -- (Ep + dir(90)*dirABp);
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dot(Ep, blue);
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label("$E'$", Ep, NW, blue);
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draw(perpABp, blue);
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pair Dp;
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Dp = intersectionpoint(perpAB, perpABp);
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dot(Dp, blue);
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label("$D'$", Dp, NW, blue);
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draw(Bp--Dp--B, red);
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draw(C--Dp--A, red);
 +
</asy>
  
 
We use the method of phantom points.
 
We use the method of phantom points.

Latest revision as of 21:01, 24 September 2020

Given a triangle $ABC$ with $AC > BC$. On the circumcircle of triangle $ABC$ there exists point $D$, which is the midpoint of arc $AB$ that contains $C$. Let $E$ be a point on $AC$ such that $DE$ is perpendicular to $AC$. Prove that $AE = EC + CB$.

Solution

[asy] size(5cm); draw(unitcircle); pair A, C, B; A = (-0.6, 0.8); C = (0.6, 0.8); B = (0.8, 0.6); dot(A); dot(C); dot(B); label("$A$", A, NW); label("$C$", C, N); label("$B$", B, SE); draw(C--B--A); pair Bp; Bp = (0.88, 0.8); dot(Bp, blue); label("$B'$", Bp, NE, blue); draw(A -- Bp); pair midAB, dirAB; midAB = (A + B)/2; dirAB = (B - A); path perpAB; perpAB = (midAB - dir(90)*dirAB) -- (midAB + dir(90)*dirAB); draw(perpAB, blue); pair Ep, dirABp; path perpABp; Ep = (A + Bp)/2; dirABp = (Bp - A); perpABp = (Ep - dir(90)*dirABp) -- (Ep + dir(90)*dirABp); dot(Ep, blue); label("$E'$", Ep, NW, blue); draw(perpABp, blue); pair Dp; Dp = intersectionpoint(perpAB, perpABp); dot(Dp, blue); label("$D'$", Dp, NW, blue); draw(Bp--Dp--B, red); draw(C--Dp--A, red); [/asy]

We use the method of phantom points.

Draw $AC$ and $BC$, and extend line $AC$ past $C$ to a point $B'$ such that $BC = B'C$. Draw point $E'$ at the midpoint of $AB'$, and $D'$ at the intersection of the perpendicular to $AC$ from $E'$ and the perpendicular bisector of $AB$.

Since $D'B = D'B', CB = CB', D'C = D'C$, we have $\triangle D'BC \cong \triangle D'B'C$ by side-side-side similarity. Then $\angle D'AC = \angle D'AE' = \angle D'B'E' = \angle D'B'C = \angle D'BC$, so $ADCB$ is cyclic.

In particular, since we have $AD' = B'D' = BD'$, we know that $D'$ must be the midpoint of the arc of the circumcircle of $\triangle ABC$ that contains point $C$, and since $D'$ was on the perpendicular to $AC$ from $E'$, we must have that $E'$ is the foot of the perpendicular of $D'$ to $AC$. But this uniquely identifies $D' = D, E' = E$, and we are done.

See also

2002 Indonesia MO (Problems)
Preceded by
Problem 3 		1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 Followed by
Problem 5
All Indonesia MO Problems and Solutions
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