AMC 10 · 2018 · #20
Grade 10 geometry-2dPick an answer.
The area condition sits on △ EMI but the question asks about CI, a length on a completely different segment, so nothing connects until the problem is broken into pieces. Tool #7 (Identify Subproblems) sets the route: first decide what shape △ EMI is, then convert its area into an actual length MI, then use that length to pin I down on AC. Tool #1 (Draw a Diagram) supplies the one auxiliary line that makes the last step trivial — the perpendicular from M to AC, whose foot is the midpoint of AC by symmetry. Tool #4 (Introduce a Variable) names AI and AE so the cyclic and area conditions become equations. Finally Tool #3 (Eliminate Possibilities) handles the last hurdle: the geometry allows two mirror positions for I, and the given AI ＞ AE eliminates one of them.
Read the symmetry at A
The isosceles right angle gives forty-five degree symmetry.
In an isosceles right triangle the median to the hypotenuse is the mirror line, so it splits the right angle into two equal 45° halves.
10.G-CO.C.10Draw A DiagramCyclic forces a 45-45-90
Lying on one circle forces an isosceles right triangle.
A circle carries an angle from one vertex to another: two points on the same arc see the same chord at the same angle, so the 45° at A reappears at both E and I.
Two points on the same arc see the same chord at the same angle, so the angle carries from one vertex to another.
▸ Why?
An angle at the circle measures the far arc, and the same chord always cuts off the same arc.
▸ Why?
Every point of the circle is one radius from the centre, which is what makes equal chords cut equal arcs.
Turn the area into a length
An area of two gives the side length.
Once you know the triangle is right-angled and isosceles, its area is just half of a leg squared, so the area hands you the leg directly.
6.G.A.1Identify SubproblemsDrop a perpendicular from M
Drop a perpendicular from the midpoint.
One extra perpendicular converts a slanted distance MI into a right triangle whose other two sides are half-lengths you already know.
10.G-CO.C.10Draw A DiagramPythagoras gives GI
Pythagoras gives the remaining length.
A known hypotenuse and one known leg pin the third side exactly, so MI=2 tells you precisely how far I strays from the midpoint.
10.G-SRT.C.8Identify SubproblemsUse AI ＞ AE to pick the root
Choosing the root and adding gives 12.
Symmetry leaves two mirror-image placements, and the single inequality AI ＞ AE is exactly the extra information needed to say which one the problem means.
9.A-CED.A.3Eliminate PossibilitiesA circle through four points copies angles from one corner to another, so the 45° at A turns △ EMI into a 45-45-90 triangle whose area instantly hands you a length.
- Read the symmetry at A
- Cyclic forces a 45-45-90
- Turn the area into a length
- Drop a perpendicular from M
- Pythagoras gives GI
- Use AI ＞ AE to pick the root
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