AMC 10 · 2010 · #22
Grade 11 geometry-2dnumber-theoryPick an answer.
Tool #14 (Extreme Principle): the question asks for a largest value, so the job is two-sided — build one quadrilateral that is big, and prove nothing bigger exists. Tool #1 (Draw a Diagram): drawing BD splits ABCD into two triangles that share it, which is where the whole solution starts. Tool #7 (Identify Subproblems): apply the Law of Cosines separately in each triangle, then link them through the circle. Tool #4 (Introduce a Variable): name the sides a,b,c,d; the unknown angle turns out to be a variable you can delete rather than compute. Tool #3 (Eliminate Possibilities): once the geometry collapses to 2BD²=a²+b²+c²+d², the search over side lengths is finished by ruling out 11, 13, and the case where 14 is unused.
Split the quadrilateral on BD
The diagonal splits the shape into two triangles.
BD is the one segment both triangles own, so it is the quantity worth writing down twice.
6.EE.A.2Draw A DiagramUse the circle on the angles
The circle makes the two angles supplementary.
The two opposite corners see complementary arcs, so between them they account for the full circle.
The two opposite corners see complementary arcs, so between them they account for the whole circle.
▸ Why?
An angle at the circle measures the far arc, and the two far arcs together make up the entire circle.
▸ Why?
A full turn is a fixed total, so the two angles are forced to add to a straight angle.
Law of Cosines in each triangle
Each triangle gives one cosine equation.
Two triangles sharing a side give two readings of the same length, and the circle makes their angles move together.
11.G-SRT.D.10Identify SubproblemsAdd, and the angle dies
Adding them makes the angle vanish.
ad=bc is precisely the condition under which the unknown angle cancels instead of having to be found.
6.EE.A.3Introduce A VariableRule out 11 and 13
Large primes are ruled out as sides.
A big prime side needs a partner that is a multiple of it, and the next multiple already breaks the limit of 14.
4.OA.B.4Eliminate PossibilitiesSet the bar without 14
Without the top side the total falls short.
Grabbing the four biggest leftovers is the most generous thing that could happen without 14, so it bounds the whole case.
6.EE.A.2Eliminate PossibilitiesWith 14 in, everything is forced
With it, the other three are forced.
Once 14 drags a 7 in with it, the last two sides must be d and 2d, so pushing d to its ceiling finishes the search.
4.OA.B.4Extreme PrincipleCheck the shape really exists
That quadrilateral really exists.
The longest side has to be reachable by walking the other three, or the four lengths never meet up on a circle.
7.G.A.2Draw A DiagramTurn the maximum into a length
The diagonal is the root of half that total.
One square root converts the biggest sum of squares into the longest diagonal.
8.EE.A.2Extreme PrincipleThe odd-looking condition BC · CD=AB · DA exists to make the unknown angle cancel; after it does, 2BD² is just the sum of the four squared sides, and the geometry problem becomes a hunt for the biggest four distinct numbers under 15 whose two pair-products agree.
- Split the quadrilateral on BD
- Use the circle on the angles
- Law of Cosines in each triangle
- Add, and the angle dies
- Rule out 11 and 13
- Set the bar without 14
- With 14 in, everything is forced
- Check the shape really exists
- Turn the maximum into a length