AMC 10 · 2006 · #23
Grade 9 geometry-2d
Pick an answer.
Three distances from one hidden point are hard to handle synthetically, so convert the whole picture to algebra (Tool #13). Placing the right angle at the origin (Tool #1) makes the perpendicular legs and their equal length automatic, and naming P=(x,y) (Tool #4) turns PA, PB, PC into three equations. Those three equations share the same x²+y², so subtracting pairs collapses them into one quadratic in s². That quadratic has two roots, and here is the part worth care: neither root is algebraic junk, because each one really does produce a point at distances 6, 11, 7 from the vertices. The word "inside" is the only thing that separates them, so Tool #3 is used with an interior test, not with a hand wave. Tool #17 is held in reserve for an independent check: rotating the figure 90° about C reproduces the same two candidates as two possible angles at P.
Put the right angle at the origin
Placing the right angle at the origin gives clean coordinates.
Choosing the axes along the legs spends the right angle and the equal sides immediately, leaving only the three distances to encode.
8.G.B.8Draw A DiagramTurn the three distances into equations
The three distances become three equations.
The distance formula is the Pythagorean theorem in disguise, so each distance becomes one clean equation.
8.G.B.8Convert To AlgebraSubtract to kill the squares
Subtracting pairs kills the squared terms.
All three equations carry the identical block x²+y², so subtracting throws the hard part away and what is left is linear.
All three equations carry the identical squared block, so subtracting throws the hard part away and leaves straight lines.
▸ Why?
Subtracting two quantities that share the same piece removes that piece entirely.
▸ Why?
Each distance is the hypotenuse of a right triangle of coordinate gaps, which is where the shared squares come from.
Feed the coordinates back into PC = 6
Feeding the coordinates back gives one quadratic.
The one equation left over is the one that finally pins the size of the triangle down.
9.A-CED.A.1Introduce A VariableSolve the quadratic in s squared
It has two roots, both real.
The √2 promised by the answer's shape was hiding inside the discriminant, since 14112=84² · 2.
9.A-REI.B.4Introduce A VariableBoth roots are honest points
One puts the point outside, so only one survives.
The equations only know the three distances, so the word "inside" is extra information, and here it is the information that decides.
9.A-CED.A.3Eliminate PossibilitiesCheck the surviving root really works
Checking the survivor confirms the point is inside.
Eliminating the rival only leaves a candidate; showing the point truly lands inside is what makes it the answer.
9.A-CED.A.3Eliminate PossibilitiesRead off a and b
Reading off the two numbers gives 127, choice (E).
A number can be written as integer plus integer times √2 in only one way, so a and b are forced, not chosen.
9.N-RN.B.3Introduce A VariablePut the right angle at the origin, subtract the distance equations to make the squares cancel, and remember that the algebra offers two real points — the word "inside" is what picks the right one.
- Put the right angle at the origin
- Turn the three distances into equations
- Subtract to kill the squares
- Feed the coordinates back into PC = 6
- Solve the quadratic in s squared
- Both roots are honest points
- Check the surviving root really works
- Read off a and b