AMC 10 · 2020 · #10

Grade 10 geometry-2d
Archetype Similar-Triangle Length / Area Ratio └ Sub-archetype Synthetic Similar-Triangle Length or Area
coordinate-geometryinscribed-anglesimilar-trianglespythagorean-theorem identify-subproblemsconvert-to-algebra ↑ Prerequisites: pythagorean-theoremsimilar-triangles
📏 Medium solution 💡 3 insights
Problem
A square has sides of length 1. The circle inscribed in the square touches all four sides, and it meets one side at a point. The segment from the opposite vertex to that point crosses the circle a second time. Find the length from that vertex to the second crossing.

Pick an answer.

(A)
$\frac{\sqrt5}{12}$
(B)
$\frac{\sqrt5}{10}$
(C)
$\frac{\sqrt5}{9}$
(D)
$\frac{\sqrt5}{8}$
(E)
$\frac{2\sqrt5}{15}$
How to solve
Strategy Draw a Diagram

The phrase "meets the circle again" gives no length to hold on to, so the first job is to pin the picture down: coordinates fix the center, the radius, and the tangency point M exactly. The second job is one auxiliary point. Drawing in the far end of the diameter through M makes the angle at P a right angle, and a right angle turns the vague second intersection into the foot of a perpendicular, which similar right triangles measure in a single ratio. No trigonometry and no circle equation are needed once that one extra point is on the page.

1STEP 1

Put it on the axes

Write the square and circle in coordinates.

A=(0,0), B=(1,0), C=(1,1), D=(0,1), O=(1/2,1/2), r=1/2, M=(1/2,1)
2STEP 2

Measure the whole segment

Measure the whole segment first.

AM=√((1/2)²+1²)=√(5/4)=√(5)/2
3STEP 3

Draw the far end of the diameter

The angle on the diameter is right.

N=(1/2,0), MN is a diameter → ∠ MPN = 90^° → NP ⊥ AM
4STEP 4

Spot the similar triangles

Two right triangles are similar.

△ APN ∼ △ ANM → AP/AN=AN/AM → AP=AN²/AM
5STEP 5

Put in the numbers

The proportion gives root five over ten.

AP=AN²/AM=1/4/√(5)/2=1/2√(5)=√(5)/10
Answer
√5/10
Numerically √(5)/10 is about 0.224. Walking from A along the ray toward M, the first circle point cannot be closer than AO-r=√(2)/2-1/2, about 0.207, so AP has to be a little more than 0.207; 0.224 fits, while choice (A), about 0.186, would put P inside the empty gap and is impossible. The leftover chord is PM=AM-AP≈ 1.118-0.224=0.894, comfortably less than the diameter 1, as every chord must be. A separate check by the power of the point A confirms the value exactly: AP · AM=AO²-r²=1/2-1/4=1/4, so AP=1/4÷√(5)/2=√(5)/10.
💡Key takeaway

When a line hits a circle at a second, unnamed point, draw the diameter from the point you do know: the angle on a diameter is right, so the unknown point becomes a perpendicular foot you can measure with similar triangles.

  • Put the square on the axes
  • Measure the whole segment AM
  • Draw in the far end of the diameter
  • Spot the two similar right triangles
  • Put the numbers in and clear the root

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