AMC 10 · 2008 · #20

Grade 10 geometry-2d

inradiusangle-bisector-theoremarea-triangles identify-subproblemsconvert-to-algebra ↑ Prerequisites: pythagorean-theoremarea-triangles

📏 Long solution 💡 3 insights

Problem

A segment from the right-angle vertex bisects it and splits a 3-4-5 triangle into two pieces. Each piece has its own inscribed circle. Find the ratio of the two radii.

Pick an answer.

(A)

$\frac{1}{28}\left(10-\sqrt{2}\right)$

(B)

$\frac{3}{56}\left(10-\sqrt{2}\right)$

(C)

$\frac{1}{14}\left(10-\sqrt{2}\right)$

(D)

$\frac{5}{56}\left(10-\sqrt{2}\right)$

(E)

$\frac{3}{28}\left(10-\sqrt{2}\right)$

How to solve

Strategy Identify Subproblems

An inradius is not a length you can read off a picture, so the first job is to trade it for something measurable. The identity [△]=rs does that: it rewrites r_a/r_b as an area ratio times a reversed semiperimeter ratio. Both of those need the same three numbers, AD, BD, and CD, so the whole problem collapses into one subproblem: locate D. That subproblem has a trigonometry-free handle, because a point on an angle bisector is equally far from both sides of the angle. Naming that one distance t produces the areas, the cevian, and the two base lengths in a few lines. A size estimate at the end keeps the algebra honest.

1STEP 1

Find where the right angle is

The side lengths reveal a right angle and the whole area.

3²+4²=25=5² ⟹ ∠ ACB=90°, [ABC]=1/2 · 3 · 4=6

2STEP 2

Read the bisector as an equal-distance line

The bisector keeps an equal distance to both sides.

△ CPD ≅ △ CQD ⟹ DP=DQ=t

3STEP 3

Let the areas pin down t

The areas then pin that distance down.

3t/2+4t/2=6 ⟹ t=12/7, [ACD]=18/7, [BCD]=24/7

4STEP 4

Get CD as the diagonal of a square

That makes the cut a square's diagonal.

CD=√(t²+t²)=t√(2)=12√(2)/7

5STEP 5

Split AB with the same areas

The same areas split the long side into two pieces.

h=(2 · 6)/5=12/5, AD=36/7·5/12=15/7, BD=48/7·5/12=20/7, AD+BD=5

6STEP 6

Trade each inradius for area over semiperimeter

Each inradius is an area over a semiperimeter.

r=[△]/s; r_a=(18/7)/((18+6√(2))/7)=3/(3+√(2)), r_b=(24/7)/((24+6√(2))/7)=4/(4+√(2))

7STEP 7

Divide and clear the radical

Dividing and clearing the radical gives the ratio.

r_a/r_b=(3(4+√(2)))/(4(3+√(2)))=(12+3√(2))/(12+4√(2))·(12-4√(2))/(12-4√(2))=(120-12√(2))/112=3/28(10-√(2))

8STEP 8

Trap the size, then read the choice

Bounding its size confirms choice (C).

11/12 ＜ r_a/r_b ＜ 1, 3/28(10-√(2))≈ 0.9199

Answer

3/28(10-√(2))

Coordinates give an independent check on D. Put C at the origin with A=(0,3) and B=(4,0); the bisector of the right angle is the line y=x, and AB is 3x+4y=12, so D=(12/7,12/7). Both coordinates equal the t found in step 3, and the distance from the origin is 12√(2)/7, matching CD. Numerically r_a=3/(3+√(2))≈ 0.6796 and r_b=4/(4+√(2))≈ 0.7388, so the ratio is 0.9199, exactly 3/28(10-√(2)). The ordering is also the right shape: △ ACD is the smaller piece, so its incircle should be smaller, but only slightly, since the two triangles share the long side CD and both have a 45° angle at C. Choice (D), about 0.767, is the trap for anyone who stops at the base ratio AD/BD=3/4 and grabs the nearest listed number; the semiperimeter factor s_b/s_a≈ 1.227 is exactly what the shared side contributes and is what pushes the true answer up to 0.92.

💡Key takeaway

An inradius is just area divided by half the perimeter, so a ratio of two inradii is a ratio of areas times a ratio of perimeters, and one well-chosen distance gives you all of them.

Find where the right angle is
Read the bisector as an equal-distance line
Let the areas pin down t
Get CD as the diagonal of a square
Split AB with the same areas
Trade each inradius for area over semiperimeter
Divide and clear the radical
Trap the size, then read the choice