AMC 10 Step by Step

2007 AMC 10B Problem 21

P21-25~5 minPrint

Right △ABC\triangle ABC has AB=3,BC=4,AB=3, BC=4, and AC=5.AC=5. Square XYZWXYZW is inscribed in △ABC\triangle ABC with XX and YY on AC‾,W\overline{AC}, W on AB‾,\overline{AB}, and ZZ on BC‾.\overline{BC}. What is the side length of the square?

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Key insight

The small triangle above the square has the same shape as the big one, so compare width to height: s over 5 equals (height minus s) over height.

Solution

Here the square sits on the long side ACAC (length 55), with its top two corners WW and ZZ touching the two short sides. Let the side of the square be ss.

1. Height of the big triangle over ACAC. The area of the right triangle is 12×3×4=6\frac12 \times 3 \times 4 = 6. Using AC=5AC = 5 as the base instead, 12×5×h=6\frac12 \times 5 \times h = 6, so
h=125.h = \frac{12}{5}.
2. The small triangle on top. The top side WZWZ of the square is parallel to ACAC. So triangle BWZBWZ has the same shape as triangle BACBAC, just smaller.
3. Compare base and height. Triangle BWZBWZ has base WZ=sWZ = s. Its height is the part of hh above the square, h−sh - s. In same-shaped triangles, base divided by height is the same:
sh−s=5h.\frac{s}{h - s} = \frac{5}{h}.
4. Put in h=125h = \frac{12}{5} and solve. Cross-multiply: s×h=5(h−s)s \times h = 5(h - s), so
125s=12−5s.\frac{12}{5}s = 12 - 5s.
Multiply by 55: 12s=60−25s12s = 60 - 25s, so 37s=6037s = 60 and
s=6037.s = \frac{60}{37}.
5. Check the size: 6037≈1.62\frac{60}{37} \approx 1.62, a bit smaller than 127≈1.71\frac{12}{7} \approx 1.71 from the corner square, which makes sense.

The answer is (B) 6037\boxed{\textbf{(B)}\ \frac{60}{37}}.

Why this works

A line parallel to one side of a triangle cuts off a smaller triangle of the same shape, and the cleanest comparison is base against height. For a square standing on a side of length bb whose opposite height is hh, this always gives s=bhb+hs = \frac{bh}{b + h}; here 5×12/55+12/5=1237/5=6037\frac{5 \times 12/5}{5 + 12/5} = \frac{12}{37/5} = \frac{60}{37}. When a figure is built on the long side of a right triangle, find the height to that side first.

Alternative approach

Walk along the long side ACAC. It is split into three pieces: AXAX, the square's bottom XY=sXY = s, and YCYC.

1. Triangle AXWAXW has a right angle at XX and the same shape as the big triangle (they share angle AA). In the big triangle, the side across from AA is 44 and the side next to AA is 33. So WX:AX=4:3WX : AX = 4 : 3, which means AX=34sAX = \frac{3}{4}s.
2. In the same way, triangle CYZCYZ shares angle CC, where the ratio is 3:43 : 4, so CY=43sCY = \frac{4}{3}s.
3. Add up the pieces:
34s+s+43s=9+12+1612s=3712s=5,\frac34 s + s + \frac43 s = \frac{9 + 12 + 16}{12}s = \frac{37}{12}s = 5,
so s=6037s = \frac{60}{37}.

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

The square sits on the hypotenuse ACAC, with its top side WZWZ parallel to ACAC and its top corners on the legs. Let the side length be ss.

Draw the altitude from BB to ACAC. Its length is
h=AB⋅BCAC=3⋅45=125.h = \frac{AB \cdot BC}{AC} = \frac{3 \cdot 4}{5} = \frac{12}{5}.

Because WZ∥ACWZ \parallel AC, triangle BWZBWZ is similar to triangle BACBAC. The altitude of BWZBWZ from BB is the part of hh above the square, namely h−sh - s, and its base is WZ=sWZ = s. Corresponding bases and altitudes are in the same ratio:
s5=h−sh.\frac{s}{5} = \frac{h - s}{h}.
Cross-multiplying, sh=5h−5ssh = 5h - 5s, so s(h+5)=5hs(h + 5) = 5h and
s=5hh+5=5⋅125125+5=12375=6037.s = \frac{5h}{h + 5} = \frac{5 \cdot \frac{12}{5}}{\frac{12}{5} + 5} = \frac{12}{\frac{37}{5}} = \frac{60}{37}.

The answer is (B) 6037\boxed{\textbf{(B)}\ \frac{60}{37}}.

Review in the AoPS books

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Problem © Mathematical Association of America (MAA), American Mathematics Competitions. Reproduced for non-commercial educational use. Solution and commentary are original to this site.