Track / Stage 4 / 318 of 340 #578 of 1964
Problem 578 AMC 12 late, AIME early Difficulty 5.0 Multiple choice
17 ⋅ 177 17 \cdot 177 17 ⋅ 177 a , b a, b a , b c c c h h h c + h , a + b , h c+h, a+b, h c + h , a + b , h
Pick one
A a right triangle B an acute triangle C an obtuse triangle D cannot be determined, the shape depends on the sizes of a , b , c a, b, c a , b , c
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Official solution [Solution] According to the Pythagorean theorem, we have a 2 + b 2 = c 2 a^{2}+b^{2}=c^{2} a 2 + b 2 = c 2 S = 1 2 a b = 1 2 c h .
S=\frac{1}{2} a b=\frac{1}{2} c h .
S = 2 1 ab = 2 1 c h .
Then( c + h ) 2 = c 2 + 2 c h + h 2 .
(c+h)^{2}=c^{2}+2 c h+h^{2} \text {. }
( c + h ) 2 = c 2 + 2 c h + h 2 .
Also, ( a + b ) 2 = a 2 + 2 a b + b 2 = a 2 + 2 c h + c 2 − a 2 = c 2 + 2 c h (a+b)^{2}=a^{2}+2 a b+b^{2}=a^{2}+2 c h+c^{2}-a^{2}=c^{2}+2 c h ( a + b ) 2 = a 2 + 2 ab + b 2 = a 2 + 2 c h + c 2 − a 2 = c 2 + 2 c h ∴ \therefore ∴ ( c + h ) 2 = ( a + b ) 2 + h 2 (c+h)^{2}=(a+b)^{2}+h^{2} ( c + h ) 2 = ( a + b ) 2 + h 2
Thus, the newly formed triangle is a right triangle.( A ) (A) ( A )
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Source: NuminaMath-1.5 ,
licensed Apache-2.0 .
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