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1. PROBLEMA 2
(1) 𝐸𝑛 𝑒𝑙 𝑡𝑟𝑖á𝑛𝑔𝑢𝑙𝑜 𝐵𝑀𝑅, 𝑚𝑒𝑑𝑖𝑎𝑛𝑡𝑒 𝑡𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑒𝑙 𝑐𝑜𝑠𝑒𝑛𝑜:
𝐵𝑅2
= 𝑎2
+ (
𝑎√2
2
)
2
− 2𝑎.
𝑎√2
2
𝑐𝑜𝑠135
𝐵𝑅 =
𝑎√10
2
…(𝛼)
(2) 𝐴𝑛á𝑙𝑜𝑔𝑎𝑚𝑒𝑛𝑡𝑒 𝑒𝑛 𝑒𝑙 𝑡𝑟𝑖á𝑛𝑔𝑢𝑙𝑜 𝐴𝑅𝐶:
𝑅𝐶2
= (
𝑎√2
2
)
2
+ 1 −
2𝑎√2
2
𝑐𝑜𝑠45
𝑅𝐶 =
√2(𝑎 − 1)2 + 2
2
…(𝛽)
(3) 𝑀𝑒𝑑𝑖𝑎𝑛𝑡𝑒 𝑒𝑙 𝑡𝑒𝑜𝑟𝑒𝑚𝑎 𝑑𝑒 𝑃𝑖𝑡á𝑔𝑜𝑟𝑎𝑠, 𝑎𝑝𝑙𝑖𝑐𝑎𝑑𝑜 𝑎𝑙 𝑡𝑟𝑖á𝑛𝑔𝑢𝑙𝑜 𝐴𝐵𝐶, 𝑐𝑜𝑛 𝐵𝐶 𝑒𝑥𝑝𝑟𝑒𝑠𝑎𝑑𝑜
𝑝𝑜𝑟 𝑙𝑎 𝑎𝑑𝑖𝑐𝑖ó𝑛 𝑑𝑒 ( 𝛼) 𝑦 ( 𝛽):
(
𝑎√10
2
+
√2(𝑎 − 1)2 + 2
2
)
2
= 1 + 4𝑎2
⇔ 4𝑎4
− 12𝑎3
+ 9𝑎2
= 0
⇔ 𝑎2
(2𝑎 − 3)2
= 0
𝑑𝑒 𝑑𝑜𝑛𝑑𝑒: 𝑎 =
3
2
. 𝐶𝑜𝑚𝑜 𝐴𝐵 = 2𝑎, 𝑒𝑛𝑡𝑜𝑛𝑐𝑒𝑠 𝐴𝐵 𝑚𝑖𝑑𝑒 3 𝑢.
A
C
R
M
B
a
a
1
𝑎√2
2
𝑎√2
2
45º 45º
45º
135º