Track / Stage 3 / 139 of 260 #139 of 2000

Problem 139

AMC 10/12, early questions

Geometry Difficulty 3.5 Find the answer 67. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje · Slovenia

Problem:

V trikotniku $ABC$ je $\angle ACB = 32^\circ$ . Na nosilki stranice $AB$ ležita točki $D$ in $E$ , za kateri velja $|AD| = |AC|$ in $|BE| = |BC|$ (glej sliko). Koliko stopinj je velikost kota $\angle DCE$ ?

(A) 90
(B) 96
(C) 100
(D) 106
(E) 116

Your answer

Official solution

Solution:

Označimo $\angle BAC = \alpha$ . Tedaj je $\angle CBA = 180^\circ - 32^\circ - \alpha = 148^\circ - \alpha$ . Torej je $\angle CAD = 180^\circ - \angle BAC = 180^\circ - \alpha$ in $\angle EBC = 180^\circ - \angle CBA = \alpha + 32^\circ$ . Trikotnika $CAD$ in $EBC$ sta enakokraka z vrhoma pri $A$ in $B$ , zato je $\angle DCA = \frac{180^\circ - \angle CAD}{2} = \frac{\alpha}{2}$ in $\angle BCE = \frac{180^\circ - \angle EBC}{2} = 74^\circ - \frac{\alpha}{2}$ . Od tod izračunamo $\angle DCE = \angle DCA + \angle ACB + \angle BCE = 32^\circ + 74^\circ = 106^\circ$ .

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