Elementary school competition questions stumped middle school students, seriously exceeding the curr

Mondo Education Updated on 2024-02-14

Two primary school math competition questions were previously released:Only one inside angle of the triangle is known, and the angle of the other inside angle is foundIt is very difficult, and it is seriously out of the curriculum, and middle school students may not be able to do it, and even most middle school students will be stumped!

The solution of these two competition problems has two things in common:

All are requiredSuper class knowledge: in a right triangle, the middle line of the hypotenuse and the corresponding edge of the 30° angle are equal to half of the hypotenuse!

Structural proof is requiredConstruct an equilateral triangle

Fifth grade competition questions, [Bei Xiao Question Collection] Question 527:As shown in Figure 1,

d is a point on the bc side of abc, abc=45°, adc=60°, cd=2bd, find acd.

Superclass Analysis: Constructing Equilateral Triangles DEF!

Pass the point C as the perpendicular line CE of AD, and remember the midpoint of CD as F, which is connected to EF. See Figure 2.

de=df=ef, thereforedef is an equilateral triangle

bd=df=de, so bde is an isosceles triangle, i.e., there is ebd= bed=30°.

ebd= dce=30°, so be=ec.

From ebd=30°, ABE=BAE=15°, i.e., BE=AE=CE.

Therefore, ace is an isosceles right triangle, i.e., ace=45°,∠acd=75°

Question 520 of the 6th grade competition question [Bei Xiao Question Collection]:As shown in Figure 3,

Point D is the midpoint of the BC at the bottom edge of ABC, b=15°, c=30°, find bad.

Superclass Analysis: Constructing Equilateral Triangle BDE!

The perpendicular line of the CA extension line BE, which is connected to the DE. As shown in Figure 4.

In a right-angled triangle, the middle line of the hypotenuse and the 30° angle correspond to the right-angled side as half of the hypotenuse, so be=bd=de, that isBDE is an equilateral triangle

ABE=60°-15°=45°=BAE, so AEB is an isosceles right triangle.

be=ae=de, so aed is an isosceles triangle, i.e., dae= ade.

Note that aed=30°, so dae= ade=75°,∠bad=∠dae-∠bae=75°-45°

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