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24 Aug 2021, 08:00
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D
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Difficulty:
75%
(hard)
Question Stats:
53% (01:48) correct
47%
(02:42)
wrong
based on 149
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Date
Time
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Not Attempted Yet
In the square ABCD shown above, what is the measure of an yellow angle ?
A. 15°
B. 20°
C. 25°
D. 30°
E. 45°
M37-25
|
M37-25
Attachment:
Untitled12.png
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22 Nov 2021, 05:51
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Official Solution:
In the square ABCD shown above, what is the measure of an yellow angle ?
A. \(15°\)
B. \(20°\)
C. \(25°\)
D. \(30°\)
E. \(45°\)
Construct triangle BFC congruent to triangle DEC on side BC (simply put draw triangle BFC on BC the same way DEC is drawn on DC). Notice that \(\angle BFC =180°-(15°+15°)=150°\).
Since BFC and DEC are congruent, then \(CE = CF\).
Since \(\angle DCE = \angle BCF = 15°\), then \(\angle ECF = 90°-(\angle DCE + \angle BCF) = 60°\)
Thus, triangle EFC is isosceles (\(CE = CF\)) and \(\angle ECF =60°\), which means that triangle EFC is equilateral and all its angles are \(60°\).
Next, since the sum of the angles around point F must be \(360°\), then \(\angle BFE = 360° - (\angle BFC + \angle CFE)=360° - (150° + 60°)=150°\)
Now, consider triangles EFB and CFB. In these triangles, two sides and angle between them are equal (BF is shared side, FC = FE because EFC is equilateral, and \(\angle BFE = \angle BFC=150°\), thus EFB and CFB are congruent.
Finally, since EFB and CFB are congruent, then \(\angle FBC=\angle FBE=15°\) and therefore, \(\angle CBE = \angle FBC+\angle FBE=30°\)
Answer: D
In the square ABCD shown above, what is the measure of an yellow angle ?
A. \(15°\)
B. \(20°\)
C. \(25°\)
D. \(30°\)
E. \(45°\)
Construct triangle BFC congruent to triangle DEC on side BC (simply put draw triangle BFC on BC the same way DEC is drawn on DC). Notice that \(\angle BFC =180°-(15°+15°)=150°\).
Since BFC and DEC are congruent, then \(CE = CF\).
Since \(\angle DCE = \angle BCF = 15°\), then \(\angle ECF = 90°-(\angle DCE + \angle BCF) = 60°\)
Thus, triangle EFC is isosceles (\(CE = CF\)) and \(\angle ECF =60°\), which means that triangle EFC is equilateral and all its angles are \(60°\).
Next, since the sum of the angles around point F must be \(360°\), then \(\angle BFE = 360° - (\angle BFC + \angle CFE)=360° - (150° + 60°)=150°\)
Now, consider triangles EFB and CFB. In these triangles, two sides and angle between them are equal (BF is shared side, FC = FE because EFC is equilateral, and \(\angle BFE = \angle BFC=150°\), thus EFB and CFB are congruent.
Finally, since EFB and CFB are congruent, then \(\angle FBC=\angle FBE=15°\) and therefore, \(\angle CBE = \angle FBC+\angle FBE=30°\)
Answer: D
General Discussion
24 Aug 2021, 08:38
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Bunuel
Wow, It's a famous problem;
There are 2 ways to solve it 1) Via Construction and pure geometry or 2) Via Trignometry;
The trigonometry method is a little complex; Although when I did it for the first time I used trigonometry.
Let's construct a triangle similar to DEC with BC as its base.[Please refer to the attachment]
So <OCB =<OBC =15';
That makes <ECO = 60';
As we know OC and EC are the same; So it's isosceles triangle with an angle 60'; So It is an equilateral triangle.
Now <EOC=60';
So <EOB=150' (Sum of all angles is 360' around a point)
As OE=OB <EBO=15';
Hence <EBC=30';
IMO D
Attachments
PB20.png [ 26.54 KiB | Viewed 4616 times ]
