2019 AMC 10 B

Complete problem set with solutions and individual problem pages

Problem 16 Medium

In \triangle ABC with a right angle at C, point D lies in the interior of \overline{AB} and point E lies in the interior of \overline {BC} so that AC=CD, DE=EB, and the ratio AC:DE=4:3. What is the ratio AD:DB? (2019 AMC 10B Problem, Question#16)

  • A.

    2:3

  • B.

    2: \sqrt{5}

  • C.

    1:1

  • D.

    3: \sqrt{5}

  • E.

    3:2

Answer:A

Without loss of generality, let AC=CD=4 and DE=EB=3.

Let \angle A=\alpha and \angle B=\beta=90^\circ -\alpha . As \triangle ACD and \triangle DEB are isosceles, \angle ADC=\alpha and \angle BDE=\beta , Then \angle CDE=180^\circ-\alpha -\beta =90^\circ, so \triangle CDE is a 3-4-5 triang|e with CE=5.

Then CB=5+3=8, and \triangle ABC is a 1-2-\sqrt{5} triangle.

In isosceles triangles \triangle ACD and \triangle DEB, drop altitudes from C and E onto AB; denote the feet of these altitudes by P_C and P_E respectively. Then\triangle ACP_C\sim \triangle ABC by \rm AAA similarity, so we get that AP_{C}=P_{C}D= \frac{4}{\sqrt{5}}, and AD=2 \times \frac{4}{\sqrt{5}}, Similarly we get BD=2 \times \frac{6}{\sqrt{5}}, and AD:DB= \text {(A)}2:3.

Let AC=CD=4x, and DE=EB=3x. (For this solution, A is above C, and B is to the right of C). Also let \angle A=t^\circ, so \angle ACD=(180-2t)^\circ, which implies \angle DCB=(2t-90)^\circ.

Similarly, \angle B=(90-t)^\circ, which implies \angle BED=2t^\circ. This further implies that \angle DEC=(180-2t)^\circ

Now we see

that \angle CDE=180^\circ-\angle ECD-\angle DEC=180^\circ -2t^\circ +90^\circ -180^\circ +2t^\circ=90^\circ.

Thus \triangle CDE is a right triangle, with side lengths of 3x, 4x, and 5x (by the Pythagorean Theorem, or simply the Pythagorean triple 3-4-5). Therefore AC=4x (by definition)，BC=5x+3x=8x，

and AB=4 \sqrt{5}x, Hence \cos(2t^\circ)=2cos2t^\circ-1 (by the double angle fomua)，

giving 2(\frac {1}{\sqrt 5})^2-1=- \frac{3}{5},

By the Law of Cosines in \triangle BED, if BD=d, we have

d^{2}=(3x)^{2}+(3x)^{2}-2 \cdot \frac{-3}{5}(3x)(3x)

\Rightarrow d^{2}=18x^{2}+ \frac{54x^{2}}{5}= \frac{144x^{2}}{5}

\Rightarrow d= \frac{12x}{\sqrt{5}},

Now AD=AB-BD=4x\sqrt{5}- \frac{12x}{\sqrt{5}}= \frac{8x}{\sqrt{5}}.

Thus the answer is \frac{(\dfrac{8x}{\sqrt{5}})}{(\dfrac{12x}{\sqrt{5}})}= \frac{8}{12}= \text {(A)}2:3.

Draw a nice big diagram and measure. The answers to this problem are not very close, so it is quite easy to get to the correct answer by simply drawing a diagram.(Note: this strategy should only be used as a last resort!).

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