2019 AMC 10A Problem 13

P11-15~4 minPrint

Circles Angles & Polygons Triangle Centers & Cevians Set up the equation/formula and compute; no special trick needed

Let $\triangle ABC$ be an isosceles triangle with $BC = AC$ and $\angle ACB = 40^{\circ}$ . Construct the circle with diameter $\overline{BC}$ , and let $D$ and $E$ be the other intersection points of the circle with the sides $\overline{AC}$ and $\overline{AB}$ , respectively. Let $F$ be the intersection of the diagonals of the quadrilateral $BCDE$ . What is the degree measure of $\angle BFC ?$

Show solution without answering

Key insight

Any point on a circle sees the diameter at a right angle, so BD and CE are both heights of the triangle. Then triangle BFC has angles 50 and 20.

Solution

1. Find the triangle's angles. Since $BC = AC$ , the angles at $A$ and $B$ are equal. They share $180^\circ - 40^\circ = 140^\circ$ , so $\angle A = \angle B = 70^\circ$ .
2. Use the circle. A key fact: if $BC$ is a diameter of a circle and $X$ is any other point on that circle, then $\angle BXC = 90^\circ$ . Both $D$ and $E$ are on this circle, so
$\angle BDC = 90^\circ \quad\text{and}\quad \angle BEC = 90^\circ.$
3. What this means. $D$ is on side $AC$ and $BD$ meets $AC$ at a right angle, so $BD$ is the height from $B$ . In the same way, $CE$ is the height from $C$ to side $AB$ .
4. Find the diagonals. Go around the quadrilateral in the order $B \to C \to D \to E$ . The diagonals join opposite corners: $B$ with $D$ , and $C$ with $E$ . So $F$ is where $BD$ and $CE$ cross.
5. Angle $FBC$ . In the right triangle $BDC$ , the angle at $C$ is $40^\circ$ , so the angle at $B$ is $90^\circ - 40^\circ = 50^\circ$ .
6. Angle $FCB$ . In the right triangle $BEC$ , the angle at $B$ is $70^\circ$ , so the angle at $C$ is $90^\circ - 70^\circ = 20^\circ$ .
7. Finish in triangle $BFC$ .
$\angle BFC = 180^\circ - 50^\circ - 20^\circ = 110^\circ.$

The answer is $\boxed{\textbf{(D)}\ 110}$ .

Why this works

A circle drawn on a side as its diameter is a signal for right angles: every other point of that circle sees the side at $90^\circ$ . So the extra points $D$ and $E$ are the feet of two heights of the triangle. After that it is just the fact that angles in a triangle add to $180^\circ$ , used three times.

Alternative approach

Look at the quadrilateral $AEFD$ . It has right angles at $E$ and $D$ (from the heights) and the angle $70^\circ$ at $A$ . The four angles of a quadrilateral add to $360^\circ$ , so $\angle EFD = 360^\circ - 90^\circ - 90^\circ - 70^\circ = 110^\circ$ . The angle $\angle BFC$ is directly opposite $\angle EFD$ where the two diagonals cross, so it is also $110^\circ$ .

Concise solution

A shorter write-up, the way contest solutions are usually written. It may take a different route from the solution above.

Since $BC = AC$ , the base angles at $A$ and $B$ are equal: $\angle A = \angle B = \frac{180^\circ - 40^\circ}{2} = 70^\circ$ .

$D$ and $E$ lie on the circle with diameter $BC$ , so $\angle BDC = 90^\circ$ and $\angle BEC = 90^\circ$ . Thus $BD$ is the altitude from $B$ to $AC$ , and $CE$ is the altitude from $C$ to $AB$ .

The quadrilateral $BCDE$ has vertices in the order $B, C, D, E$ , so its diagonals are $BD$ and $CE$ , and $F$ is where the two altitudes meet.

In right triangle $BDC$ , $\angle DBC = 90^\circ - 40^\circ = 50^\circ$ . In right triangle $BEC$ , $\angle ECB = 90^\circ - 70^\circ = 20^\circ$ . In triangle $BFC$ ,
$\angle BFC = 180^\circ - 50^\circ - 20^\circ = 110^\circ.$

The answer is $\boxed{\textbf{(D)}\ 110}$ .

Review in the AoPS books

Introduction to Geometry, Chapter 11: Circles
Introduction to Geometry, Chapter 7: Special Parts of a Triangle
Introduction to Geometry, Chapter 2: Angles
Introduction to Geometry, Chapter 12: Circles and Angles